From<EMAIL_ADDRESS>(Steve Summit) Newsgroups: comp.lang.c,comp.lang.c.moderated,alt.comp.lang.learn.c-c++,comp.answers,alt.answers,news.answers Subject: comp.lang.c Answers to Frequently Asked Questions (FAQ List) Followup-To: poster Date: 1 Oct 1999 10:00:17 GMT Organization: better late than never Expires: 3 Nov 1999 00:00:00 GMT Message-ID<EMAIL_ADDRESS>Reply-To<EMAIL_ADDRESS>X-Last-Modified: February 7, 1999 X-Archive-Name: C-faq/faq X-Version: 3.5 X-URL: http://www.eskimo.com/~scs/C-faq/top.html X-PGP-Signature: Version: 2.6.2 iQCSAwUBNr5mkN6sm4I1rmP1AQEr/APoniUefFSgXsFWaMy+nDcCCzvH9phH7BVx 0CwFcGKz/udQ6DsXSynb3d9i50DUeRUXP2RcY69dmV41SZrBraQmXjjlwAulRlqB mDzL9zAhlQOSaS33s6zYmUB4A4kq+61XRYGGFwBc5dSWCzTlzxbl1nWo5Uru+azJ 9MKtfOg= =f2Ny Archive-name: C-faq/faq Comp-lang-c-archive-name: C-FAQ-list URL: http://www.eskimo.com/~scs/C-faq/top.html [Last modified February 7, 1999 by scs.] This article is Copyright 1990-1999 by Steve Summit. Content from the book _C Programming FAQs: Frequently Asked Questions_ is made available here by permission of the author and the publisher as a service to the community. It is intended to complement the use of the published text and is protected by international copyright laws. The content is made available here and may be accessed freely for personal use but may not be republished without permission. Certain topics come up again and again on this newsgroup. They are good questions, and the answers may not be immediately obvious, but each time they recur, much net bandwidth and reader time is wasted on repetitive responses, and on tedious corrections to the incorrect answers which are inevitably posted. This article, which is posted monthly, attempts to answer these common questions definitively and succinctly, so that net discussion can move on to more constructive topics without continual regression to first principles. No mere newsgroup article can substitute for thoughtful perusal of a full-length tutorial or language reference manual. Anyone interested enough in C to be following this newsgroup should also be interested enough to read and study one or more such manuals, preferably several times. Some C books and compiler manuals are unfortunately inadequate; a few even perpetuate some of the myths which this article attempts to refute. Several noteworthy books on C are listed in this article's bibliography; see also questions 18.9 and 18.10. Many of the questions and answers are cross-referenced to these books, for further study by the interested and dedicated reader. If you have a question about C which is not answered in this article, first try to answer it by checking a few of the referenced books, or by asking knowledgeable colleagues, before posing your question to the net at large. There are many people on the net who are happy to answer questions, but the volume of repetitive answers posted to one question, as well as the growing number of questions as the net attracts more readers, can become oppressive. If you have questions or comments prompted by this article, please reply by mail rather than following up -- this article is meant to decrease net traffic, not increase it. Besides listing frequently-asked questions, this article also summarizes frequently-posted answers. Even if you know all the answers, it's worth skimming through this list once in a while, so that when you see one of its questions unwittingly posted, you won't have to waste time answering. (However, this is a large and heavy document, so don't assume that everyone on the newsgroup has managed to read all of it in detail, and please don't roll it up and thwack people over the head with it just because they missed their answer in it.) This article was last modified on February 7, 1999, and its travels may have taken it far from its original home on Usenet. It may, however, be out-of-date, particularly if you are looking at a printed copy or one retrieved from a tertiary archive site or CD-ROM. You should be able to obtain the most up-to-date copy on the web at http://www.eskimo.com/~scs/C-faq/top.html or http://www.faqs.org/faqs/ , or from one of the ftp sites mentioned in question 20.40. Since this list is modified from time to time, its question numbers may not match those in older or newer copies which are in circulation; be careful when referring to FAQ list entries by number alone. This article was produced for free redistribution. You should not need to pay anyone for a copy of it. Other versions of this document are also available. Posted along with it are an abridged version and (when there are changes) a list of differences with respect to the previous version. A hypertext version is available on the web at the aforementioned URL. Finally, for those who might prefer a bound, hardcopy version (and even longer answers to even more questions!), a book-length version has been published by Addison-Wesley (ISBN 0-201-84519-9). This article is always being improved. Your input is welcomed. Send your comments to<EMAIL_ADDRESS>. The questions answered here are divided into several categories: 1. Declarations and Initializations 2. Structures, Unions, and Enumerations 3. Expressions 4. Pointers 5. Null Pointers 6. Arrays and Pointers 7. Memory Allocation 8. Characters and Strings 9. Boolean Expressions and Variables 10. C Preprocessor 11. ANSI/ISO Standard C 12. Stdio 13. Library Functions 14. Floating Point 15. Variable-Length Argument Lists 16. Strange Problems 17. Style 18. Tools and Resources 19. System Dependencies 20. Miscellaneous Bibliography Acknowledgements (The question numbers within each section are not always continuous, because they are aligned with the aforementioned book-length version, which contains even more questions.) Herewith, some frequently-asked questions and their answers: Section 1. Declarations and Initializations 1.1: How do you decide which integer type to use? A: If you might need large values (above 32,767 or below -32,767), use long. Otherwise, if space is very important (i.e. if there are large arrays or many structures), use short. Otherwise, use int. If well-defined overflow characteristics are important and negative values are not, or if you want to steer clear of sign- extension problems when manipulating bits or bytes, use one of the corresponding unsigned types. (Beware when mixing signed and unsigned values in expressions, though.) Although character types (especially unsigned char) can be used as "tiny" integers, doing so is sometimes more trouble than it's worth, due to unpredictable sign extension and increased code size. (Using unsigned char can help; see question 12.1 for a related problem.) A similar space/time tradeoff applies when deciding between float and double. None of the above rules apply if the address of a variable is taken and must have a particular type. If for some reason you need to declare something with an *exact* size (usually the only good reason for doing so is when attempting to conform to some externally-imposed storage layout, but see question 20.5), be sure to encapsulate the choice behind an appropriate typedef. References: K&R1 Sec. 2.2 p. 34; K&R2 Sec. 2.2 p. 36, Sec. A4.2 pp. 195-6, Sec. B11 p. 257; ISO Sec. <IP_ADDRESS>.1, Sec. <IP_ADDRESS>; H&S Secs. 5.1,5.2 pp. 110-114. 1.4: What should the 64-bit type on a machine that can support it? A: The forthcoming revision to the C Standard (C9X) specifies type long long as effectively being at least 64 bits, and this type has been implemented by a number of compilers for some time. (Others have implemented extensions such as __longlong.) On the other hand, there's no theoretical reason why a compiler couldn't implement type short int as 16, int as 32, and long int as 64 bits, and some compilers do indeed choose this arrangement. See also question 18.15d. References: C9X Sec. <IP_ADDRESS>.1, Sec. <IP_ADDRESS>. 1.7: What's the best way to declare and define global variables and functions? A: First, though there can be many "declarations" (and in many translation units) of a single "global" (strictly speaking, "external") variable or function, there must be exactly one "definition". (The definition is the declaration that actually allocates space, and provides an initialization value, if any.) The best arrangement is to place each definition in some relevant .c file, with an external declaration in a header (".h") file, which is #included wherever the declaration is needed. The .c file containing the definition should also #include the same header file, so that the compiler can check that the definition matches the declarations. This rule promotes a high degree of portability: it is consistent with the requirements of the ANSI C Standard, and is also consistent with most pre-ANSI compilers and linkers. (Unix compilers and linkers typically use a "common model" which allows multiple definitions, as long as at most one is initialized; this behavior is mentioned as a "common extension" by the ANSI Standard, no pun intended. A few very odd systems may require an explicit initializer to distinguish a definition from an external declaration.) It is possible to use preprocessor tricks to arrange that a line like DEFINE(int, i); need only be entered once in one header file, and turned into a definition or a declaration depending on the setting of some macro, but it's not clear if this is worth the trouble. It's especially important to put global declarations in header files if you want the compiler to catch inconsistent declarations for you. In particular, never place a prototype for an external function in a .c file: it wouldn't generally be checked for consistency with the definition, and an incompatible prototype is worse than useless. See also questions 10.6 and 18.8. References: K&R1 Sec. 4.5 pp. 76-7; K&R2 Sec. 4.4 pp. 80-1; ISO Sec. <IP_ADDRESS>, Sec. 6.7, Sec. 6.7.2, Sec. G.5.11; Rationale Sec. <IP_ADDRESS>; H&S Sec. 4.8 pp. 101-104, Sec. 9.2.3 p. 267; CT&P Sec. 4.2 pp. 54-56. 1.11: What does extern mean in a function declaration? A: It can be used as a stylistic hint to indicate that the function's definition is probably in another source file, but there is no formal difference between extern int f(); and int f(); References: ISO Sec. <IP_ADDRESS>, Sec. 6.5.1; Rationale Sec. <IP_ADDRESS>; H&S Secs. 4.3,4.3.1 pp. 75-6. 1.12: What's the auto keyword good for? A: Nothing; it's archaic. See also question 20.37. References: K&R1 Sec. A8.1 p. 193; ISO Sec. <IP_ADDRESS>, Sec. 6.5.1; H&S Sec. 4.3 p. 75, Sec. 4.3.1 p. 76. 1.14: I can't seem to define a linked list successfully. I tried typedef struct { char *item; NODEPTR next; } *NODEPTR; but the compiler gave me error messages. Can't a structure in C contain a pointer to itself? A: Structures in C can certainly contain pointers to themselves; the discussion and example in section 6.5 of K&R make this clear. The problem with the NODEPTR example is that the typedef has not been defined at the point where the "next" field is declared. To fix this code, first give the structure a tag ("struct node"). Then, declare the "next" field as a simple "struct node *", or disentangle the typedef declaration from the structure definition, or both. One corrected version would be struct node { char *item; struct node *next; }; typedef struct node *NODEPTR; and there are at least three other equivalently correct ways of arranging it. A similar problem, with a similar solution, can arise when attempting to declare a pair of typedef'ed mutually referential structures. See also question 2.1. References: K&R1 Sec. 6.5 p. 101; K&R2 Sec. 6.5 p. 139; ISO Sec. 6.5.2, Sec. <IP_ADDRESS>; H&S Sec. 5.6.1 pp. 132-3. 1.21: How do I declare an array of N pointers to functions returning pointers to functions returning pointers to characters? A: The first part of this question can be answered in at least three ways: 1. char *(*(*a[N])())(); 2. Build the declaration up incrementally, using typedefs: typedef char *pc; /* pointer to char */ typedef pc fpc(); /* function returning pointer to char */ typedef fpc *pfpc; /* pointer to above */ typedef pfpc fpfpc(); /* function returning... */ typedef fpfpc *pfpfpc; /* pointer to... */ pfpfpc a[N]; /* array of... */ 3. Use the cdecl program, which turns English into C and vice versa: cdecl> declare a as array of pointer to function returning pointer to function returning pointer to char char *(*(*a[])())() cdecl can also explain complicated declarations, help with casts, and indicate which set of parentheses the arguments go in (for complicated function definitions, like the one above). See question 18.1. Any good book on C should explain how to read these complicated C declarations "inside out" to understand them ("declaration mimics use"). The pointer-to-function declarations in the examples above have not included parameter type information. When the parameters have complicated types, declarations can *really* get messy. (Modern versions of cdecl can help here, too.) References: K&R2 Sec. 5.12 p. 122; ISO Sec. 6.5ff (esp. Sec. 6.5.4); H&S Sec. 4.5 pp. 85-92, Sec. 5.10.1 pp. 149-50. 1.22: How can I declare a function that can return a pointer to a function of the same type? I'm building a state machine with one function for each state, each of which returns a pointer to the function for the next state. But I can't find a way to declare the functions. A: You can't quite do it directly. Either have the function return a generic function pointer, with some judicious casts to adjust the types as the pointers are passed around; or have it return a structure containing only a pointer to a function returning that structure. 1.25: My compiler is complaining about an invalid redeclaration of a function, but I only define it once and call it once. A: Functions which are called without a declaration in scope (perhaps because the first call precedes the function's definition) are assumed to be declared as returning int (and without any argument type information), leading to discrepancies if the function is later declared or defined otherwise. Non-int functions must be declared before they are called. Another possible source of this problem is that the function has the same name as another one declared in some header file. See also questions 11.3 and 15.1. References: K&R1 Sec. 4.2 p. 70; K&R2 Sec. 4.2 p. 72; ISO Sec. <IP_ADDRESS>; H&S Sec. 4.7 p. 101. 1.25b: What's the right declaration for main()? Is void main() correct? A: See questions 11.12a to 11.15. (But no, it's not correct.) 1.30: What am I allowed to assume about the initial values of variables which are not explicitly initialized? If global variables start out as "zero", is that good enough for null pointers and floating-point zeroes? A: Uninitialized variables with "static" duration (that is, those declared outside of functions, and those declared with the storage class static), are guaranteed to start out as zero, as if the programmer had typed "= 0". Therefore, such variables are implicitly initialized to the null pointer (of the correct type; see also section 5) if they are pointers, and to 0.0 if they are floating-point. Variables with "automatic" duration (i.e. local variables without the static storage class) start out containing garbage, unless they are explicitly initialized. (Nothing useful can be predicted about the garbage.) Dynamically-allocated memory obtained with malloc() and realloc() is also likely to contain garbage, and must be initialized by the calling program, as appropriate. Memory obtained with calloc() is all-bits-0, but this is not necessarily useful for pointer or floating-point values (see question 7.31, and section 5). References: K&R1 Sec. 4.9 pp. 82-4; K&R2 Sec. 4.9 pp. 85-86; ISO Sec. 6.5.7, Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>; H&S Sec. 4.2.8 pp. 72- 3, Sec. 4.6 pp. 92-3, Sec. 4.6.2 pp. 94-5, Sec. 4.6.3 p. 96, Sec. 16.1 p. 386. 1.31: This code, straight out of a book, isn't compiling: int f() { char a[] = "Hello, world!"; } A: Perhaps you have a pre-ANSI compiler, which doesn't allow initialization of "automatic aggregates" (i.e. non-static local arrays, structures, and unions). (As a workaround, and depending on how the variable a is used, you may be able to make it global or static, or replace it with a pointer, or initialize it by hand with strcpy() when f() is called.) See also question 11.29. 1.31b: What's wrong with this initialization? char *p = malloc(10); My compiler is complaining about an "invalid initializer", or something. A: Is the declaration of a static or non-local variable? Function calls are allowed only in initializers for automatic variables (that is, for local, non-static variables). 1.32: What is the difference between these initializations? char a[] = "string literal"; char *p = "string literal"; My program crashes if I try to assign a new value to p[i]. A: A string literal can be used in two slightly different ways. As an array initializer (as in the declaration of char a[]), it specifies the initial values of the characters in that array. Anywhere else, it turns into an unnamed, static array of characters, which may be stored in read-only memory, which is why you can't safely modify it. In an expression context, the array is converted at once to a pointer, as usual (see section 6), so the second declaration initializes p to point to the unnamed array's first element. (For compiling old code, some compilers have a switch controlling whether strings are writable or not.) See also questions 1.31, 6.1, 6.2, and 6.8. References: K&R2 Sec. 5.5 p. 104; ISO Sec. 6.1.4, Sec. 6.5.7; Rationale Sec. 3.1.4; H&S Sec. 2.7.4 pp. 31-2. 1.34: I finally figured out the syntax for declaring pointers to functions, but now how do I initialize one? A: Use something like extern int func(); int (*fp)() = func; When the name of a function appears in an expression like this, it "decays" into a pointer (that is, it has its address implicitly taken), much as an array name does. An explicit declaration for the function is normally needed, since implicit external function declaration does not happen in this case (because the function name in the initialization is not part of a function call). See also questions 1.25 and 4.12. Section 2. Structures, Unions, and Enumerations 2.1: What's the difference between these two declarations? struct x1 { ... }; typedef struct { ... } x2; A: The first form declares a "structure tag"; the second declares a "typedef". The main difference is that you subsequently refer to the first type as "struct x1" and the second simply as "x2". That is, the second declaration is of a slightly more abstract type -- its users don't necessarily know that it is a structure, and the keyword struct is not used when declaring instances of it. 2.2: Why doesn't struct x { ... }; x thestruct; work? A: C is not C++. Typedef names are not automatically generated for structure tags. See also question 2.1 above. 2.3: Can a structure contain a pointer to itself? A: Most certainly. See question 1.14. 2.4: What's the best way of implementing opaque (abstract) data types in C? A: One good way is for clients to use structure pointers (perhaps additionally hidden behind typedefs) which point to structure types which are not publicly defined. 2.6: I came across some code that declared a structure like this: struct name { int namelen; char namestr[1]; }; and then did some tricky allocation to make the namestr array act like it had several elements. Is this legal or portable? A: This technique is popular, although Dennis Ritchie has called it "unwarranted chumminess with the C implementation." An official interpretation has deemed that it is not strictly conforming with the C Standard, although it does seem to work under all known implementations. (Compilers which check array bounds carefully might issue warnings.) Another possibility is to declare the variable-size element very large, rather than very small; in the case of the above example: ... char namestr[MAXSIZE]; where MAXSIZE is larger than any name which will be stored. However, it looks like this technique is disallowed by a strict interpretation of the Standard as well. Furthermore, either of these "chummy" structures must be used with care, since the programmer knows more about their size than the compiler does. (In particular, they can generally only be manipulated via pointers.) C9X will introduce the concept of a "flexible array member", which will allow the size of an array to be omitted if it is the last member in a structure, thus providing a well-defined solution. References: Rationale Sec. <IP_ADDRESS>; C9X Sec. <IP_ADDRESS>. 2.7: I heard that structures could be assigned to variables and passed to and from functions, but K&R1 says not. A: What K&R1 said (though this was quite some time ago by now) was that the restrictions on structure operations would be lifted in a forthcoming version of the compiler, and in fact structure assignment and passing were fully functional in Ritchie's compiler even as K&R1 was being published. A few ancient C compilers may have lacked these operations, but all modern compilers support them, and they are part of the ANSI C standard, so there should be no reluctance to use them. (Note that when a structure is assigned, passed, or returned, the copying is done monolithically; the data pointed to by any pointer fields is *not* copied.) References: K&R1 Sec. 6.2 p. 121; K&R2 Sec. 6.2 p. 129; ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>, Sec. 6.3.16; H&S Sec. 5.6.2 p. 133. 2.8: Is there a way to compare structures automatically? A: No. There is no single, good way for a compiler to implement implicit structure comparison (i.e. to support the == operator for structures) which is consistent with C's low-level flavor. A simple byte-by-byte comparison could founder on random bits present in unused "holes" in the structure (such padding is used to keep the alignment of later fields correct; see question 2.12). A field-by-field comparison might require unacceptable amounts of repetitive code for large structures. If you need to compare two structures, you'll have to write your own function to do so, field by field. References: K&R2 Sec. 6.2 p. 129; Rationale Sec. 3.3.9; H&S Sec. 5.6.2 p. 133. 2.10: How can I pass constant values to functions which accept structure arguments? A: As of this writing, C has no way of generating anonymous structure values. You will have to use a temporary structure variable or a little structure-building function. The C9X Standard will introduce "compound literals"; one form of compound literal will allow structure constants. For example, to pass a constant coordinate pair to a plotpoint() function which expects a struct point, you will be able to call plotpoint((struct point){1, 2}); Combined with "designated initializers" (another C9X feature), it will also be possible to specify member values by name: plotpoint((struct point){.x=1, .y=2}); See also question 4.10. References: C9X Sec. <IP_ADDRESS>, Sec. 6.5.8. 2.11: How can I read/write structures from/to data files? A: It is relatively straightforward to write a structure out using fwrite(): fwrite(&somestruct, sizeof somestruct, 1, fp); and a corresponding fread invocation can read it back in. However, data files so written will *not* be portable (see questions 2.12 and 20.5). Note also that if the structure contains any pointers, only the pointer values will be written, and they are most unlikely to be valid when read back in. Finally, note that for widespread portability you must use the "b" flag when fopening the files; see question 12.38. A more portable solution, though it's a bit more work initially, is to write a pair of functions for writing and reading a structure, field-by-field, in a portable (perhaps even human- readable) way. References: H&S Sec. 15.13 p. 381. 2.12: My compiler is leaving holes in structures, which is wasting space and preventing "binary" I/O to external data files. Can I turn off the padding, or otherwise control the alignment of structure fields? A: Your compiler may provide an extension to give you this control (perhaps a #pragma; see question 11.20), but there is no standard method. See also question 20.5. References: K&R2 Sec. 6.4 p. 138; H&S Sec. 5.6.4 p. 135. 2.13: Why does sizeof report a larger size than I expect for a structure type, as if there were padding at the end? A: Structures may have this padding (as well as internal padding), if necessary, to ensure that alignment properties will be preserved when an array of contiguous structures is allocated. Even when the structure is not part of an array, the end padding remains, so that sizeof can always return a consistent size. See also question 2.12 above. References: H&S Sec. 5.6.7 pp. 139-40. 2.14: How can I determine the byte offset of a field within a structure? A: ANSI C defines the offsetof() macro, which should be used if available; see <stddef.h>. If you don't have it, one possible implementation is #define offsetof(type, mem) ((size_t) \ ((char *)&((type *)0)->mem - (char *)(type *)0)) This implementation is not 100% portable; some compilers may legitimately refuse to accept it. See question 2.15 below for a usage hint. References: ISO Sec. 7.1.6; Rationale Sec. <IP_ADDRESS>; H&S Sec. 11.1 pp. 292-3. 2.15: How can I access structure fields by name at run time? A: Build a table of names and offsets, using the offsetof() macro. The offset of field b in struct a is offsetb = offsetof(struct a, b) If structp is a pointer to an instance of this structure, and field b is an int (with offset as computed above), b's value can be set indirectly with *(int *)((char *)structp + offsetb) = value; 2.18: This program works correctly, but it dumps core after it finishes. Why? struct list { char *item; struct list *next; } /* Here is the main program. */ main(argc, argv) { ... } A: A missing semicolon causes main() to be declared as returning a structure. (The connection is hard to see because of the intervening comment.) Since structure-valued functions are usually implemented by adding a hidden return pointer, the generated code for main() tries to accept three arguments, although only two are passed (in this case, by the C start-up code). See also questions 10.9 and 16.4. References: CT&P Sec. 2.3 pp. 21-2. 2.20: Can I initialize unions? A: The current C Standard allows an initializer for the first-named member of a union. C9X will introduce "designated initializers" which can be used to initialize any member. References: K&R2 Sec. 6.8 pp. 148-9; ISO Sec. 6.5.7; C9X Sec. 6.5.8; H&S Sec. 4.6.7 p. 100. 2.22: What is the difference between an enumeration and a set of preprocessor #defines? A: At the present time, there is little difference. The C Standard says that enumerations may be freely intermixed with other integral types, without errors. (If, on the other hand, such intermixing were disallowed without explicit casts, judicious use of enumerations could catch certain programming errors.) Some advantages of enumerations are that the numeric values are automatically assigned, that a debugger may be able to display the symbolic values when enumeration variables are examined, and that they obey block scope. (A compiler may also generate nonfatal warnings when enumerations and integers are indiscriminately mixed, since doing so can still be considered bad style even though it is not strictly illegal.) A disadvantage is that the programmer has little control over those nonfatal warnings; some programmers also resent not having control over the sizes of enumeration variables. References: K&R2 Sec. 2.3 p. 39, Sec. A4.2 p. 196; ISO Sec. <IP_ADDRESS>, Sec. 6.5.2, Sec. <IP_ADDRESS>, Annex F; H&S Sec. 5.5 pp. 127-9, Sec. 5.11.2 p. 153. 2.24: Is there an easy way to print enumeration values symbolically? A: No. You can write a little function to map an enumeration constant to a string. (For debugging purposes, a good debugger should automatically print enumeration constants symbolically.) Section 3. Expressions 3.1: Why doesn't this code: a[i] = i++; work? A: The subexpression i++ causes a side effect -- it modifies i's value -- which leads to undefined behavior since i is also referenced elsewhere in the same expression, and there's no way to determine whether the reference (in a[i] on the left-hand side) should be to the old or the new value. (Note that although the language in K&R suggests that the behavior of this expression is unspecified, the C Standard makes the stronger statement that it is undefined -- see question 11.33.) References: K&R1 Sec. 2.12; K&R2 Sec. 2.12; ISO Sec. 6.3; H&S Sec. 7.12 pp. 227-9. 3.2: Under my compiler, the code int i = 7; printf("%d\n", i++ * i++); prints 49. Regardless of the order of evaluation, shouldn't it print 56? A: Although the postincrement and postdecrement operators ++ and -- perform their operations after yielding the former value, the implication of "after" is often misunderstood. It is *not* guaranteed that an increment or decrement is performed immediately after giving up the previous value and before any other part of the expression is evaluated. It is merely guaranteed that the update will be performed sometime before the expression is considered "finished" (before the next "sequence point," in ANSI C's terminology; see question 3.8). In the example, the compiler chose to multiply the previous value by itself and to perform both increments afterwards. The behavior of code which contains multiple, ambiguous side effects has always been undefined. (Loosely speaking, by "multiple, ambiguous side effects" we mean any combination of ++, --, =, +=, -=, etc. in a single expression which causes the same object either to be modified twice or modified and then inspected. This is a rough definition; see question 3.8 for a precise one, and question 11.33 for the meaning of "undefined.") Don't even try to find out how your compiler implements such things (contrary to the ill-advised exercises in many C textbooks); as K&R wisely point out, "if you don't know *how* they are done on various machines, that innocence may help to protect you." References: K&R1 Sec. 2.12 p. 50; K&R2 Sec. 2.12 p. 54; ISO Sec. 6.3; H&S Sec. 7.12 pp. 227-9; CT&P Sec. 3.7 p. 47; PCS Sec. 9.5 pp. 120-1. 3.3: I've experimented with the code int i = 3; i = i++; on several compilers. Some gave i the value 3, and some gave 4. Which compiler is correct? A: There is no correct answer; the expression is undefined. See questions 3.1, 3.8, 3.9, and 11.33. (Also, note that neither i++ nor ++i is the same as i+1. If you want to increment i, use i=i+1, i+=1, i++, or ++i, not some combination. See also question 3.12.) 3.3b: Here's a slick expression: a ^= b ^= a ^= b It swaps a and b without using a temporary. A: Not portably, it doesn't. It attempts to modify the variable a twice between sequence points, so its behavior is undefined. For example, it has been reported that when given the code int a = 123, b = 7654; a ^= b ^= a ^= b; the SCO Optimizing C compiler (icc) sets b to 123 and a to 0. See also questions 3.1, 3.8, 10.3, and 20.15c. 3.4: Can I use explicit parentheses to force the order of evaluation I want? Even if I don't, doesn't precedence dictate it? A: Not in general. Operator precedence and explicit parentheses impose only a partial ordering on the evaluation of an expression. In the expression f() + g() * h() although we know that the multiplication will happen before the addition, there is no telling which of the three functions will be called first. When you need to ensure the order of subexpression evaluation, you may need to use explicit temporary variables and separate statements. References: K&R1 Sec. 2.12 p. 49, Sec. A.7 p. 185; K&R2 Sec. 2.12 pp. 52-3, Sec. A.7 p. 200. 3.5: But what about the && and || operators? I see code like "while((c = getchar()) != EOF && c != '\n')" ... A: There is a special "short-circuiting" exception for those operators. The right-hand side is not evaluated if the left- hand side determines the outcome (i.e. is true for || or false for &&). Therefore, left-to-right evaluation is guaranteed, as it also is for the comma operator. Furthermore, all of these operators (along with ?:) introduce an extra internal sequence point (see question 3.8). References: K&R1 Sec. 2.6 p. 38, Secs. A7.11-12 pp. 190-1; K&R2 Sec. 2.6 p. 41, Secs. A7.14-15 pp. 207-8; ISO Sec. 6.3.13, Sec. 6.3.14, Sec. 6.3.15; H&S Sec. 7.7 pp. 217-8, Sec. 7.8 pp. 218-20, Sec. 7.12.1 p. 229; CT&P Sec. 3.7 pp. 46-7. 3.8: How can I understand these complex expressions? What's a "sequence point"? A: A sequence point is a point in time (at the end of the evaluation of a full expression, or at the ||, &&, ?:, or comma operators, or just before a function call) at which the dust has settled and all side effects are guaranteed to be complete. The ANSI/ISO C Standard states that Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be accessed only to determine the value to be stored. The second sentence can be difficult to understand. It says that if an object is written to within a full expression, any and all accesses to it within the same expression must be for the purposes of computing the value to be written. This rule effectively constrains legal expressions to those in which the accesses demonstrably precede the modification. See also question 3.9 below. References: ISO Sec. <IP_ADDRESS>, Sec. 6.3, Sec. 6.6, Annex C; Rationale Sec. <IP_ADDRESS>; H&S Sec. 7.12.1 pp. 228-9. 3.9: So given a[i] = i++; we don't know which cell of a[] gets written to, but i does get incremented by one, right? A: *No*. Once an expression or program becomes undefined, *all* aspects of it become undefined. See questions 3.2, 3.3, 11.33, and 11.35. 3.12: If I'm not using the value of the expression, should I use i++ or ++i to increment a variable? A: Since the two forms differ only in the value yielded, they are entirely equivalent when only their side effect is needed. (However, the prefix form is preferred in C++.) See also question 3.3. References: K&R1 Sec. 2.8 p. 43; K&R2 Sec. 2.8 p. 47; ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>; H&S Sec. 7.4.4 pp. 192-3, Sec. 7.5.8 pp. 199-200. 3.14: Why doesn't the code int a = 1000, b = 1000; long int c = a * b; work? A: Under C's integral promotion rules, the multiplication is carried out using int arithmetic, and the result may overflow or be truncated before being promoted and assigned to the long int left-hand side. Use an explicit cast to force long arithmetic: long int c = (long int)a * b; Note that (long int)(a * b) would *not* have the desired effect. A similar problem can arise when two integers are divided, with the result assigned to a floating-point variable; the solution is similar, too. References: K&R1 Sec. 2.7 p. 41; K&R2 Sec. 2.7 p. 44; ISO Sec. <IP_ADDRESS>; H&S Sec. 6.3.4 p. 176; CT&P Sec. 3.9 pp. 49-50. 3.16: I have a complicated expression which I have to assign to one of two variables, depending on a condition. Can I use code like this? ((condition) ? a : b) = complicated_expression; A: No. The ?: operator, like most operators, yields a value, and you can't assign to a value. (In other words, ?: does not yield an "lvalue".) If you really want to, you can try something like *((condition) ? &a : &b) = complicated_expression; although this is admittedly not as pretty. References: ISO Sec. 6.3.15; H&S Sec. 7.1 pp. 179-180. Section 4. Pointers 4.2: I'm trying to declare a pointer and allocate some space for it, but it's not working. What's wrong with this code? char *p; *p = malloc(10); A: The pointer you declared is p, not *p. To make a pointer point somewhere, you just use the name of the pointer: p = malloc(10); It's when you're manipulating the pointed-to memory that you use * as an indirection operator: *p = 'H'; See also questions 1.21, 7.1, 7.3c, and 8.3. References: CT&P Sec. 3.1 p. 28. 4.3: Does *p++ increment p, or what it points to? A: Postfix ++ essentially has higher precedence than the prefix unary operators. Therefore, *p++ is equivalent to *(p++); it increments p, and returns the value which p pointed to before p was incremented. To increment the value pointed to by p, use (*p)++ (or perhaps ++*p, if the order of the side effect doesn't matter). References: K&R1 Sec. 5.1 p. 91; K&R2 Sec. 5.1 p. 95; ISO Sec. 6.3.2, Sec. 6.3.3; H&S Sec. 7.4.4 pp. 192-3, Sec. 7.5 p. 193, Secs. 7.5.7,7.5.8 pp. 199-200. 4.5: I have a char * pointer that happens to point to some ints, and I want to step it over them. Why doesn't ((int *)p)++; work? A: In C, a cast operator does not mean "pretend these bits have a different type, and treat them accordingly"; it is a conversion operator, and by definition it yields an rvalue, which cannot be assigned to, or incremented with ++. (It is either an accident or a delibrate but nonstandard extension if a particular compiler accepts expressions such as the above.) Say what you mean: use p = (char *)((int *)p + 1); or (since p is a char *) simply p += sizeof(int); Whenever possible, you should choose appropriate pointer types in the first place, instead of trying to treat one type as another. References: K&R2 Sec. A7.5 p. 205; ISO Sec. 6.3.4; Rationale Sec. <IP_ADDRESS>; H&S Sec. 7.1 pp. 179-80. 4.8: I have a function which accepts, and is supposed to initialize, a pointer: void f(int *ip) { static int dummy = 5; ip = &dummy; } But when I call it like this: int *ip; f(ip); the pointer in the caller remains unchanged. A: Are you sure the function initialized what you thought it did? Remember that arguments in C are passed by value. The called function altered only the passed copy of the pointer. You'll either want to pass the address of the pointer (the function will end up accepting a pointer-to-a-pointer), or have the function return the pointer. See also questions 4.9 and 4.11. 4.9: Can I use a void ** pointer as a parameter so that a function can accept a generic pointer by reference? A: Not portably. There is no generic pointer-to-pointer type in C. void * acts as a generic pointer only because conversions are applied automatically when other pointer types are assigned to and from void *'s; these conversions cannot be performed (the correct underlying pointer type is not known) if an attempt is made to indirect upon a void ** value which points at a pointer type other than void *. 4.10: I have a function extern int f(int *); which accepts a pointer to an int. How can I pass a constant by reference? A call like f(&5); doesn't seem to work. A: You can't do this directly. You will have to declare a temporary variable, and then pass its address to the function: int five = 5; f(&five); See also questions 2.10, 4.8, and 20.1. 4.11: Does C even have "pass by reference"? A: Not really. Strictly speaking, C always uses pass by value. You can simulate pass by reference yourself, by defining functions which accept pointers and then using the & operator when calling, and the compiler will essentially simulate it for you when you pass an array to a function (by passing a pointer instead, see question 6.4 et al.). However, C has nothing truly equivalent to formal pass by reference or C++ reference parameters. (On the other hand, function-like preprocessor macros can provide a form of "pass by name".) See also questions 4.8 and 20.1. References: K&R1 Sec. 1.8 pp. 24-5, Sec. 5.2 pp. 91-3; K&R2 Sec. 1.8 pp. 27-8, Sec. 5.2 pp. 95-7; ISO Sec. <IP_ADDRESS>; H&S Sec. 9.5 pp. 273-4. 4.12: I've seen different methods used for calling functions via pointers. What's the story? A: Originally, a pointer to a function had to be "turned into" a "real" function, with the * operator (and an extra pair of parentheses, to keep the precedence straight), before calling: int r, func(), (*fp)() = func; r = (*fp)(); It can also be argued that functions are always called via pointers, and that "real" function names always decay implicitly into pointers (in expressions, as they do in initializations; see question 1.34). This reasoning (which is in fact used in the ANSI standard) means that r = fp(); is legal and works correctly, whether fp is the name of a function or a pointer to one. (The usage has always been unambiguous; there is nothing you ever could have done with a function pointer followed by an argument list except call the function pointed to.) An explicit * is still allowed. See also question 1.34. References: K&R1 Sec. 5.12 p. 116; K&R2 Sec. 5.11 p. 120; ISO Sec. <IP_ADDRESS>; Rationale Sec. <IP_ADDRESS>; H&S Sec. 5.8 p. 147, Sec. 7.4.3 p. 190. Section 5. Null Pointers 5.1: What is this infamous null pointer, anyway? A: The language definition states that for each pointer type, there is a special value -- the "null pointer" -- which is distinguishable from all other pointer values and which is "guaranteed to compare unequal to a pointer to any object or function." That is, the address-of operator & will never yield a null pointer, nor will a successful call to malloc(). (malloc() does return a null pointer when it fails, and this is a typical use of null pointers: as a "special" pointer value with some other meaning, usually "not allocated" or "not pointing anywhere yet.") A null pointer is conceptually different from an uninitialized pointer. A null pointer is known not to point to any object or function; an uninitialized pointer might point anywhere. See also questions 1.30, 7.1, and 7.31. As mentioned above, there is a null pointer for each pointer type, and the internal values of null pointers for different types may be different. Although programmers need not know the internal values, the compiler must always be informed which type of null pointer is required, so that it can make the distinction if necessary (see questions 5.2, 5.5, and 5.6 below). References: K&R1 Sec. 5.4 pp. 97-8; K&R2 Sec. 5.4 p. 102; ISO Sec. <IP_ADDRESS>; Rationale Sec. <IP_ADDRESS>; H&S Sec. 5.3.2 pp. 121-3. 5.2: How do I get a null pointer in my programs? A: According to the language definition, a constant 0 in a pointer context is converted into a null pointer at compile time. That is, in an initialization, assignment, or comparison when one side is a variable or expression of pointer type, the compiler can tell that a constant 0 on the other side requests a null pointer, and generate the correctly-typed null pointer value. Therefore, the following fragments are perfectly legal: char *p = 0; if(p != 0) (See also question 5.3.) However, an argument being passed to a function is not necessarily recognizable as a pointer context, and the compiler may not be able to tell that an unadorned 0 "means" a null pointer. To generate a null pointer in a function call context, an explicit cast may be required, to force the 0 to be recognized as a pointer. For example, the Unix system call execl takes a variable-length, null-pointer-terminated list of character pointer arguments, and is correctly called like this: execl("/bin/sh", "sh", "-c", "date", (char *)0); If the (char *) cast on the last argument were omitted, the compiler would not know to pass a null pointer, and would pass an integer 0 instead. (Note that many Unix manuals get this example wrong.) When function prototypes are in scope, argument passing becomes an "assignment context," and most casts may safely be omitted, since the prototype tells the compiler that a pointer is required, and of which type, enabling it to correctly convert an unadorned 0. Function prototypes cannot provide the types for variable arguments in variable-length argument lists however, so explicit casts are still required for those arguments. (See also question 15.3.) It is probably safest to properly cast all null pointer constants in function calls, to guard against varargs functions or those without prototypes. Summary: Unadorned 0 okay: Explicit cast required: initialization function call, no prototype in scope assignment variable argument in comparison varargs function call function call, prototype in scope, fixed argument References: K&R1 Sec. A7.7 p. 190, Sec. A7.14 p. 192; K&R2 Sec. A7.10 p. 207, Sec. A7.17 p. 209; ISO Sec. <IP_ADDRESS>; H&S Sec. 4.6.3 p. 95, Sec. 6.2.7 p. 171. 5.3: Is the abbreviated pointer comparison "if(p)" to test for non- null pointers valid? What if the internal representation for null pointers is nonzero? A: When C requires the Boolean value of an expression, a false value is inferred when the expression compares equal to zero, and a true value otherwise. That is, whenever one writes if(expr) where "expr" is any expression at all, the compiler essentially acts as if it had been written as if((expr) != 0) Substituting the trivial pointer expression "p" for "expr", we have if(p) is equivalent to if(p != 0) and this is a comparison context, so the compiler can tell that the (implicit) 0 is actually a null pointer constant, and use the correct null pointer value. There is no trickery involved here; compilers do work this way, and generate identical code for both constructs. The internal representation of a null pointer does *not* matter. The boolean negation operator, !, can be described as follows: !expr is essentially equivalent to (expr)?0:1 or to ((expr) == 0) which leads to the conclusion that if(!p) is equivalent to if(p == 0) "Abbreviations" such as if(p), though perfectly legal, are considered by some to be bad style (and by others to be good style; see question 17.10). See also question 9.2. References: K&R2 Sec. A7.4.7 p. 204; ISO Sec. <IP_ADDRESS>, Sec. 6.3.9, Sec. 6.3.13, Sec. 6.3.14, Sec. 6.3.15, Sec. <IP_ADDRESS>, Sec. 6.6.5; H&S Sec. 5.3.2 p. 122. 5.4: What is NULL and how is it #defined? A: As a matter of style, many programmers prefer not to have unadorned 0's scattered through their programs. Therefore, the preprocessor macro NULL is #defined (by <stdio.h> and several other headers) with the value 0, possibly cast to (void *) (see also question 5.6). A programmer who wishes to make explicit the distinction between 0 the integer and 0 the null pointer constant can then use NULL whenever a null pointer is required. Using NULL is a stylistic convention only; the preprocessor turns NULL back into 0 which is then recognized by the compiler, in pointer contexts, as before. In particular, a cast may still be necessary before NULL (as before 0) in a function call argument. The table under question 5.2 above applies for NULL as well as 0 (an unadorned NULL is equivalent to an unadorned 0). NULL should *only* be used for pointers; see question 5.9. References: K&R1 Sec. 5.4 pp. 97-8; K&R2 Sec. 5.4 p. 102; ISO Sec. 7.1.6, Sec. <IP_ADDRESS>; Rationale Sec. 4.1.5; H&S Sec. 5.3.2 p. 122, Sec. 11.1 p. 292. 5.5: How should NULL be defined on a machine which uses a nonzero bit pattern as the internal representation of a null pointer? A: The same as on any other machine: as 0 (or some version of 0; see question 5.4). Whenever a programmer requests a null pointer, either by writing "0" or "NULL", it is the compiler's responsibility to generate whatever bit pattern the machine uses for that null pointer. Therefore, #defining NULL as 0 on a machine for which internal null pointers are nonzero is as valid as on any other: the compiler must always be able to generate the machine's correct null pointers in response to unadorned 0's seen in pointer contexts. See also questions 5.2, 5.10, and 5.17. References: ISO Sec. 7.1.6; Rationale Sec. 4.1.5. 5.6: If NULL were defined as follows: #define NULL ((char *)0) wouldn't that make function calls which pass an uncast NULL work? A: Not in general. The complication is that there are machines which use different internal representations for pointers to different types of data. The suggested definition would make uncast NULL arguments to functions expecting pointers to characters work correctly, but pointer arguments of other types would still be problematical, and legal constructions such as FILE *fp = NULL; could fail. Nevertheless, ANSI C allows the alternate definition #define NULL ((void *)0) for NULL. Besides potentially helping incorrect programs to work (but only on machines with homogeneous pointers, thus questionably valid assistance), this definition may catch programs which use NULL incorrectly (e.g. when the ASCII NUL character was really intended; see question 5.9). References: Rationale Sec. 4.1.5. 5.9: If NULL and 0 are equivalent as null pointer constants, which should I use? A: Many programmers believe that NULL should be used in all pointer contexts, as a reminder that the value is to be thought of as a pointer. Others feel that the confusion surrounding NULL and 0 is only compounded by hiding 0 behind a macro, and prefer to use unadorned 0 instead. There is no one right answer. (See also questions 9.2 and 17.10.) C programmers must understand that NULL and 0 are interchangeable in pointer contexts, and that an uncast 0 is perfectly acceptable. Any usage of NULL (as opposed to 0) should be considered a gentle reminder that a pointer is involved; programmers should not depend on it (either for their own understanding or the compiler's) for distinguishing pointer 0's from integer 0's. NULL should *not* be used when another kind of 0 is required, even though it might work, because doing so sends the wrong stylistic message. (Furthermore, ANSI allows the definition of NULL to be ((void *)0), which will not work at all in non- pointer contexts.) In particular, do not use NULL when the ASCII null character (NUL) is desired. Provide your own definition #define NUL '\0' if you must. References: K&R1 Sec. 5.4 pp. 97-8; K&R2 Sec. 5.4 p. 102. 5.10: But wouldn't it be better to use NULL (rather than 0), in case the value of NULL changes, perhaps on a machine with nonzero internal null pointers? A: No. (Using NULL may be preferable, but not for this reason.) Although symbolic constants are often used in place of numbers because the numbers might change, this is *not* the reason that NULL is used in place of 0. Once again, the language guarantees that source-code 0's (in pointer contexts) generate null pointers. NULL is used only as a stylistic convention. See questions 5.5 and 9.2. 5.12: I use the preprocessor macro #define Nullptr(type) (type *)0 to help me build null pointers of the correct type. A: This trick, though popular and superficially attractive, does not buy much. It is not needed in assignments or comparisons; see question 5.2. (It does not even save keystrokes.) See also questions 9.1 and 10.2. 5.13: This is strange. NULL is guaranteed to be 0, but the null pointer is not? A: When the term "null" or "NULL" is casually used, one of several things may be meant: 1. The conceptual null pointer, the abstract language concept defined in question 5.1. It is implemented with... 2. The internal (or run-time) representation of a null pointer, which may or may not be all-bits-0 and which may be different for different pointer types. The actual values should be of concern only to compiler writers. Authors of C programs never see them, since they use... 3. The null pointer constant, which is a constant integer 0 (see question 5.2). It is often hidden behind... 4. The NULL macro, which is #defined to be 0 (see question 5.4). Finally, as red herrings, we have... 5. The ASCII null character (NUL), which does have all bits zero, but has no necessary relation to the null pointer except in name; and... 6. The "null string," which is another name for the empty string (""). Using the term "null string" can be confusing in C, because an empty string involves a null ('\0') character, but *not* a null pointer, which brings us full circle... This article uses the phrase "null pointer" (in lower case) for sense 1, the character "0" or the phrase "null pointer constant" for sense 3, and the capitalized word "NULL" for sense 4. 5.14: Why is there so much confusion surrounding null pointers? Why do these questions come up so often? A: C programmers traditionally like to know more than they might need to about the underlying machine implementation. The fact that null pointers are represented both in source code, and internally to most machines, as zero invites unwarranted assumptions. The use of a preprocessor macro (NULL) may seem to suggest that the value could change some day, or on some weird machine. The construct "if(p == 0)" is easily misread as calling for conversion of p to an integral type, rather than 0 to a pointer type, before the comparison. Finally, the distinction between the several uses of the term "null" (listed in question 5.13 above) is often overlooked. One good way to wade out of the confusion is to imagine that C used a keyword (perhaps "nil", like Pascal) as a null pointer constant. The compiler could either turn "nil" into the appropriate type of null pointer when it could unambiguously determine that type from the source code, or complain when it could not. Now in fact, in C the keyword for a null pointer constant is not "nil" but "0", which works almost as well, except that an uncast "0" in a non-pointer context generates an integer zero instead of an error message, and if that uncast 0 was supposed to be a null pointer constant, the code may not work. 5.15: I'm confused. I just can't understand all this null pointer stuff. A: Here are two simple rules you can follow: 1. When you want a null pointer constant in source code, use "0" or "NULL". 2. If the usage of "0" or "NULL" is an argument in a function call, cast it to the pointer type expected by the function being called. The rest of the discussion has to do with other people's misunderstandings, with the internal representation of null pointers (which you shouldn't need to know), and with the complexities of function prototypes. (Taking those complexities into account, we find that rule 2 is conservative, of course; but it doesn't hurt.) Understand questions 5.1, 5.2, and 5.4, and consider 5.3, 5.9, 5.13, and 5.14, and you'll do fine. 5.16: Given all the confusion surrounding null pointers, wouldn't it be easier simply to require them to be represented internally by zeroes? A: If for no other reason, doing so would be ill-advised because it would unnecessarily constrain implementations which would otherwise naturally represent null pointers by special, nonzero bit patterns, particularly when those values would trigger automatic hardware traps for invalid accesses. Besides, what would such a requirement really accomplish? Proper understanding of null pointers does not require knowledge of the internal representation, whether zero or nonzero. Assuming that null pointers are internally zero does not make any code easier to write (except for a certain ill-advised usage of calloc(); see question 7.31). Known-zero internal pointers would not obviate casts in function calls, because the *size* of the pointer might still be different from that of an int. (If "nil" were used to request null pointers, as mentioned in question 5.14 above, the urge to assume an internal zero representation would not even arise.) 5.17: Seriously, have any actual machines really used nonzero null pointers, or different representations for pointers to different types? A: The Prime 50 series used segment 07777, offset 0 for the null pointer, at least for PL/I. Later models used segment 0, offset 0 for null pointers in C, necessitating new instructions such as TCNP (Test C Null Pointer), evidently as a sop to all the extant poorly-written C code which made incorrect assumptions. Older, word-addressed Prime machines were also notorious for requiring larger byte pointers (char *'s) than word pointers (int *'s). The Eclipse MV series from Data General has three architecturally supported pointer formats (word, byte, and bit pointers), two of which are used by C compilers: byte pointers for char * and void *, and word pointers for everything else. Some Honeywell-Bull mainframes use the bit pattern 06000 for (internal) null pointers. The CDC Cyber 180 Series has 48-bit pointers consisting of a ring, segment, and offset. Most users (in ring 11) have null pointers of 0xB00000000000. It was common on old CDC ones- complement machines to use an all-one-bits word as a special flag for all kinds of data, including invalid addresses. The old HP 3000 series uses a different addressing scheme for byte addresses than for word addresses; like several of the machines above it therefore uses different representations for char * and void * pointers than for other pointers. The Symbolics Lisp Machine, a tagged architecture, does not even have conventional numeric pointers; it uses the pair <NIL, 0> (basically a nonexistent <object, offset> handle) as a C null pointer. Depending on the "memory model" in use, 8086-family processors (PC compatibles) may use 16-bit data pointers and 32-bit function pointers, or vice versa. Some 64-bit Cray machines represent int * in the lower 48 bits of a word; char * additionally uses the upper 16 bits to indicate a byte address within a word. References: K&R1 Sec. A14.4 p. 211. 5.20: What does a run-time "null pointer assignment" error mean? How can I track it down? A: This message, which typically occurs with MS-DOS compilers, means that you've written, via a null (perhaps because uninitialized) pointer, to an invalid location (probably offset 0 in the default data segment). A debugger may let you set a data watchpoint on location 0. Alternatively, you could write a bit of code to stash away a copy of 20 or so bytes from location 0, and periodically check that the memory at location 0 hasn't changed. See also question 16.8. Section 6. Arrays and Pointers 6.1: I had the definition char a[6] in one source file, and in another I declared extern char *a. Why didn't it work? A: In one source file you defind an array of characters and in the other you declared a pointer to characters. The declaration extern char *a simply does not match the actual definition. The type pointer-to-type-T is not the same as array-of-type-T. Use extern char a[]. References: ISO Sec. <IP_ADDRESS>; CT&P Sec. 3.3 pp. 33-4, Sec. 4.5 pp. 64-5. 6.2: But I heard that char a[] was identical to char *a. A: Not at all. (What you heard has to do with formal parameters to functions; see question 6.4.) Arrays are not pointers. The array declaration char a[6] requests that space for six characters be set aside, to be known by the name "a". That is, there is a location named "a" at which six characters can sit. The pointer declaration char *p, on the other hand, requests a place which holds a pointer, to be known by the name "p". This pointer can point almost anywhere: to any char, or to any contiguous array of chars, or nowhere (see also questions 5.1 and 1.30). As usual, a picture is worth a thousand words. The declarations char a[] = "hello"; char *p = "world"; would initialize data structures which could be represented like this: +---+---+---+---+---+---+ a: | h | e | l | l | o |\0 | +---+---+---+---+---+---+ +-----+ +---+---+---+---+---+---+ p: | *======> | w | o | r | l | d |\0 | +-----+ +---+---+---+---+---+---+ It is important to realize that a reference like x[3] generates different code depending on whether x is an array or a pointer. Given the declarations above, when the compiler sees the expression a[3], it emits code to start at the location "a", move three past it, and fetch the character there. When it sees the expression p[3], it emits code to start at the location "p", fetch the pointer value there, add three to the pointer, and finally fetch the character pointed to. In other words, a[3] is three places past (the start of) the object *named* a, while p[3] is three places past the object *pointed to* by p. In the example above, both a[3] and p[3] happen to be the character 'l', but the compiler gets there differently. (The essential difference is that the values of an array like a and a pointer like p are computed differently *whenever* they appear in expressions, whether or not they are being subscripted, as explained further in the next question.) References: K&R2 Sec. 5.5 p. 104; CT&P Sec. 4.5 pp. 64-5. 6.3: So what is meant by the "equivalence of pointers and arrays" in C? A: Much of the confusion surrounding arrays and pointers in C can be traced to a misunderstanding of this statement. Saying that arrays and pointers are "equivalent" means neither that they are identical nor even interchangeable. What it means is that array and pointer arithmetic is defined such that a pointer can be conveniently used to access an array or to simulate an array. Specifically, the cornerstone of the equivalence is this key definition: An lvalue of type array-of-T which appears in an expression decays (with three exceptions) into a pointer to its first element; the type of the resultant pointer is pointer-to-T. That is, whenever an array appears in an expression, the compiler implicitly generates a pointer to the array's first element, just as if the programmer had written &a[0]. (The exceptions are when the array is the operand of a sizeof or & operator, or is a string literal initializer for a character array.) As a consequence of this definition, the compiler doesn't apply the array subscripting operator [] that differently to arrays and pointers, after all. In an expression of the form a[i], the array decays into a pointer, following the rule above, and is then subscripted just as would be a pointer variable in the expression p[i] (although the eventual memory accesses will be different, as explained in question 6.2). If you were to assign the array's address to the pointer: p = a; then p[3] and a[3] would access the same element. See also questions 6.8 and 6.14. References: K&R1 Sec. 5.3 pp. 93-6; K&R2 Sec. 5.3 p. 99; ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>, Sec. 6.3.6; H&S Sec. 5.4.1 p. 124. 6.4: Then why are array and pointer declarations interchangeable as function formal parameters? A: It's supposed to be a convenience. Since arrays decay immediately into pointers, an array is never actually passed to a function. Allowing pointer parameters to be declared as arrays is a simply a way of making it look as though an array was being passed, perhaps because the parameter will be used within the function as if it were an array. Specifically, any parameter declarations which "look like" arrays, e.g. void f(char a[]) { ... } are treated by the compiler as if they were pointers, since that is what the function will receive if an array is passed: void f(char *a) { ... } This conversion holds only within function formal parameter declarations, nowhere else. If the conversion bothers you, avoid it; many programmers have concluded that the confusion it causes outweighs the small advantage of having the declaration "look like" the call or the uses within the function. See also question 6.21. References: K&R1 Sec. 5.3 p. 95, Sec. A10.1 p. 205; K&R2 Sec. 5.3 p. 100, Sec. A8.6.3 p. 218, Sec. A10.1 p. 226; ISO Sec. <IP_ADDRESS>, Sec. 6.7.1, Sec. 6.9.6; H&S Sec. 9.3 p. 271; CT&P Sec. 3.3 pp. 33-4. 6.7: How can an array be an lvalue, if you can't assign to it? A: The ANSI C Standard defines a "modifiable lvalue," which an array is not. References: ISO Sec. <IP_ADDRESS>; Rationale Sec. <IP_ADDRESS>; H&S Sec. 7.1 p. 179. 6.8: Practically speaking, what is the difference between arrays and pointers? A: Arrays automatically allocate space, but can't be relocated or resized. Pointers must be explicitly assigned to point to allocated space (perhaps using malloc), but can be reassigned (i.e. pointed at different objects) at will, and have many other uses besides serving as the base of blocks of memory. Due to the so-called equivalence of arrays and pointers (see question 6.3), arrays and pointers often seem interchangeable, and in particular a pointer to a block of memory assigned by malloc is frequently treated (and can be referenced using []) exactly as if it were a true array. See questions 6.14 and 6.16. (Be careful with sizeof, though.) See also questions 1.32 and 20.14. 6.9: Someone explained to me that arrays were really just constant pointers. A: This is a bit of an oversimplification. An array name is "constant" in that it cannot be assigned to, but an array is *not* a pointer, as the discussion and pictures in question 6.2 should make clear. See also questions 6.3 and 6.8. 6.11: I came across some "joke" code containing the "expression" 5["abcdef"] . How can this be legal C? A: Yes, Virginia, array subscripting is commutative in C. This curious fact follows from the pointer definition of array subscripting, namely that a[e] is identical to *((a)+(e)), for *any* two expressions a and e, as long as one of them is a pointer expression and one is integral. This unsuspected commutativity is often mentioned in C texts as if it were something to be proud of, but it finds no useful application outside of the Obfuscated C Contest (see question 20.36). References: Rationale Sec. <IP_ADDRESS>; H&S Sec. 5.4.1 p. 124, Sec. 7.4.1 pp. 186-7. 6.12: Since array references decay into pointers, if arr is an array, what's the difference between arr and &arr? A: The type. In Standard C, &arr yields a pointer, of type pointer-to-array- of-T, to the entire array. (In pre-ANSI C, the & in &arr generally elicited a warning, and was generally ignored.) Under all C compilers, a simple reference (without an explicit &) to an array yields a pointer, of type pointer-to-T, to the array's first element. (See also questions 6.3, 6.13, and 6.18.) References: ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>; Rationale Sec. <IP_ADDRESS>; H&S Sec. 7.5.6 p. 198. 6.13: How do I declare a pointer to an array? A: Usually, you don't want to. When people speak casually of a pointer to an array, they usually mean a pointer to its first element. Instead of a pointer to an array, consider using a pointer to one of the array's elements. Arrays of type T decay into pointers to type T (see question 6.3), which is convenient; subscripting or incrementing the resultant pointer will access the individual members of the array. True pointers to arrays, when subscripted or incremented, step over entire arrays, and are generally useful only when operating on arrays of arrays, if at all. (See question 6.18.) If you really need to declare a pointer to an entire array, use something like "int (*ap)[N];" where N is the size of the array. (See also question 1.21.) If the size of the array is unknown, N can in principle be omitted, but the resulting type, "pointer to array of unknown size," is useless. See also question 6.12 above. References: ISO Sec. <IP_ADDRESS>. 6.14: How can I set an array's size at run time? How can I avoid fixed-sized arrays? A: The equivalence between arrays and pointers (see question 6.3) allows a pointer to malloc'ed memory to simulate an array quite effectively. After executing #include <stdlib.h> int *dynarray; dynarray = malloc(10 * sizeof(int)); (and if the call to malloc succeeds), you can reference dynarray[i] (for i from 0 to 9) almost as if dynarray were a conventional, statically-allocated array (int a[10]). The only difference is that sizeof will not give the size of the "array". See also questions 1.31b, 6.16, and 7.7. 6.15: How can I declare local arrays of a size matching a passed-in array? A: Until recently, you couldn't. Array dimensions in C traditionally had to be compile-time constants. C9X will introduce variable-length arrays (VLA's) which will solve this problem; local arrays may have sizes set by variables or other expressions, perhaps involving function parameters. (gcc has provided parameterized arrays as an extension for some time.) If you can't use C9X or gcc, you'll have to use malloc(), and remember to call free() before the function returns. See also questions 6.14, 6.16, 6.19, 7.22, and maybe 7.32. References: ISO Sec. 6.4, Sec. <IP_ADDRESS>; C9X Sec. <IP_ADDRESS>. 6.16: How can I dynamically allocate a multidimensional array? A: The traditional solution is to allocate an array of pointers, and then initialize each pointer to a dynamically-allocated "row." Here is a two-dimensional example: #include <stdlib.h> int **array1 = malloc(nrows * sizeof(int *)); for(i = 0; i < nrows; i++) array1[i] = malloc(ncolumns * sizeof(int)); (In real code, of course, all of malloc's return values would be checked.) You can keep the array's contents contiguous, at the cost of making later reallocation of individual rows more difficult, with a bit of explicit pointer arithmetic: int **array2 = malloc(nrows * sizeof(int *)); array2[0] = malloc(nrows * ncolumns * sizeof(int)); for(i = 1; i < nrows; i++) array2[i] = array2[0] + i * ncolumns; In either case, the elements of the dynamic array can be accessed with normal-looking array subscripts: arrayx[i][j] (for 0 <= i < nrows and 0 <= j < ncolumns). If the double indirection implied by the above schemes is for some reason unacceptable, you can simulate a two-dimensional array with a single, dynamically-allocated one-dimensional array: int *array3 = malloc(nrows * ncolumns * sizeof(int)); However, you must now perform subscript calculations manually, accessing the i,jth element with array3[i * ncolumns + j]. (A macro could hide the explicit calculation, but invoking it would require parentheses and commas which wouldn't look exactly like multidimensional array syntax, and the macro would need access to at least one of the dimensions, as well. See also question 6.19.) Yet another option is to use pointers to arrays: int (*array4)[NCOLUMNS] = malloc(nrows * sizeof(*array4)); but the syntax starts getting horrific and at most one dimension may be specified at run time. With all of these techniques, you may of course need to remember to free the arrays (which may take several steps; see question 7.23) when they are no longer needed, and you cannot necessarily intermix dynamically-allocated arrays with conventional, statically-allocated ones (see question 6.20, and also question 6.18). Finally, in C9X you can use a variable-length array. All of these techniques can also be extended to three or more dimensions. References: C9X Sec. <IP_ADDRESS>. 6.17: Here's a neat trick: if I write int realarray[10]; int *array = &realarray[-1]; I can treat "array" as if it were a 1-based array. A: Although this technique is attractive (and was used in old editions of the book _Numerical Recipes in C_), it is not strictly conforming to the C Standard. Pointer arithmetic is defined only as long as the pointer points within the same allocated block of memory, or to the imaginary "terminating" element one past it; otherwise, the behavior is undefined, *even if the pointer is not dereferenced*. The code above could fail if, while subtracting the offset, an illegal address were generated (perhaps because the address tried to "wrap around" past the beginning of some memory segment). References: K&R2 Sec. 5.3 p. 100, Sec. 5.4 pp. 102-3, Sec. A7.7 pp. 205-6; ISO Sec. 6.3.6; Rationale Sec. <IP_ADDRESS>. 6.18: My compiler complained when I passed a two-dimensional array to a function expecting a pointer to a pointer. A: The rule (see question 6.3) by which arrays decay into pointers is not applied recursively. An array of arrays (i.e. a two- dimensional array in C) decays into a pointer to an array, not a pointer to a pointer. Pointers to arrays can be confusing, and must be treated carefully; see also question 6.13. If you are passing a two-dimensional array to a function: int array[NROWS][NCOLUMNS]; f(array); the function's declaration must match: void f(int a[][NCOLUMNS]) { ... } or void f(int (*ap)[NCOLUMNS]) /* ap is a pointer to an array */ { ... } In the first declaration, the compiler performs the usual implicit parameter rewriting of "array of array" to "pointer to array" (see questions 6.3 and 6.4); in the second form the pointer declaration is explicit. Since the called function does not allocate space for the array, it does not need to know the overall size, so the number of rows, NROWS, can be omitted. The width of the array is still important, so the column dimension NCOLUMNS (and, for three- or more dimensional arrays, the intervening ones) must be retained. If a function is already declared as accepting a pointer to a pointer, it is almost certainly meaningless to pass a two- dimensional array directly to it. See also questions 6.12 and 6.15. References: K&R1 Sec. 5.10 p. 110; K&R2 Sec. 5.9 p. 113; H&S Sec. 5.4.3 p. 126. 6.19: How do I write functions which accept two-dimensional arrays when the width is not known at compile time? A: It's not always easy. One way is to pass in a pointer to the [0][0] element, along with the two dimensions, and simulate array subscripting "by hand": void f2(int *aryp, int nrows, int ncolumns) { ... array[i][j] is accessed as aryp[i * ncolumns + j] ... } This function could be called with the array from question 6.18 as f2(&array[0][0], NROWS, NCOLUMNS); It must be noted, however, that a program which performs multidimensional array subscripting "by hand" in this way is not in strict conformance with the ANSI C Standard; according to an official interpretation, the behavior of accessing (&array[0][0])[x] is not defined for x >= NCOLUMNS. C9X will allow variable-length arrays, and once compilers which accept C9X's extensions become widespread, this will probably become the preferred solution. (gcc has supported variable- sized arrays for some time.) When you want to be able to use a function on multidimensional arrays of various sizes, one solution is to simulate all the arrays dynamically, as in question 6.16. See also questions 6.18, 6.20, and 6.15. References: ISO Sec. 6.3.6; C9X Sec. <IP_ADDRESS>. 6.20: How can I use statically- and dynamically-allocated multidimensional arrays interchangeably when passing them to functions? A: There is no single perfect method. Given the declarations int array[NROWS][NCOLUMNS]; int **array1; /* ragged */ int **array2; /* contiguous */ int *array3; /* "flattened" */ int (*array4)[NCOLUMNS]; with the pointers initialized as in the code fragments in question 6.16, and functions declared as void f1a(int a[][NCOLUMNS], int nrows, int ncolumns); void f1b(int (*a)[NCOLUMNS], int nrows, int ncolumns); void f2(int *aryp, int nrows, int ncolumns); void f3(int **pp, int nrows, int ncolumns); where f1a() and f1b() accept conventional two-dimensional arrays, f2() accepts a "flattened" two-dimensional array, and f3() accepts a pointer-to-pointer, simulated array (see also questions 6.18 and 6.19), the following calls should work as expected: f1a(array, NROWS, NCOLUMNS); f1b(array, NROWS, NCOLUMNS); f1a(array4, nrows, NCOLUMNS); f1b(array4, nrows, NCOLUMNS); f2(&array[0][0], NROWS, NCOLUMNS); f2(*array, NROWS, NCOLUMNS); f2(*array2, nrows, ncolumns); f2(array3, nrows, ncolumns); f2(*array4, nrows, NCOLUMNS); f3(array1, nrows, ncolumns); f3(array2, nrows, ncolumns); The following calls would probably work on most systems, but involve questionable casts, and work only if the dynamic ncolumns matches the static NCOLUMNS: f1a((int (*)[NCOLUMNS])(*array2), nrows, ncolumns); f1a((int (*)[NCOLUMNS])(*array2), nrows, ncolumns); f1b((int (*)[NCOLUMNS])array3, nrows, ncolumns); f1b((int (*)[NCOLUMNS])array3, nrows, ncolumns); It must again be noted that passing &array[0][0] (or, equivalently, *array) to f2() is not strictly conforming; see question 6.19. If you can understand why all of the above calls work and are written as they are, and if you understand why the combinations that are not listed would not work, then you have a *very* good understanding of arrays and pointers in C. Rather than worrying about all of this, one approach to using multidimensional arrays of various sizes is to make them *all* dynamic, as in question 6.16. If there are no static multidimensional arrays -- if all arrays are allocated like array1 or array2 in question 6.16 -- then all functions can be written like f3(). 6.21: Why doesn't sizeof properly report the size of an array when the array is a parameter to a function? A: The compiler pretends that the array parameter was declared as a pointer (see question 6.4), and sizeof reports the size of the pointer. References: H&S Sec. 7.5.2 p. 195. Section 7. Memory Allocation 7.1: Why doesn't this fragment work? char *answer; printf("Type something:\n"); gets(answer); printf("You typed \"%s\"\n", answer); A: The pointer variable answer, which is handed to gets() as the location into which the response should be stored, has not been set to point to any valid storage. That is, we cannot say where the pointer answer points. (Since local variables are not initialized, and typically contain garbage, it is not even guaranteed that answer starts out as a null pointer. See questions 1.30 and 5.1.) The simplest way to correct the question-asking program is to use a local array, instead of a pointer, and let the compiler worry about allocation: #include <stdio.h> #include <string.h> char answer[100], *p; printf("Type something:\n"); fgets(answer, sizeof answer, stdin); if((p = strchr(answer, '\n')) != NULL) *p = '\0'; printf("You typed \"%s\"\n", answer); This example also uses fgets() instead of gets(), so that the end of the array cannot be overwritten. (See question 12.23. Unfortunately for this example, fgets() does not automatically delete the trailing \n, as gets() would.) It would also be possible to use malloc() to allocate the answer buffer. 7.2: I can't get strcat() to work. I tried char *s1 = "Hello, "; char *s2 = "world!"; char *s3 = strcat(s1, s2); but I got strange results. A: As in question 7.1 above, the main problem here is that space for the concatenated result is not properly allocated. C does not provide an automatically-managed string type. C compilers only allocate memory for objects explicitly mentioned in the source code (in the case of strings, this includes character arrays and string literals). The programmer must arrange for sufficient space for the results of run-time operations such as string concatenation, typically by declaring arrays, or by calling malloc(). strcat() performs no allocation; the second string is appended to the first one, in place. Therefore, one fix would be to declare the first string as an array: char s1[20] = "Hello, "; Since strcat() returns the value of its first argument (s1, in this case), the variable s3 is superfluous; after the call to strcat(), s1 contains the result. The original call to strcat() in the question actually has two problems: the string literal pointed to by s1, besides not being big enough for any concatenated text, is not necessarily writable at all. See question 1.32. References: CT&P Sec. 3.2 p. 32. 7.3: But the man page for strcat() says that it takes two char *'s as arguments. How am I supposed to know to allocate things? A: In general, when using pointers you *always* have to consider memory allocation, if only to make sure that the compiler is doing it for you. If a library function's documentation does not explicitly mention allocation, it is usually the caller's problem. The Synopsis section at the top of a Unix-style man page or in the ANSI C standard can be misleading. The code fragments presented there are closer to the function definitions used by an implementor than the invocations used by the caller. In particular, many functions which accept pointers (e.g. to structures or strings) are usually called with a pointer to some object (a structure, or an array -- see questions 6.3 and 6.4) which the caller has allocated. Other common examples are time() (see question 13.12) and stat(). 7.3b: I just tried the code char *p; strcpy(p, "abc"); and it worked. How? Why didn't it crash? A: You got lucky, I guess. The memory pointed to by the unitialized pointer p happened to be writable by you, and apparently was not already in use for anything vital. 7.3c: How much memory does a pointer variable allocate? A: That's a pretty misleading question. When you declare a pointer variable, as in char *p; you (or, more properly, the compiler) have allocated only enough memory to hold the pointer itself; that is, in this case you have allocated sizeof(char *) bytes of memory. But you have not yet allocated *any* memory for the pointer to point to. See also questions 7.1 and 7.2. 7.5a: I have a function that is supposed to return a string, but when it returns to its caller, the returned string is garbage. A: Make sure that the pointed-to memory is properly allocated. For example, make sure you have *not* done something like char *itoa(int n) { char retbuf[20]; /* WRONG */ sprintf(retbuf, "%d", n); return retbuf; /* WRONG */ } One fix (which is imperfect, especially if the function in question is called recursively, or if several of its return values are needed simultaneously) would be to declare the return buffer as static char retbuf[20]; See also questions 7.5b, 12.21, and 20.1. References: ISO Sec. <IP_ADDRESS>. 7.5b: So what's the right way to return a string or other aggregate? A: The returned pointer should be to a statically-allocated buffer, or to a buffer passed in by the caller, or to memory obtained with malloc(), but *not* to a local (automatic) array. See also question 20.1. 7.6: Why am I getting "warning: assignment of pointer from integer lacks a cast" for calls to malloc()? A: Have you #included <stdlib.h>, or otherwise arranged for malloc() to be declared properly? See also question 1.25. References: H&S Sec. 4.7 p. 101. 7.7: Why does some code carefully cast the values returned by malloc to the pointer type being allocated? A: Before ANSI/ISO Standard C introduced the void * generic pointer type, these casts were typically required to silence warnings (and perhaps induce conversions) when assigning between incompatible pointer types. Under ANSI/ISO Standard C, these casts are no longer necessary, and in fact modern practice discourages them, since they can camouflage important warnings which would otherwise be generated if malloc() happened not to be declared correctly; see question 7.6 above. (However, the casts are typically seen in C code which for one reason or another is intended to be compatible with C++, where explicit casts from void * are required.) References: H&S Sec. 16.1 pp. 386-7. 7.8: I see code like char *p = malloc(strlen(s) + 1); strcpy(p, s); Shouldn't that be malloc((strlen(s) + 1) * sizeof(char))? A: It's never necessary to multiply by sizeof(char), since sizeof(char) is, by definition, exactly 1. (On the other hand, multiplying by sizeof(char) doesn't hurt, and in some circumstances may help by introducing a size_t into the expression.) See also question 8.9. References: ISO Sec. <IP_ADDRESS>; H&S Sec. 7.5.2 p. 195. 7.14: I've heard that some operating systems don't actually allocate malloc'ed memory until the program tries to use it. Is this legal? A: It's hard to say. The Standard doesn't say that systems can act this way, but it doesn't explicitly say that they can't, either. References: ISO Sec. 7.10.3. 7.16: I'm allocating a large array for some numeric work, using the line double *array = malloc(300 * 300 * sizeof(double)); malloc() isn't returning null, but the program is acting strangely, as if it's overwriting memory, or malloc() isn't allocating as much as I asked for, or something. A: Notice that 300 x 300 is 90,000, which will not fit in a 16-bit int, even before you multiply it by sizeof(double). If you need to allocate this much memory, you'll have to be careful. If size_t (the type accepted by malloc()) is a 32-bit type on your machine, but int is 16 bits, you might be able to get away with writing 300 * (300 * sizeof(double)) (see question 3.14). Otherwise, you'll have to break your data structure up into smaller chunks, or use a 32-bit machine or compiler, or use some nonstandard memory allocation functions. See also question 19.23. 7.17: I've got 8 meg of memory in my PC. Why can I only seem to malloc 640K or so? A: Under the segmented architecture of PC compatibles, it can be difficult to use more than 640K with any degree of transparency, especially under MS-DOS. See also question 19.23. 7.19: My program is crashing, apparently somewhere down inside malloc, but I can't see anything wrong with it. Is there a bug in malloc()? A: It is unfortunately very easy to corrupt malloc's internal data structures, and the resulting problems can be stubborn. The most common source of problems is writing more to a malloc'ed region than it was allocated to hold; a particularly common bug is to malloc(strlen(s)) instead of strlen(s) + 1. Other problems may involve using pointers to memory that has been freed, freeing pointers twice, freeing pointers not obtained from malloc, or trying to realloc a null pointer (see question 7.30). See also questions 7.26, 16.8, and 18.2. 7.20: You can't use dynamically-allocated memory after you free it, can you? A: No. Some early documentation for malloc() stated that the contents of freed memory were "left undisturbed," but this ill- advised guarantee was never universal and is not required by the C Standard. Few programmers would use the contents of freed memory deliberately, but it is easy to do so accidentally. Consider the following (correct) code for freeing a singly-linked list: struct list *listp, *nextp; for(listp = base; listp != NULL; listp = nextp) { nextp = listp->next; free(listp); } and notice what would happen if the more-obvious loop iteration expression listp = listp->next were used, without the temporary nextp pointer. References: K&R2 Sec. 7.8.5 p. 167; ISO Sec. 7.10.3; Rationale Sec. <IP_ADDRESS>; H&S Sec. 16.2 p. 387; CT&P Sec. 7.10 p. 95. 7.21: Why isn't a pointer null after calling free()? How unsafe is it to use (assign, compare) a pointer value after it's been freed? A: When you call free(), the memory pointed to by the passed pointer is freed, but the value of the pointer in the caller probably remains unchanged, because C's pass-by-value semantics mean that called functions never permanently change the values of their arguments. (See also question 4.8.) A pointer value which has been freed is, strictly speaking, invalid, and *any* use of it, even if is not dereferenced, can theoretically lead to trouble, though as a quality of implementation issue, most implementations will probably not go out of their way to generate exceptions for innocuous uses of invalid pointers. References: ISO Sec. 7.10.3; Rationale Sec. <IP_ADDRESS>. 7.22: When I call malloc() to allocate memory for a pointer which is local to a function, do I have to explicitly free() it? A: Yes. Remember that a pointer is different from what it points to. Local variables are deallocated when the function returns, but in the case of a pointer variable, this means that the pointer is deallocated, *not* what it points to. Memory allocated with malloc() always persists until you explicitly free it. In general, for every call to malloc(), there should be a corresponding call to free(). 7.23: I'm allocating structures which contain pointers to other dynamically-allocated objects. When I free a structure, do I also have to free each subsidiary pointer? A: Yes. In general, you must arrange that each pointer returned from malloc() be individually passed to free(), exactly once (if it is freed at all). A good rule of thumb is that for each call to malloc() in a program, you should be able to point at the call to free() which frees the memory allocated by that malloc() call. See also question 7.24. 7.24: Must I free allocated memory before the program exits? A: You shouldn't have to. A real operating system definitively reclaims all memory and other resources when a program exits. Nevertheless, some personal computers are said not to reliably recover memory, and all that can be inferred from the ANSI/ISO C Standard is that this is a "quality of implementation issue." References: ISO Sec. <IP_ADDRESS>. 7.25: I have a program which mallocs and later frees a lot of memory, but I can see from the operating system that memory usage doesn't actually go back down. A: Most implementations of malloc/free do not return freed memory to the operating system, but merely make it available for future malloc() calls within the same program. 7.26: How does free() know how many bytes to free? A: The malloc/free implementation remembers the size of each block as it is allocated, so it is not necessary to remind it of the size when freeing. 7.27: So can I query the malloc package to find out how big an allocated block is? A: Unfortunately, there is no standard or portable way. 7.30: Is it legal to pass a null pointer as the first argument to realloc()? Why would you want to? A: ANSI C sanctions this usage (and the related realloc(..., 0), which frees), although several earlier implementations do not support it, so it may not be fully portable. Passing an initially-null pointer to realloc() can make it easier to write a self-starting incremental allocation algorithm. References: ISO Sec. <IP_ADDRESS>; H&S Sec. 16.3 p. 388. 7.31: What's the difference between calloc() and malloc()? Is it safe to take advantage of calloc's zero-filling? Does free() work on memory allocated with calloc(), or do you need a cfree()? A: calloc(m, n) is essentially equivalent to p = malloc(m * n); memset(p, 0, m * n); The zero fill is all-bits-zero, and does *not* therefore guarantee useful null pointer values (see section 5 of this list) or floating-point zero values. free() is properly used to free the memory allocated by calloc(). References: ISO Sec. 7.10.3 to <IP_ADDRESS>; H&S Sec. 16.1 p. 386, Sec. 16.2 p. 386; PCS Sec. 11 pp. 141,142. 7.32: What is alloca() and why is its use discouraged? A: alloca() allocates memory which is automatically freed when the function which called alloca() returns. That is, memory allocated with alloca is local to a particular function's "stack frame" or context. alloca() cannot be written portably, and is difficult to implement on machines without a conventional stack. Its use is problematical (and the obvious implementation on a stack-based machine fails) when its return value is passed directly to another function, as in fgets(alloca(100), 100, stdin). For these reasons, alloca() is not Standard and cannot be used in programs which must be widely portable, no matter how useful it might be. See also question 7.22. References: Rationale Sec. 4.10.3. Section 8. Characters and Strings 8.1: Why doesn't strcat(string, '!'); work? A: There is a very real difference between characters and strings, and strcat() concatenates *strings*. Characters in C are represented by small integers corresponding to their character set values (see also question 8.6 below). Strings are represented by arrays of characters; you usually manipulate a pointer to the first character of the array. It is never correct to use one when the other is expected. To append a ! to a string, use strcat(string, "!"); See also questions 1.32, 7.2, and 16.6. References: CT&P Sec. 1.5 pp. 9-10. 8.2: I'm checking a string to see if it matches a particular value. Why isn't this code working? char *string; ... if(string == "value") { /* string matches "value" */ ... } A: Strings in C are represented as arrays of characters, and C never manipulates (assigns, compares, etc.) arrays as a whole. The == operator in the code fragment above compares two pointers -- the value of the pointer variable string and a pointer to the string literal "value" -- to see if they are equal, that is, if they point to the same place. They probably don't, so the comparison never succeeds. To compare two strings, you generally use the library function strcmp(): if(strcmp(string, "value") == 0) { /* string matches "value" */ ... } 8.3: If I can say char a[] = "Hello, world!"; why can't I say char a[14]; a = "Hello, world!"; A: Strings are arrays, and you can't assign arrays directly. Use strcpy() instead: strcpy(a, "Hello, world!"); See also questions 1.32, 4.2, and 7.2. 8.6: How can I get the numeric (character set) value corresponding to a character, or vice versa? A: In C, characters are represented by small integers corresponding to their values (in the machine's character set), so you don't need a conversion function: if you have the character, you have its value. 8.9: I think something's wrong with my compiler: I just noticed that sizeof('a') is 2, not 1 (i.e. not sizeof(char)). A: Perhaps surprisingly, character constants in C are of type int, so sizeof('a') is sizeof(int) (though this is another area where C++ differs). See also question 7.8. References: ISO Sec. <IP_ADDRESS>; H&S Sec. 2.7.3 p. 29. Section 9. Boolean Expressions and Variables 9.1: What is the right type to use for Boolean values in C? Why isn't it a standard type? Should I use #defines or enums for the true and false values? A: C does not provide a standard Boolean type, in part because picking one involves a space/time tradeoff which can best be decided by the programmer. (Using an int may be faster, while using char may save data space. Smaller types may make the generated code bigger or slower, though, if they require lots of conversions to and from int.) The choice between #defines and enumeration constants for the true/false values is arbitrary and not terribly interesting (see also questions 2.22 and 17.10). Use any of #define TRUE 1 #define YES 1 #define FALSE 0 #define NO 0 enum bool {false, true}; enum bool {no, yes}; or use raw 1 and 0, as long as you are consistent within one program or project. (An enumeration may be preferable if your debugger shows the names of enumeration constants when examining variables.) Some people prefer variants like #define TRUE (1==1) #define FALSE (!TRUE) or define "helper" macros such as #define Istrue(e) ((e) != 0) These don't buy anything (see question 9.2 below; see also questions 5.12 and 10.2). 9.2: Isn't #defining TRUE to be 1 dangerous, since any nonzero value is considered "true" in C? What if a built-in logical or relational operator "returns" something other than 1? A: It is true (sic) that any nonzero value is considered true in C, but this applies only "on input", i.e. where a Boolean value is expected. When a Boolean value is generated by a built-in operator, it is guaranteed to be 1 or 0. Therefore, the test if((a == b) == TRUE) would work as expected (as long as TRUE is 1), but it is obviously silly. In fact, explicit tests against TRUE and FALSE are generally inappropriate, because some library functions (notably isupper(), isalpha(), etc.) return, on success, a nonzero value which is not necessarily 1. (Besides, if you believe that "if((a == b) == TRUE)" is an improvement over "if(a == b)", why stop there? Why not use "if(((a == b) == TRUE) == TRUE)"?) A good rule of thumb is to use TRUE and FALSE (or the like) only for assignment to a Boolean variable or function parameter, or as the return value from a Boolean function, but never in a comparison. The preprocessor macros TRUE and FALSE (and, of course, NULL) are used for code readability, not because the underlying values might ever change. (See also questions 5.3 and 5.10.) Although the use of macros like TRUE and FALSE (or YES and NO) seems clearer, Boolean values and definitions can be sufficiently confusing in C that some programmers feel that TRUE and FALSE macros only compound the confusion, and prefer to use raw 1 and 0 instead. (See also question 5.9.) References: K&R1 Sec. 2.6 p. 39, Sec. 2.7 p. 41; K&R2 Sec. 2.6 p. 42, Sec. 2.7 p. 44, Sec. A7.4.7 p. 204, Sec. A7.9 p. 206; ISO Sec. <IP_ADDRESS>, Sec. 6.3.8, Sec. 6.3.9, Sec. 6.3.13, Sec. 6.3.14, Sec. 6.3.15, Sec. <IP_ADDRESS>, Sec. 6.6.5; H&S Sec. 7.5.4 pp. 196-7, Sec. 7.6.4 pp. 207-8, Sec. 7.6.5 pp. 208-9, Sec. 7.7 pp. 217-8, Sec. 7.8 pp. 218-9, Sec. 8.5 pp. 238-9, Sec. 8.6 pp. 241-4; "What the Tortoise Said to Achilles". 9.3: Is if(p), where p is a pointer, a valid conditional? A: Yes. See question 5.3. Section 10. C Preprocessor 10.2: Here are some cute preprocessor macros: #define begin { #define end } What do y'all think? A: Bleah. See also section 17. 10.3: How can I write a generic macro to swap two values? A: There is no good answer to this question. If the values are integers, a well-known trick using exclusive-OR could perhaps be used, but it will not work for floating-point values or pointers, or if the two values are the same variable. (See questions 3.3b and 20.15c.) If the macro is intended to be used on values of arbitrary type (the usual goal), it cannot use a temporary, since it does not know what type of temporary it needs (and would have a hard time picking a name for it if it did), and standard C does not provide a typeof operator. The best all-around solution is probably to forget about using a macro, unless you're willing to pass in the type as a third argument. 10.4: What's the best way to write a multi-statement macro? A: The usual goal is to write a macro that can be invoked as if it were a statement consisting of a single function call. This means that the "caller" will be supplying the final semicolon, so the macro body should not. The macro body cannot therefore be a simple brace-enclosed compound statement, because syntax errors would result if it were invoked (apparently as a single statement, but with a resultant extra semicolon) as the if branch of an if/else statement with an explicit else clause. The traditional solution, therefore, is to use #define MACRO(arg1, arg2) do { \ /* declarations */ \ stmt1; \ stmt2; \ /* ... */ \ } while(0) /* (no trailing ; ) */ When the caller appends a semicolon, this expansion becomes a single statement regardless of context. (An optimizing compiler will remove any "dead" tests or branches on the constant condition 0, although lint may complain.) If all of the statements in the intended macro are simple expressions, with no declarations or loops, another technique is to write a single, parenthesized expression using one or more comma operators. (For an example, see the first DEBUG() macro in question 10.26.) This technique also allows a value to be "returned." References: H&S Sec. 3.3.2 p. 45; CT&P Sec. 6.3 pp. 82-3. 10.6: I'm splitting up a program into multiple source files for the first time, and I'm wondering what to put in .c files and what to put in .h files. (What does ".h" mean, anyway?) A: As a general rule, you should put these things in header (.h) files: macro definitions (preprocessor #defines) structure, union, and enumeration declarations typedef declarations external function declarations (see also question 1.11) global variable declarations It's especially important to put a declaration or definition in a header file when it will be shared between several other files. (In particular, never put external function prototypes in .c files. See also question 1.7.) On the other hand, when a definition or declaration should remain private to one .c file, it's fine to leave it there. See also questions 1.7 and 10.7. References: K&R2 Sec. 4.5 pp. 81-2; H&S Sec. 9.2.3 p. 267; CT&P Sec. 4.6 pp. 66-7. 10.7: Is it acceptable for one header file to #include another? A: It's a question of style, and thus receives considerable debate. Many people believe that "nested #include files" are to be avoided: the prestigious Indian Hill Style Guide (see question 17.9) disparages them; they can make it harder to find relevant definitions; they can lead to multiple-definition errors if a file is #included twice; and they make manual Makefile maintenance very difficult. On the other hand, they make it possible to use header files in a modular way (a header file can #include what it needs itself, rather than requiring each #includer to do so); a tool like grep (or a tags file) makes it easy to find definitions no matter where they are; a popular trick along the lines of: #ifndef HFILENAME_USED #define HFILENAME_USED ...header file contents... #endif (where a different bracketing macro name is used for each header file) makes a header file "idempotent" so that it can safely be #included multiple times; and automated Makefile maintenance tools (which are a virtual necessity in large projects anyway; see question 18.1) handle dependency generation in the face of nested #include files easily. See also question 17.10. References: Rationale Sec. 4.1.2. 10.8a: What's the difference between #include <> and #include "" ? A: The <> syntax is typically used with Standard or system-supplied headers, while "" is typically used for a program's own header files. 10.8b: What are the complete rules for header file searching? A: The exact behavior is implementation-defined (which means that it is supposed to be documented; see question 11.33). Typically, headers named with <> syntax are searched for in one or more standard places. Header files named with "" syntax are first searched for in the "current directory," then (if not found) in the same standard places. Traditionally (especially under Unix compilers), the current directory is taken to be the directory containing the file containing the #include directive. Under other compilers, however, the current directory (if any) is the directory in which the compiler was initially invoked. Check your compiler documentation. References: K&R2 Sec. A12.4 p. 231; ISO Sec. 6.8.2; H&S Sec. 3.4 p. 55. 10.9: I'm getting strange syntax errors on the very first declaration in a file, but it looks fine. A: Perhaps there's a missing semicolon at the end of the last declaration in the last header file you're #including. See also questions 2.18, 11.29, and 16.1b. 10.10b: I'm #including the right header file for the library function I'm using, but the linker keeps saying it's undefined. A: See question 13.25. 10.11: I seem to be missing the system header file <sgtty.h>. Can someone send me a copy? A: Standard headers exist in part so that definitions appropriate to your compiler, operating system, and processor can be supplied. You cannot just pick up a copy of someone else's header file and expect it to work, unless that person is using exactly the same environment. Ask your compiler vendor why the file was not provided (or to send a replacement copy). 10.12: How can I construct preprocessor #if expressions which compare strings? A: You can't do it directly; preprocessor #if arithmetic uses only integers. An alternative is to #define several macros with symbolic names and distinct integer values, and implement conditionals on those. See also question 20.17. References: K&R2 Sec. 4.11.3 p. 91; ISO Sec. 6.8.1; H&S Sec. 7.11.1 p. 225. 10.13: Does the sizeof operator work in preprocessor #if directives? A: No. Preprocessing happens during an earlier phase of compilation, before type names have been parsed. Instead of sizeof, consider using the predefined constants in ANSI's <limits.h>, if applicable, or perhaps a "configure" script. (Better yet, try to write code which is inherently insensitive to type sizes; see also question 1.1.) References: ISO Sec. <IP_ADDRESS>, Sec. 6.8.1; H&S Sec. 7.11.1 p. 225. 10.14: Can I use an #ifdef in a #define line, to define something two different ways? A: No. You can't "run the preprocessor on itself," so to speak. What you can do is use one of two completely separate #define lines, depending on the #ifdef setting. References: ISO Sec. 6.8.3, Sec. <IP_ADDRESS>; H&S Sec. 3.2 pp. 40-1. 10.15: Is there anything like an #ifdef for typedefs? A: Unfortunately, no. You may have to keep sets of preprocessor macros (e.g. MY_TYPE_DEFINED) recording whether certain typedefs have been declared. (See also question 10.13.) References: ISO Sec. <IP_ADDRESS>, Sec. 6.8.1; H&S Sec. 7.11.1 p. 225. 10.16: How can I use a preprocessor #if expression to tell if a machine is big-endian or little-endian? A: You probably can't. (Preprocessor arithmetic uses only long integers, and there is no concept of addressing.) Are you sure you need to know the machine's endianness explicitly? Usually it's better to write code which doesn't care. See also question 20.9. References: ISO Sec. 6.8.1; H&S Sec. 7.11.1 p. 225. 10.18: I inherited some code which contains far too many #ifdef's for my taste. How can I preprocess the code to leave only one conditional compilation set, without running it through the preprocessor and expanding all of the #include's and #define's as well? A: There are programs floating around called unifdef, rmifdef, and scpp ("selective C preprocessor") which do exactly this. See question 18.16. 10.19: How can I list all of the predefined identifiers? A: There's no standard way, although it is a common need. gcc provides a -dM option which works with -E, and other compilers may provide something similar. If the compiler documentation is unhelpful, the most expedient way is probably to extract printable strings from the compiler or preprocessor executable with something like the Unix strings utility. Beware that many traditional system-specific predefined identifiers (e.g. "unix") are non-Standard (because they clash with the user's namespace) and are being removed or renamed. 10.20: I have some old code that tries to construct identifiers with a macro like #define Paste(a, b) a/**/b but it doesn't work any more. A: It was an undocumented feature of some early preprocessor implementations (notably John Reiser's) that comments disappeared entirely and could therefore be used for token pasting. ANSI affirms (as did K&R1) that comments are replaced with white space. However, since the need for pasting tokens was demonstrated and real, ANSI introduced a well-defined token- pasting operator, ##, which can be used like this: #define Paste(a, b) a##b See also question 11.17. References: ISO Sec. <IP_ADDRESS>; Rationale Sec. <IP_ADDRESS>; H&S Sec. 3.3.9 p. 52. 10.22: Why is the macro #define TRACE(n) printf("TRACE: %d\n", n) giving me the warning "macro replacement within a string literal"? It seems to be expanding TRACE(count); as printf("TRACE: %d\count", count); A: See question 11.18. 10.23-4: I'm having trouble using macro arguments inside string literals, using the `#' operator. A: See questions 11.17 and 11.18. 10.25: I've got this tricky preprocessing I want to do and I can't figure out a way to do it. A: C's preprocessor is not intended as a general-purpose tool. (Note also that it is not guaranteed to be available as a separate program.) Rather than forcing it to do something inappropriate, consider writing your own little special-purpose preprocessing tool, instead. You can easily get a utility like make(1) to run it for you automatically. If you are trying to preprocess something other than C, consider using a general-purpose preprocessor. (One older one available on most Unix systems is m4.) 10.26: How can I write a macro which takes a variable number of arguments? A: One popular trick is to define and invoke the macro with a single, parenthesized "argument" which in the macro expansion becomes the entire argument list, parentheses and all, for a function such as printf(): #define DEBUG(args) (printf("DEBUG: "), printf args) if(n != 0) DEBUG(("n is %d\n", n)); The obvious disadvantage is that the caller must always remember to use the extra parentheses. gcc has an extension which allows a function-like macro to accept a variable number of arguments, but it's not standard. Other possible solutions are to use different macros (DEBUG1, DEBUG2, etc.) depending on the number of arguments, or to play tricky games with commas: #define DEBUG(args) (printf("DEBUG: "), printf(args)) #define _ , DEBUG("i = %d" _ i) C9X will introduce formal support for function-like macros with variable-length argument lists. The notation ... will appear at the end of the macro "prototype" (just as it does for varargs functions), and the pseudomacro __VA_ARGS__ in the macro definition will be replaced by the variable arguments during invocation. Finally, you can always use a bona-fide function, which can take a variable number of arguments in a well-defined way. See questions 15.4 and 15.5. (If you needed a macro replacement, try using a function plus a non-function-like macro, e.g. #define printf myprintf .) References: C9X Sec. 6.8.3, Sec. <IP_ADDRESS>. Section 11. ANSI/ISO Standard C 11.1: What is the "ANSI C Standard?" A: In 1983, the American National Standards Institute (ANSI) commissioned a committee, X3J11, to standardize the C language. After a long, arduous process, including several widespread public reviews, the committee's work was finally ratified as ANS X3.159-1989 on December 14, 1989, and published in the spring of 1990. For the most part, ANSI C standardizes existing practice, with a few additions from C++ (most notably function prototypes) and support for multinational character sets (including the controversial trigraph sequences). The ANSI C standard also formalizes the C run-time library support routines. More recently, the Standard has been adopted as an international standard, ISO/IEC 9899:1990, and this ISO Standard replaces the earlier X3.159 even within the United States (where it is known as ANSI/ISO 9899-1990 [1992]). As an ISO Standard, it is subject to ongoing revision through the release of Technical Corrigenda and Normative Addenda. In 1994, Technical Corrigendum 1 (TC1) amended the Standard in about 40 places, most of them minor corrections or clarifications, and Normative Addendum 1 (NA1) added about 50 pages of new material, mostly specifying new library functions for internationalization. In 1995, TC2 added a few more minor corrections. As of this writing, a complete revision of the Standard is in its final stages. The new Standard is nicknamed "C9X" on the assumption that it will be finished by the end of 1999. (Many of this article's answers have been updated to reflect new C9X features.) The original ANSI Standard included a "Rationale," explaining many of its decisions, and discussing a number of subtle points, including several of those covered here. (The Rationale was "not part of ANSI Standard X3.159-1989, but... included for information only," and is not included with the ISO Standard. A new one is being prepared for C9X.) 11.2: How can I get a copy of the Standard? A: Copies are available in the United States from American National Standards Institute 11 W. 42nd St., 13th floor New York, NY 10036 USA (+1)<PHONE_NUMBER> and Global Engineering Documents 15 Inverness Way E Englewood, CO 80112 USA (+1)<PHONE_NUMBER><PHONE_NUMBER> (U.S. & Canada) In other countries, contact the appropriate national standards body, or ISO in Geneva at: ISO Sales Case Postale 56 CH-1211 Geneve 20 Switzerland (or see URL http://www.iso.ch or check the comp.std.internat FAQ list, Standards.Faq). The last time I checked, the cost was $130.00 from ANSI or $400.50 from Global. Copies of the original X3.159 (including the Rationale) may still be available at $205.00 from ANSI or $162.50 from Global. Note that ANSI derives revenues to support its operations from the sale of printed standards, so electronic copies are *not* available. In the U.S., it may be possible to get a copy of the original ANSI X3.159 (including the Rationale) as "FIPS PUB 160" from National Technical Information Service (NTIS) U.S. Department of Commerce Springfield, VA 22161<PHONE_NUMBER> The mistitled _Annotated ANSI C Standard_, with annotations by Herbert Schildt, contains most of the text of ISO 9899; it is published by Osborne/McGraw-Hill, ISBN 0-07-881952-0, and sells in the U.S. for approximately $40. It has been suggested that the price differential between this work and the official standard reflects the value of the annotations: they are plagued by numerous errors and omissions, and a few pages of the Standard itself are missing. Many people on the net recommend ignoring the annotations entirely. A review of the annotations ("annotated annotations") by Clive Feather can be found on the web at http://www.lysator.liu.se/c/schildt.html . The text of the Rationale (not the full Standard) can be obtained by anonymous ftp from ftp.uu.net (see question 18.16) in directory doc/standards/ansi/X3.159-1989, and is also available on the web at http://www.lysator.liu.se/c/rat/title.html . The Rationale has also been printed by Silicon Press, ISBN 0-929306-07-4. Public review drafts of C9X are available from ISO/IEC JTC1/SC22/WG14's web site, http://www.dkuug.dk/JTC1/SC22/WG14/ . See also question 11.2b below. 11.2b: Where can I get information about updates to the Standard? A: You can find information (including C9X drafts) at the web sites http://www.lysator.liu.se/c/index.html, http://www.dkuug.dk/JTC1/SC22/WG14/, and http://www.dmk.com/ . 11.3: My ANSI compiler complains about a mismatch when it sees extern int func(float); int func(x) float x; { ... A: You have mixed the new-style prototype declaration "extern int func(float);" with the old-style definition "int func(x) float x;". It is usually possible to mix the two styles (see question 11.4), but not in this case. Old C (and ANSI C, in the absence of prototypes, and in variable- length argument lists; see question 15.2) "widens" certain arguments when they are passed to functions. floats are promoted to double, and characters and short integers are promoted to int. (For old-style function definitions, the values are automatically converted back to the corresponding narrower types within the body of the called function, if they are declared that way there.) This problem can be fixed either by using new-style syntax consistently in the definition: int func(float x) { ... } or by changing the new-style prototype declaration to match the old-style definition: extern int func(double); (In this case, it would be clearest to change the old-style definition to use double as well, if possible.) It is arguably much safer to avoid "narrow" (char, short int, and float) function arguments and return types altogether. See also question 1.25. References: K&R1 Sec. A7.1 p. 186; K&R2 Sec. A7.3.2 p. 202; ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>; Rationale Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>; H&S Sec. 9.2 pp. 265-7, Sec. 9.4 pp. 272-3. 11.4: Can you mix old-style and new-style function syntax? A: Doing so is legal, but requires a certain amount of care (see especially question 11.3). Modern practice, however, is to use the prototyped form in both declarations and definitions. (The old-style syntax is marked as obsolescent, so official support for it may be removed some day.) References: ISO Sec. 6.7.1, Sec. 6.9.5; H&S Sec. 9.2.2 pp. 265- 7, Sec. 9.2.5 pp. 269-70. 11.5: Why does the declaration extern int f(struct x *p); give me an obscure warning message about "struct x introduced in prototype scope"? A: In a quirk of C's normal block scoping rules, a structure declared (or even mentioned) for the first time within a prototype cannot be compatible with other structures declared in the same source file (it goes out of scope at the end of the prototype). To resolve the problem, precede the prototype with the vacuous- looking declaration struct x; which places an (incomplete) declaration of struct x at file scope, so that all following declarations involving struct x can at least be sure they're referring to the same struct x. References: ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>. 11.8: I don't understand why I can't use const values in initializers and array dimensions, as in const int n = 5; int a[n]; A: The const qualifier really means "read-only"; an object so qualified is a run-time object which cannot (normally) be assigned to. The value of a const-qualified object is therefore *not* a constant expression in the full sense of the term. (C is unlike C++ in this regard.) When you need a true compile- time constant, use a preprocessor #define (or perhaps an enum). References: ISO Sec. 6.4; H&S Secs. 7.11.2,7.11.3 pp. 226-7. 11.9: What's the difference between "const char *p" and "char * const p"? A: "const char *p" (which can also be written "char const *p") declares a pointer to a constant character (you can't change the character); "char * const p" declares a constant pointer to a (variable) character (i.e. you can't change the pointer). Read these "inside out" to understand them; see also question 1.21. References: ISO Sec. <IP_ADDRESS>; Rationale Sec. <IP_ADDRESS>; H&S Sec. 4.4.4 p. 81. 11.10: Why can't I pass a char ** to a function which expects a const char **? A: You can use a pointer-to-T (for any type T) where a pointer-to- const-T is expected. However, the rule (an explicit exception) which permits slight mismatches in qualified pointer types is not applied recursively, but only at the top level. You must use explicit casts (e.g. (const char **) in this case) when assigning (or passing) pointers which have qualifier mismatches at other than the first level of indirection. References: ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>, Sec. 6.5.3; H&S Sec. 7.9.1 pp. 221-2. 11.12a: What's the correct declaration of main()? A: Either int main(), int main(void), or int main(int argc, char *argv[]) (with alternate spellings of argc and *argv[] obviously allowed). See also questions 11.12b to 11.15 below.  A: No. main() must be declared as returning an int, and as taking either zero or two arguments, of the appropriate types. If you're calling exit() but still getting warnings, you may have to insert a redundant return statement (or use some kind of "not reached" directive, if available). Declaring a function as void does not merely shut off or rearrange warnings: it may also result in a different function call/return sequence, incompatible with what the caller (in main's case, the C run-time startup code) expects. (Note that this discussion of main() pertains only to "hosted" implementations; none of it applies to "freestanding" implementations, which may not even have main(). However, freestanding implementations are comparatively rare, and if you're using one, you probably know it. If you've never heard of the distinction, you're probably using a hosted implementation, and the above rules apply.) References: ISO Sec. <IP_ADDRESS>.1, Sec. G.5.1; H&S Sec. 20.1 p. 416; CT&P Sec. 3.10 pp. 50-51. 11.13: But what about main's third argument, envp? A: It's a non-standard (though common) extension. If you really need to access the environment in ways beyond what the standard getenv() function provides, though, the global variable environ is probably a better avenue (though it's equally non-standard). References: ISO Sec. G.5.1; H&S Sec. 20.1 pp. 416-7. 11.14: I believe that declaring void main() can't fail, since I'm calling exit() instead of returning, and anyway my operating system ignores a program's exit/return status. A: It doesn't matter whether main() returns or not, or whether anyone looks at the status; the problem is that when main() is misdeclared, its caller (the runtime startup code) may not even be able to *call* it correctly (due to the potential clash of calling conventions; see question 11.12b). It has been reported that programs using void main() and compiled using BC++ 4.5 can crash. Some compilers (including DEC C V4.1 and gcc with certain warnings enabled) will complain about void main(). Your operating system may ignore the exit status, and void main() may work for you, but it is not portable and not correct. 11.15: The book I've been using, _C Programing for the Compleat Idiot_, always uses void main(). A: Perhaps its author counts himself among the target audience. Many books unaccountably use void main() in examples, and assert that it's correct. They're wrong. 11.16: Is exit(status) truly equivalent to returning the same status from main()? A: Yes and no. The Standard says that they are equivalent. However, a return from main() cannot be expected to work if data local to main() might be needed during cleanup; see also question 16.4. A few very old, nonconforming systems may once have had problems with one or the other form. (Finally, the two forms are obviously not equivalent in a recursive call to main().) References: K&R2 Sec. 7.6 pp. 163-4; ISO Sec. <IP_ADDRESS>.3. 11.17: I'm trying to use the ANSI "stringizing" preprocessing operator `#' to insert the value of a symbolic constant into a message, but it keeps stringizing the macro's name rather than its value. A: You can use something like the following two-step procedure to force a macro to be expanded as well as stringized: #define Str(x) #x #define Xstr(x) Str(x) #define OP plus char *opname = Xstr(OP); This code sets opname to "plus" rather than "OP". An equivalent circumlocution is necessary with the token-pasting operator ## when the values (rather than the names) of two macros are to be concatenated. References: ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>. 11.18: What does the message "warning: macro replacement within a string literal" mean? A: Some pre-ANSI compilers/preprocessors interpreted macro definitions like #define TRACE(var, fmt) printf("TRACE: var = fmt\n", var) such that invocations like TRACE(i, %d); were expanded as printf("TRACE: i = %d\n", i); In other words, macro parameters were expanded even inside string literals and character constants. Macro expansion is *not* defined in this way by K&R or by Standard C. When you do want to turn macro arguments into strings, you can use the new # preprocessing operator, along with string literal concatenation (another new ANSI feature): #define TRACE(var, fmt) \ printf("TRACE: " #var " = " #fmt "\n", var) See also question 11.17 above. References: H&S Sec. 3.3.8 p. 51. 11.19: I'm getting strange syntax errors inside lines I've #ifdeffed out. A: Under ANSI C, the text inside a "turned off" #if, #ifdef, or #ifndef must still consist of "valid preprocessing tokens." This means that the characters " and ' must each be paired just as in real C code, and the pairs mustn't cross line boundaries. (Note particularly that an apostrophe within a contracted word looks like the beginning of a character constant.) Therefore, natural-language comments and pseudocode should always be written between the "official" comment delimiters /* and */. (But see question 20.20, and also 10.25.) References: ISO Sec. <IP_ADDRESS>, Sec. 6.1; H&S Sec. 3.2 p. 40. 11.20: What are #pragmas and what are they good for? A: The #pragma directive provides a single, well-defined "escape hatch" which can be used for all sorts of (nonportable) implementation-specific controls and extensions: source listing control, structure packing, warning suppression (like lint's old /* NOTREACHED */ comments), etc. References: ISO Sec. 6.8.6; H&S Sec. 3.7 p. 61. 11.21: What does "#pragma once" mean? I found it in some header files. A: It is an extension implemented by some preprocessors to help make header files idempotent; it is equivalent to the #ifndef trick mentioned in question 10.7, though less portable. 11.22: Is char a[3] = "abc"; legal? What does it mean? A: It is legal in ANSI C (and perhaps in a few pre-ANSI systems), though useful only in rare circumstances. It declares an array of size three, initialized with the three characters 'a', 'b', and 'c', *without* the usual terminating '\0' character. The array is therefore not a true C string and cannot be used with strcpy, printf %s, etc. Most of the time, you should let the compiler count the initializers when initializing arrays (in the case of the initializer "abc", of course, the computed size will be 4). References: ISO Sec. 6.5.7; H&S Sec. 4.6.4 p. 98. 11.24: Why can't I perform arithmetic on a void * pointer? A: The compiler doesn't know the size of the pointed-to objects. Before performing arithmetic, convert the pointer either to char * or to the pointer type you're trying to manipulate (but see also question 4.5). References: ISO Sec. <IP_ADDRESS>, Sec. 6.3.6; H&S Sec. 7.6.2 p. 204. 11.25: What's the difference between memcpy() and memmove()? A: memmove() offers guaranteed behavior if the source and destination arguments overlap. memcpy() makes no such guarantee, and may therefore be more efficiently implementable. When in doubt, it's safer to use memmove(). References: K&R2 Sec. B3 p. 250; ISO Sec. <IP_ADDRESS>, Sec. <IP_ADDRESS>; Rationale Sec. 4.11.2; H&S Sec. 14.3 pp. 341-2; PCS Sec. 11 pp. 165-6. 11.26: What should malloc(0) do? Return a null pointer or a pointer to 0 bytes? A: The ANSI/ISO Standard says that it may do either; the behavior is implementation-defined (see question 11.33). References: ISO Sec. 7.10.3; PCS Sec. 16.1 p. 386. 11.27: Why does the ANSI Standard not guarantee more than six case- insensitive characters of external identifier significance? A: The problem is older linkers which are under the control of neither the ANSI/ISO Standard nor the C compiler developers on the systems which have them. The limitation is only that identifiers be *significant* in the first six characters, not that they be restricted to six characters in length. This limitation is marked in the Standard as "obsolescent", and will be removed in C9X. References: ISO Sec. 6.1.2, Sec. 6.9.1; Rationale Sec. 3.1.2; C9X Sec. 6.1.2; H&S Sec. 2.5 pp. 22-3. 11.29: My compiler is rejecting the simplest possible test programs, with all kinds of syntax errors. A: Perhaps it is a pre-ANSI compiler, unable to accept function prototypes and the like. See also questions 1.31, 10.9, 11.30, and 16.1b. 11.30: Why are some ANSI/ISO Standard library functions showing up as undefined, even though I've got an ANSI compiler? A: It's possible to have a compiler available which accepts ANSI syntax, but not to have ANSI-compatible header files or run-time libraries installed. (In fact, this situation is rather common when using a non-vendor-supplied compiler such as gcc.) See also questions 11.29, 13.25, and 13.26. 11.31: Does anyone have a tool for converting old-style C programs to ANSI C, or vice versa, or for automatically generating prototypes? A: Two programs, protoize and unprotoize, convert back and forth between prototyped and "old style" function definitions and declarations. (These programs do *not* handle full-blown translation between "Classic" C and ANSI C.) These programs are part of the FSF's GNU C compiler distribution; see question 18.3. The unproto program (/pub/unix/unproto5.shar.Z on ftp.win.tue.nl) is a filter which sits between the preprocessor and the next compiler pass, converting most of ANSI C to traditional C on-the-fly. The GNU GhostScript package comes with a little program called ansi2knr. Before converting ANSI C back to old-style, beware that such a conversion cannot always be made both safely and automatically. ANSI C introduces new features and complexities not found in K&R C. You'll especially need to be careful of prototyped function calls; you'll probably need to insert explicit casts. See also questions 11.3 and 11.29. Several prototype generators exist, many as modifications to lint. A program called CPROTO was posted to comp.sources.misc in March, 1992. There is another program called "cextract." Many vendors supply simple utilities like these with their compilers. See also question 18.16. (But be careful when generating prototypes for old functions with "narrow" parameters; see question 11.3.) 11.32: Why won't the Frobozz Magic C Compiler, which claims to be ANSI compliant, accept this code? I know that the code is ANSI, because gcc accepts it. A: Many compilers support a few non-Standard extensions, gcc more so than most. Are you sure that the code being rejected doesn't rely on such an extension? It is usually a bad idea to perform experiments with a particular compiler to determine properties of a language; the applicable standard may permit variations, or the compiler may be wrong. See also question 11.35. 11.33: People seem to make a point of distinguishing between implementation-defined, unspecified, and undefined behavior. What's the difference? A: Br