If you get up in the morning and think the future is going to be better, it is a bright day. Otherwise, it’s not. -Elon Musk-


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Always here, too constantly elevate to your needs.
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Phone:
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Email:
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# of Apartments
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Test input & output¶
Routines related to text input/output
-
inout.
clear_stdout
(lines=1, stream=<_io.TextIOWrapper name='<stdout>' mode='w' encoding='UTF-8'>, flush=True)[source]¶ Uses escape characters to move back number of lines and empty them
To write something on the empty lines, use :func:sys.stdout.write() and :func:sys.stdout.flush()
Parameters: - lines – the number of lines to clear
- stream – by default, stdout is cleared, but this can be replaced by another stream that uses these escape characters (stderr comes to mind)
- flush – by default, uses flush after writing, but canbe overriden if you will do it yourself soon after
-
inout.
reprint
(txt, lines=1, stream=<_io.TextIOWrapper name='<stdout>' mode='w' encoding='UTF-8'>)[source]¶ Removes lines from stdout and prints the requested text txt instead of them
Parameters: - txt – the text you want to print; works best if no more than lines lines
- stream – see
clear_stdout
- lines – the number of lines you want to use; this amount is cleared and if txt is shorter, the remainder is prepended
For example for status monitoring:
- print N newlines
- do something or check something until there is news
- use
reprint
to print the as most N lines update using lines = N - repeat step 2 and 3 until task is done
- if you want to remove the last status message, call clear_stdout using lines = N


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PicoCTF 2019 - General Skills
Let’s Warm Up [50]
Description
If I told you a word started with 0x70 in hexadecimal, what would it start with in ASCII?
Solution
A look at the ASCII manual should help!
$ man ascii
Then, it’s just a matter of finding the right line, the one that has 70
in the Hex
column.
Oct Dec Hex Char Oct Dec Hex Char
────────────────────────────────────────────────────────────────────────
[...]
060 48 30 0 160 112 70 p
Flag: picoCTF{p}
2Warm [50]
Description
Can you convert the number 42 (base 10) to binary (base 2)?
Solution
Sure thing! Python can easily do that with its bin built-in:
$ python -c "print(bin(42))"
0b101010
Flag: picoCTF{101010}
Warmed Up [50]
Description
What is 0x3D (base 16) in decimal (base 10).
Solution
Python prints anything in base 10 by default.
$ python -c "print(0x3D)"
61
Flag: picoCTF{61}
Bases [100]
Description
What does this bDNhcm5fdGgzX3IwcDM1 mean? I think it has something to do with bases.
Hint: Submit your answer in our competition’s flag format. For example, if you answer was ‘hello’, you would submit ‘picoCTF{hello}’ as the flag.
Solution
The mysterious text is encoded using base64. You can usually spot it because it uses characters from a-z, A-Z, 0-9, +
, /
and ends with one or more =
for padding. The base64
program, called with the option will try to decode any base64 input:
$ echo "bDNhcm5fdGgzX3IwcDM1" | base64 -d
l3arn_th3_r0p35
Flag: picoCTF{l3arn_th3_r0p35}
First Grep [100]
Description
Can you find the flag in file? This would be really tedious to look through manually, something tells me there is a better way. You can also find the file in /problems/first-grep_6_c2319e8af66fa6bec197edc733dd52dd on the shell server.
Hint: grep tutorial
Solution
grep is a program to find something in a file (or a bunch of files). It needs the following arguments: grep <stuff-I'm-looking-for> <file-in-which-to-search>
.  If you go over there, you might even find a flag! https://picoctf.com/resources (link)
Solution
Visit the resources pages to find the flag :)
Flag: picoCTF{r3source_pag3_f1ag}
Strings It [100]
Description
Can you find the flag in file without running it? You can also find the file in /problems/strings-it_6_f71e1644786ed3c019c38d961f6e29de on the shell server.
Hint: strings
Solution
strings
will print all the strings found in a file. Combined with grep
to filter the output:
$ strings strings | grep picoCTF
picoCTF{5tRIng5_1T_976fbd5c}
Flag: picoCTF{5tRIng5_1T_976fbd5c}
What’s a Net Cat? [100]
Description
Using netcat (nc) is going to be pretty important. Can you connect to 2019shell1.picoctf.com at port 47229 to get the flag?
Hint: nc tutorial
Solution
From the man page:
Netcat is a simple Unix utility which reads and writes data across network connections, using TCP or UDP protocol.
Note that the actual command to run netcat is nc though:
$ nc 2019shell1.picoctf.com 47229
You're on your way to becoming the net cat master
picoCTF{nEtCat_Mast3ry_bfae0e4c}
First Grep: Part II [200]
Description
Can you find the flag in /problems/first-grep–part-ii_4_ca16fbcd16c92f0cb1e376a6c188d58f/files on the shell server? Remember to use grep.
Hint: grep tutorial
Solution
Similar to the first challenge, we should use grep to find the flag. Except we have a bunch of files instead of just one.
man grep
has a good option to help in that case:
-r, –recursive Read all files under each directory, recursively, following symbolic links only if they are on the command line. Note that if no file operand is given, grep searches the working directory. This is equivalent to the -d recurse option.
kylma@pico-2019-shell1:~$ cd /problems/first-grep--part-ii_4_ca16fbcd16c92f0cb1e376a6c188d58f/files
kylma@pico-2019-shell1:/problems/first-grep--part-ii_4_ca16fbcd16c92f0cb1e376a6c188d58f/files$ ls
files0 files1 files10 files2 files3 files4 files5 files6 files7 files8 files9
kylma@pico-2019-shell1:/problems/first-grep--part-ii_4_ca16fbcd16c92f0cb1e376a6c188d58f/files$ grep -nr picoCTF
files6/file5:2:picoCTF{grep_r_to_find_this_0e28f3ee}
Flag: picoCTF{grep_r_to_find_this_0e28f3ee}
Plumbing [200]
Description
Sometimes you need to handle process data outside of a file. Can you find a way to keep the output from this program and search for the flag? Connect to 2019shell1.picoctf.com 18944.
Hints:
- Remember the flag format is picoCTF{XXXX}
- What’s a pipe? No not that kind of pipe… This kind
Solution
If we connect to the given address, a big phat bunch of text greets us:
$ nc 2019shell1.picoctf.com 18944
[...]
I don't think this is a flag either
This is defintely not a flag
This is defintely not a flag
Not a flag either
I don't think this is a flag either
Again, I really don't think this is a flag
This is defintely not a flag
A pipe, represented by |
can redirect the output of one command to another command.  Drawing inspirations from previous challenges, grep
is super skilled at finding stuff! To search for the word dog, grep dog
would do the trick.  Maybe something lies in /problems/where-is-the-file_4_f26b413d005c16c61f127740ab242b35.
Hints:
- What command can see/read files?
- What’s in the manual page of ls?
Solution
The command ls
can list files:
kylma@pico-2019-shell1:~$ cd /problems/where-is-the-file_4_f26b413d005c16c61f127740ab242b35
kylma@pico-2019-shell1:/problems/where-is-the-file_4_f26b413d005c16c61f127740ab242b35$ ls
kylma@pico-2019-shell1:/problems/where-is-the-file_4_f26b413d005c16c61f127740ab242b35$
Lies! There are no files here! Actually, reading the man page of ls, there’s a trick:
-a, –all do not ignore entries starting with .
On Unix, files with a name starting with a .
are hidden files and will not appear by default. Unless you tell ls
to list them by using the option -a
.
kylma@pico-2019-shell1:/problems/where-is-the-file_4_f26b413d005c16c61f127740ab242b35$ ls -la
total 80
drwxr-xr-x 2 root root 4096 Sep 28 22:05 .
drwxr-x--x 684 root root 69632 Sep 29 03:37 ..
-rw-rw-r-- 1 hacksports hacksports 39 Sep 28 22:05 .cant_see_me
See, that directory wasn’t empty after all! Now, to print the content of the file, cat
can do so:
kylma@pico-2019-shell1:/problems/where-is-the-file_4_f26b413d005c16c61f127740ab242b35$ cat .cant_see_me
picoCTF{w3ll_that_d1dnt_w0RK_cb4a5081}
Flag: picoCTF{w3ll_that_d1dnt_w0RK_cb4a5081}
What’s the difference [200]
Description
Can you spot the difference? kitters cattos. They are also available at /problems/whats-the-difference_0_00862749a2aeb45993f36cc9cf98a47a on the shell server
Hints:
- How do you find the difference between two files?
- Dumping the data from a hex editor may make it easier to compare.
Solution
Two files are given, kitters.jpg and cattos.jpg. kitters.jpg
is a regular picture of 2 cats, cattos.jpg
is the glitchy version of the same picture.
diff
can compare 2 files and list the differences lines per line:
$ diff kitters.jpg cattos.jpg
Binary files kitters.jpg and cattos.jpg differ
Except that it doesn’t do well with binary files… xxd
can make the hexdump of a file though, it will probably be easier to compare. The -p
option enables the plaintext hexdump style.
$ xxd -p kitters.jpg kitters.xxd
$ xxd -p cattos.jpg cattos.xxd
Now we can call diff again on the two hexdumps, and use > results.diff
to store the results in a new file.
$ diff kitters.xxd cattos.xxd > results.diff
We can manually get the flag from results.diff by taking only the differences:
-0000c240: 21d3 6da6 4b70 6963 6f60 1b95 41e8 b93d !.m.Kpico`..A..=
+0000c240: 21d3 6da6 4b99 9d98 c860 1b95 41e8 b93d !.m.K....`..A..=
But I got bored of doing it manually and I wasn’t even past picoCTF :DD
Instead, I wrote a Python script to compare each byte in the two images:
#!/usr/bin/env python3
kitters = open("kitters.jpg", "rb").read()
cattos = open("cattos.jpg", "rb").read()
flag = []
for i in range(len(kitters)):
if kitters[i] != cattos[i]:
flag.append(chr(cattos[i]))
print("".join(flag))
$ ./solve.py
picoCTF{th3yr3_a5_d1ff3r3nt_4s_bu773r_4nd_j311y_aslkjfdsalkfslkflkjdsfdszmz10548}
Flag: picoCTF{th3yr3_a5_d1ff3r3nt_4s_bu773r_4nd_j311y_aslkjfdsalkfslkflkjdsfdszmz10548}
Based [200]
Description
To get truly 1337, you must understand different data encodings, such as hexadecimal or binary. Can you get the flag from this program to prove you are on the way to becoming 1337? Connect with nc 2019shell1.picoctf.com 28758.
Hints:
- I hear python can convert things.
- It might help to have multiple windows open
Solution
Let’s connect and see what the conversions look like:
$ nc 2019shell1.picoctf.com 28758
Let us see how data is stored
oven
Please give the 01101111 01110110 01100101 01101110 as a word.
First challenge uses the binary representation
Please give me the 156 165 162 163 145 as a word
Second challenge probably uses the octal representation as there is no digit > 7
Please give me the 6f76656e as a word
Third challenge is probably good old hexadecimal.
I ended up writing a Python script to do the conversions and send them back automatically. It was totally overkill, but I didn’t know there would be only 3 conversions to do, did I?
#!/usr/bin/env python
"""
nc 2019shell1.picoctf.com 28758
"""
import socket
import binascii
PORT = 28758
HOSTNAME = "2019shell1.picoctf.com"
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect(( HOSTNAME, PORT ))
def receive():
data = s.recv(1024)
data = data.decode("utf-8").split("\n")
print(data)
return data
def send(answer):
print("SENDING", answer)
s.sendall(bytes(answer + "\n", "utf-8"))
# Please give the 01101111 01110110 01100101 01101110 as a word
def level_1():
data = receive()
send(data[1])
# Please give me the 156 165 162 163 145 as a word
def level_2():
data = receive()
data = data[0].split(" ")
new_list = []
for d in data:
if d.isdigit():
d = chr(int(d, 8))
new_list.append(d)
send("".join(new_list))
# Please give me the 6f76656e as a word
def level_3():
data = receive()[0]
data = data.split(" ")[4]
data = binascii.unhexlify(data)
data = data.decode("utf-8")
send(data)
def level_4():
data = receive()
level_1()
level_2()
level_3()
level_4()
$ ./solve.py
['Let us see how data is stored', 'light', 'Please give the 01101100 01101001 01100111 01101000 01110100 as a word.', '...', 'you have 45 seconds.....', '', 'Input:', '']
SENDING light
['Please give me the 143 157 154 157 162 141 144 157 as a word.', 'Input:', '']
SENDING colorado
['Please give me the<PHONE_NUMBER> as a word.', 'Input:', '']
SENDING chair
["You've beaten the challenge", 'Flag: picoCTF{learning_about_converting_values_4b4e293e}', '']
Flag: picoCTF{learning_about_converting_values_4b4e293e}
flag_shop [300]
Description
There’s a flag shop selling stuff, can you buy a flag? Source. Connect with nc 2019shell1.picoctf.com 3967.
Hint: Two’s compliment can do some weird things when numbers get really big!
#include <stdio.h>
#include <stdlib.h>
int main()
{
setbuf(stdout, NULL);
int con;
con = 0;
int account_balance = 1100;
while(con == 0){
printf("Welcome to the flag exchange\n");
printf("We sell flags\n");
printf("\n1. Check Account Balance\n");
printf("\n2. Buy Flags\n");
printf("\n3. Exit\n");
int menu;
printf("\n Enter a menu selection\n");
fflush(stdin);
scanf("%d", &menu);
if(menu == 1){
printf("\n\n\n Balance: %d \n\n\n", account_balance);
}
else if(menu == 2){
printf("Currently for sale\n");
printf("1. Defintely not the flag Flag\n");
printf("2. 1337 Flag\n");
int auction_choice;
fflush(stdin);
scanf("%d", &auction_choice);
if(auction_choice == 1){
printf("These knockoff Flags cost 900 each, enter desired quantity\n");
int number_flags = 0;
fflush(stdin);
scanf("%d", &number_flags);
if(number_flags > 0){
int total_cost = 0;
total_cost = 900*number_flags;
printf("\nThe final cost is: %d\n", total_cost);
if(total_cost <= account_balance){
account_balance = account_balance - total_cost;
printf("\nYour current balance after transaction: %d\n\n", account_balance);
}
else{
printf("Not enough funds to complete purchase\n");
}
}
}
else if(auction_choice == 2){
printf("1337 flags cost 100000 dollars, and we only have 1 in stock\n");
printf("Enter 1 to buy one");
int bid = 0;
fflush(stdin);
scanf("%d", &bid);
if(bid == 1){
if(account_balance > 100000){
FILE *f = fopen("flag.txt", "r");
if(f == NULL){
printf("flag not found: please run this on the server\n");
exit(0);
}
char buf[64];
fgets(buf, 63, f);
printf("YOUR FLAG IS: %s\n", buf);
}
else{
printf("\nNot enough funds for transaction\n\n\n");
}}
}
}
else{
con = 1;
}
}
return 0;
}
Solution
So, flagshop is a tiny program which sells flag (I bet you didn’t guess from the name right?). Let’s poke it a bit to see how it works:
$ gcc store.c -o flagshop
$ ./flagshop
Welcome to the flag exchange
We sell flags
1. Check Account Balance
2. Buy Flags
3. Exit
Enter a menu selection
2
Currently for sale
1. Defintely not the flag Flag
2. 1337 Flag
1
These knockoff Flags cost 900 each, enter desired quantity
1
The final cost is: 900
Your current balance after transaction: 200
Welcome to the flag exchange
We sell flags
1. Check Account Balance
2. Buy Flags
3. Exit
Enter a menu selection
2
Currently for sale
1. Defintely not the flag Flag
2. 1337 Flag
2
1337 flags cost 100000 dollars, and we only have 1 in stock
Enter 1 to buy one
Not enough funds for transaction
You can buy two types of flags:
- “not the flag” flag for 900$
- 1337 flag for 100000$
Obviously, we want the 1337 flag :D It’s the only one we cannot buy with our starting balance of 1100$ :((
Since the code for 1337 flag is straight-forward (have enough money > get the flag), let’s have a look at the code handling the “not the flag” flags:
int account_balance = 1100;
[...]
if(auction_choice == 1){
printf("These knockoff Flags cost 900 each, enter desired quantity\n");
int number_flags = 0;
fflush(stdin);
scanf("%d", &number_flags);
if(number_flags > 0){
int total_cost = 0;
total_cost = 900*number_flags;
printf("\nThe final cost is: %d\n", total_cost);
if(total_cost <= account_balance){
account_balance = account_balance - total_cost;
printf("\nYour current balance after transaction: %d\n\n", account_balance);
}
else{
printf("Not enough funds to complete purchase\n");
}
There is one important thing to notice here: all the numbers handling the money amounts are signed integers. Which means that we can probably trigger an integer overflow. So, if we order a large amount of flags, total_cost
should hit the limit of INT_MAX and wrap around and become a negative number:
total_cost = 900*number_flags;
We can verify it using the numpy.int32
type in Python, let’s buy 200000000 flags:
>>> import numpy
>>> total_cost = numpy.int32(900*200000000)
-388626432
It means that we will actually gain money if we buy a large amount of flags:
account_balance = account_balance - total_cost;
Test with Python:
>>> account_balance = 1100 - numpy.int32(900*200000000)
388627532
We now have enough $$$ to buy the 1337 flag ~~
$ nc 2019shell1.picoctf.com 3967
Welcome to the flag exchange
We sell flags
1. Check Account Balance
2. Buy Flags
3. Exit
Enter a menu selection
2
Currently for sale
1. Defintely not the flag Flag
2. 1337 Flag
1
These knockoff Flags cost 900 each, enter desired quantity
200000000
The final cost is: -388626432
Your current balance after transaction: 388627532
Welcome to the flag exchange
We sell flags
1. Check Account Balance
2. Buy Flags
3. Exit
Enter a menu selection
2
Currently for sale
1. Defintely not the flag Flag
2. 1337 Flag
2
1337 flags cost 100000 dollars, and we only have 1 in stock
Enter 1 to buy one1
YOUR FLAG IS: picoCTF{m0n3y_bag5_cd0ead78}
Flag: picoCTF{m0n3y_bag5_cd0ead78}
Mus1c [300]
Description
I wrote you a song. Put it in the picoCTF{} flag format
Hint: Do you think you can master rockstar?
Solution
It does look like some esoteric programming language, lots of shout
, put
, knock <var> down
, build <var> down
… Let’s ask the Internet! It’s a programming language called rockstar and it even has an online interpreter!
114
114
114
111
99
107
110
114
110
48
49
49
51
114
Program completed in 125 ms
The output looks like ASCII, let’s convert it to a flag:
#!/usr/bin/python3
DECODER_OUTPUT = [114, 114, 114, 111, 99, 107, 110, 114, 110, 48, 49, 49, 51, 114]
DECODED = [chr(c) for c in DECODER_OUTPUT]
print("".join(DECODED))
Flag: picoCTF{rrrocknrn0113r}
1_wanna_b3_a_r0ck5tar [350]
Description
I wrote you another song. Put the flag in the picoCTF{} flag format
Rocknroll is right
Silence is wrong
A guitar is a six-string // guitar = 10
Tommy's been down
Music is a billboard-burning razzmatazz! // music = 170
Listen to the music // get input
If the music is a guitar // if input == a guitar
Say "Keep on rocking!"
Listen to the rhythm
If the rhythm without Music is nothing
Tommy is rockin guitar
Shout Tommy!
Music is amazing sensation
Jamming is awesome presence
Scream Music!
Scream Jamming!
Tommy is playing rock
Scream Tommy!
They are dazzled audiences
Shout it!
Rock is electric heaven
Scream it!
Tommy is jukebox god
Say it!
Break it down
Shout "Bring on the rock!"
Else Whisper "That ain't it, Chief"
Break it downs
Solution
This challenge is a follow-up to the challenge Mus1c. Let’s try the same trick: run the program in the online intepreter :
Meh. Where’s the flag .__.
Since I’m not planning on really learning to code some rockstar, I asked the internet for a way to convert rockstar to something I understand… like… Python :D The godz of the Internet nudged me to the following project: https://github.com/yyyyyyyan/rockstar-py
Let’s run it:
$ python3 -W ignore -m rockstarpy --input ../lyrics.txt --output lyrics.py
Rocknroll = true
Silence = 5 // Silence = false
a_guitar = 19
Tommy = 44
Music = 160
the_music = input()
if the_music == a_guitar:
print("Keep on rocking!")
the_rhythm = input()
if the_rhythm - Music == False:
Tommy = 66
print(Tommy!)
Music = 79
Jamming = 78
print(Music!)
print(Jamming!)
Tommy = 74
print(Tommy!)
They are dazzled audiences
print(it!)
Rock = 86
print(it!)
Tommy = 73
print(it!)
break
print("Bring on the rock!")
Else print("That ain't it, Chief")
break
Unfortunately, the generated Python code doesn’t run :/
File "lyrics.py", line 12
print(Tommy!)
^
SyntaxError: invalid syntax
I tried to fix the errors but eventually gave up, the generated code is still useful to understand what the code is doing!
the_music = input()
if the_music == a_guitar:
print("Keep on rocking!")
So the first condition to pass is entering a value equal to the variable a_guitar
, which is defined earlier as a_guitar = 19
. So, entering 19
as input value in the online interpreter should print Keep on rocking!
.
Except it doesn’t work :((
Seems like I’m gonna have to learn some rockstar after all, let’s go read bits of The Rockstar Language Specification
This section should help understanding what value should a_guitar
be initialized with.
Poetic Number Literals
A poetic number literal begins with a variable name, followed by the keyword `is`, or the aliases `was` or `were`. As long as the next symbol is not a Literal Word, the rest of the line is treated as a decimal number in which the values of consecutive digits are given by the lengths of the subsequent barewords, up until the end of the line. To allow the digit zero, and to compensate for a lack of suitably rock’n’roll 1- and 2-letter words, word lengths are parsed modulo 10. A period (.) character denotes a decimal place. Other than the first period, any non-alphabetical characters are ignored.
- `Tommy was a lovestruck ladykiller` initialises `Tommy` with the value `100`
- `Sweet Lucy was a dancer` - initialises `Sweet Lucy` with the value 16
- `A killer is on the loose` - initialises `a killer` with the value 235.
- `My dreams were ice. A life unfulfilled; wakin' everybody up, taking booze and pills` - initialises `my dreams` with the value `3.1415926535`
- `Tommy was without` initialises `Tommy` with the value `7` because `without` is a Reserved Keyword, but not a Literal Word.
- Note that poetic literals **can** include Reserved Keywords, as with `taking` in this example.
- The hyphen (`-`) is counted as a letter – so you can use terms like ‘all-consuming’ (13 letters > 3) and ‘power-hungry’ (12 letters > 2) instead of having to think of 12- and 13-letter words.
- The semi-colon, comma, apostrophe and any other non-alphabetical characters are ignored.
Comparing the initialisation of a_guitar
in the original rockstar source code:
A guitar is a six-string
a_guitar
value should actually be:
a
= 1six-string
= 10 modulo 10 = 0
The final value of a guitar
is therefore 10, not 19.
Yay, we passed he first check! Back to the Python code:
if the_rhythm - Music == False:
Tommy = 66
print(Tommy!)
Music = 79
Jamming = 78
print(Music!)
print(Jamming!)
[...]
After that, a loop checks character one by one. However, the interesting thing is that each character is new variable, for example Tommy
, it gets assigned and is printed. If we get rid of that loop, all the characters would be printed then!
Remove this line and run in the interpreter again:
If the rhythm without Music is nothing
Only one last thing to turn the output back to a flag:
#!/usr/bin/python3
DECODER_OUTPUT = [66, 79, 78, 74, 79, 86, 73]
DECODED = [chr(c) for c in DECODER_OUTPUT]
print("".join(DECODED))
Flag: picoCTF{BONJOVI}
Pifou, that was a long writeup ~~


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The string/to_lower.h
header defines an algorithm that converts a string to lowercase.
#include <chaos/preprocessor/string/to_lower.h>
The CHAOS_PP_TO_LOWER
macro converts a character to lowercase.
CHAOS_PP_TO_LOWER
(c)
The character to be converted to lowercase.
Valid values are the numerals (0
through 9
), the alphabetic characters in the basic source character set (a
through z
and A
through Z
), and the underscore (_
).
If c is an uppercase letter, this macro expands to the letter's lowercase equivalent. Otherwise, it expands to c.
#include<chaos/preprocessor/string/to_lower.h>
CHAOS_PP_TO_LOWER
(a) // aCHAOS_PP_TO_LOWER
(A) // aCHAOS_PP_TO_LOWER
(0) // 0CHAOS_PP_TO_LOWER
(_) // _
The CHAOS_PP_TO_LOWER_ID
macro indirectly expands to CHAOS_PP_TO_LOWER
.
CHAOS_PP_TO_LOWER_ID
()
This macro effectively allows CHAOS_PP_TO_LOWER
to pass through itself without becoming disabled.
The CHAOS_PP_TO_LOWER_
macro expands to a λ expression representing CHAOS_PP_TO_LOWER
.
CHAOS_PP_TO_LOWER_
This macro is a shorthand form of the following λ binding:
The CHAOS_PP_STRING_TO_LOWER
macro converts its argument to lowercase.
CHAOS_PP_STRING_TO_LOWER
(string)
The string to be converted to lowercase.
CHAOS_PP_STRING_TO_LOWER
(string)
The string to be converted to lowercase.
This macro expands to a string containing the elements of string converted to lowercase. For example,
CHAOS_PP_STRING_TO_LOWER
(C H A O S) // c h a o s
This macro uses bypass semantics.
If bypass semantics are already in effect, CHAOS_PP_STRING_TO_LOWER_BYPASS
must be used instead.
#include<chaos/preprocessor/algorithm/concat.h>
#include<chaos/preprocessor/string/core.h>
#include<chaos/preprocessor/string/to_lower.h>
#include<chaos/preprocessor/string/to_upper.h>
CHAOS_PP_CONCAT
( (CHAOS_PP_STRING
)CHAOS_PP_STRING_TO_LOWER
( A B c _ 1 0 ) ) // abc_10 #define PROPER(string) \ PROPER_I(CHAOS_PP_STRING_TO_LOWER
(string)) \ /**/ #define PROPER_I(string) \CHAOS_PP_TO_UPPER
(CHAOS_PP_STRING_HEAD
(string)) \CHAOS_PP_STRING_TAIL
(string) \ /**/ PROPER(C H A O S) // C h a o s
The CHAOS_PP_STRING_TO_LOWER_ID
macro indirectly expands to CHAOS_PP_STRING_TO_LOWER
.
CHAOS_PP_STRING_TO_LOWER_ID
()
This macro effectively allows CHAOS_PP_STRING_TO_LOWER
to pass through itself without becoming disabled.
The CHAOS_PP_STRING_TO_LOWER_
macro expands to a λ expression representing CHAOS_PP_STRING_TO_LOWER
.
CHAOS_PP_STRING_TO_LOWER_
This macro is a shorthand form of the following λ binding:
The CHAOS_PP_STRING_TO_LOWER_BYPASS
macro behaves identically to CHAOS_PP_STRING_TO_LOWER
except that it is parametized by the bypass recursion state.
CHAOS_PP_STRING_TO_LOWER_BYPASS
(s, string)
The current bypass recursion state.
The string to be converted to lowercase.
CHAOS_PP_STRING_TO_LOWER_BYPASS
(s, string)
The current bypass recursion state.
The string to be converted to lowercase.
If bypass semantics are already in effect, this macro must be used instead of CHAOS_PP_STRING_TO_LOWER
.
The CHAOS_PP_STRING_TO_LOWER_BYPASS_ID
macro indirectly expands to CHAOS_PP_STRING_TO_LOWER_BYPASS
.
CHAOS_PP_STRING_TO_LOWER_BYPASS_ID
()
This macro effectively allows CHAOS_PP_STRING_TO_LOWER_BYPASS
to pass through itself without becoming disabled.
The CHAOS_PP_STRING_TO_LOWER_BYPASS_
macro expands to a λ expression representing CHAOS_PP_STRING_TO_LOWER_BYPASS
.
CHAOS_PP_STRING_TO_LOWER_BYPASS_
This macro is a shorthand form of the following λ binding:


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You are given N counters, initially set to 0, and you have two possible operations on them:
- increase(X) − counter X is increased by 1,
- max counter − all counters are set to the maximum value of any counter.
A non-empty zero-indexed array A of M integers is given. This array represents consecutive operations:
- if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),
- if A[K] = N + 1 then operation K is max counter.
For example, given integer N = 5 and array A such that:
A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4the values of the counters after each consecutive operation will be:
(0, 0, 1, 0, 0) (0, 0, 1, 1, 0) (0, 0, 1, 2, 0) (2, 2, 2, 2, 2) (3, 2, 2, 2, 2) (3, 2, 2, 3, 2) (3, 2, 2, 4, 2)The goal is to calculate the value of every counter after all operations.
Write a function:
def solution(N, A)
that, given an integer N and a non-empty zero-indexed array A consisting of M integers, returns a sequence of integers representing the values of the counters.
The sequence should be returned as:
- a structure Results (in C), or
- a vector of integers (in C++), or
- a record Results (in Pascal), or
- an array of integers (in any other programming language).
For example, given:
A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4the function should return [3, 2, 2, 4, 2], as explained above.
Assume that:
- N and M are integers within the range [1..100,000];
- each element of array A is an integer within the range [1..N + 1].
Complexity:
- expected worst-case time complexity is O(N+M);
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).


Samsung Music now supports Android Auto
The latest update also brings support for Android 10
by Aisha MalikThe latest update for Samsung Music includes support for Android Auto allowing it work with Google’s in-car software experience.
Prior to the update, users had to resort to the Spotify app on Android Auto since the in-car software experience didn’t support Samsung Music. However, users still had the option to integrate Spotfiy with Samsung Music.
Now, the newest update allows for playback and control on the users’ dashboard directly through Samsung Music.
Once you download the update, Samsung Music should appear on the Android Auto home screen in your vehicle. The app works similar to the other apps on the platform and lets you search for songs, playlists, albums and artists.
Android Police notes that if you’re using the old Auto interface on your device, you can still select the music player from the audio drop-down menu in the bottom bar.
Source: Samsung Music Via: Android Police


Plugin
Github
Doc
Grails Admin Interface
Administer out-of-the-box your Grails applications
Try a sample application!
Install the plugin
Configure with a few lines
Enjoy!
Easy!
In a few minutes you will have a working admin enviroment
Secure!
Secured by default, you configure what roles can access!
Configurable!
You configure what domain classes are exposed, what fields, aggrupations, order...
Customizable!
You can select what widgets will be used for your fields!
Extensible!
Do you want more? You can build your own widgets easily!
Open source!
Grails Admin Interface is open source, you can check it at
github!
Grails Admin Interface is a product by Kaleidos
Quick links
Official plugin page
Github source
Documentation
Sample application


--- layout: page title: "About" description: "This is what I do." header-img: "img/home-bg-2.jpg" ---
Lazy to say anything here. If you are still interested, refer to my old school CV here : )


Ant Trip
问题描述 :
Ant Country consist of N towns.There are M roads connecting the towns.
Ant Tony,together with his friends,wants to go through every part of the country.
They intend to visit every road , and every road must be visited for exact one time.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now tony wants to know what is the least groups of ants that needs to form to achieve their goal.
输入:
Input contains multiple cases.Test cases are separated by several blank lines. Each test case starts with two integer N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns and M roads in Ant Country.Followed by M lines,each line contains two integers a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town b.No two roads will be the same,and there is no road connecting the same town.
输出:
Input contains multiple cases.Test cases are separated by several blank lines. Each test case starts with two integer N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns and M roads in Ant Country.Followed by M lines,each line contains two integers a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town b.No two roads will be the same,and there is no road connecting the same town.
样例输入:
3 3 1 2 2 3 1 3 4 2 1 2 3 4
样例输出:
1 2HintNew ~~~ Notice: if there are no road connecting one town ,tony may forget about the town. In sample 1,tony and his friends just form one group,they can start at either town 1,2,or 3. In sample 2,tony and his friends must form two group.
HDU-3018-Ant Trip
http://acm.hdu.edu.cn/showproblem.php?pid=3018
这道题就是求一个无向图能够最小几笔画的问题，对于每一个连通分图，如果是欧拉回路，一笔可以画成；否则就是奇数次数的点的个数的一半
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<vector> using namespace std; int f[100005],hash[100005]; int du[100005],count[100005]; //du为每个节点的次数,count为每个分图奇数次数的点的个数 vector<int> a; int n,m; void init() { int i; a.clear(); memset(hash,0,sizeof(hash)); memset(du,0,sizeof(du)); memset(count,0,sizeof(count)); for(i=1;i<=n;i++) f[i]=i; } int find(int x) { int r=x; while(f[r]!=r) r=f[r]; f[x]=r; return r; } void Union(int x,int y) { int fx,fy; fx=find(x); fy=find(y); if(fx!=fy) f[fx]=fy; } int main() { int a1,a2; int k,i,sum; while(scanf("%d%d",&n,&m)!=EOF) { init(); while(m--) { scanf("%d %d",&a1,&a2); du[a1]++; du[a2]++; Union(a1,a2); } for(i=1;i<=n;i++) { k=find(i); if(!hash[k]) { a.push_back(k); //k为一个连通分图的代表元素 hash[k]=1; } if(du[i]%2==1)//计算奇数次数的点的个数 count[k]++; } sum=0; for(i=0;i<a.size();i++) { k=a[i]; if(du[k]==0) //孤立结点 continue; if(count[k]==0)//欧拉回路一笔即可画成 sum++; else sum+=(count[k]/2);//图中次数为奇数的点一定为偶数个 } printf("%d\n",sum); } return 0; }
参考：http://blog.csdn.net/comeace/article/details/8082953
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遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
遇到鸡就说超载鸡，遇到黑框就是蛤蟆？
不错，现在就缺少新的资料
总数为2n，遍历到第n个，就是中间的位置
看见的晚了，只能提交一次唉
能不能说一下tmp数组和solve都是干嘛的？
值得分享！
比较基础，找些自信
原来如此 了解了
都是一些基础题啊
#pragma pack(2) 这个用法太坑爹
第一题还是比较考验智商~
示例程序中的代码不够稳健，没有考虑到输入了非法字符或者不合理输入的情况。
有一点问题。。后面动态规划的程序中
int dp[n+1][W+1];
会报错 提示表达式必须含有常量值。该怎么修改呢。。
<EMAIL_ADDRESS>算法是程序的灵魂，算法分简单和复杂，如果不搞大数据类，程序员了解一下简单点的算法也是可以的，但是会算法的一定要会编程才行，程序员不一定要会算法，利于自己项目需要的可以简单了解。


"""distutils.command.build_scripts Implements the Distutils 'build_scripts' command.""" import os, re from stat import ST_MODE from distutils import sysconfig from distutils.core import Command from distutils.dep_util import newer from distutils.util import convert_path, Mixin2to3 from distutils import log import tokenize # check if Python is called on the first line with this expression first_line_re = re.compile(b'^#!.*python[0-9.]*([ \t].*)?$') class build_scripts(Command): description = "\"build\" scripts (copy and fixup #! line)" user_options = [ ('build-dir=', 'd', "directory to \"build\" (copy) to"), ('force', 'f', "forcibly build everything (ignore file timestamps"), ('executable=', 'e', "specify final destination interpreter path"), ] boolean_options = ['force'] def initialize_options(self): self.build_dir = None self.scripts = None self.force = None self.executable = None self.outfiles = None def finalize_options(self): self.set_undefined_options('build', ('build_scripts', 'build_dir'), ('force', 'force'), ('executable', 'executable')) self.scripts = self.distribution.scripts def get_source_files(self): return self.scripts def run(self): if not self.scripts: return self.copy_scripts() def copy_scripts(self): r"""Copy each script listed in 'self.scripts'; if it's marked as a Python script in the Unix way (first line matches 'first_line_re', ie. starts with "\#!" and contains "python"), then adjust the first line to refer to the current Python interpreter as we copy. """ self.mkpath(self.build_dir) outfiles = [] updated_files = [] for script in self.scripts: adjust = False script = convert_path(script) outfile = os.path.join(self.build_dir, os.path.basename(script)) outfiles.append(outfile) if not self.force and not newer(script, outfile): log.debug("not copying %s (up-to-date)", script) continue # Always open the file, but ignore failures in dry-run mode -- # that way, we'll get accurate feedback if we can read the # script. try: f = open(script, "rb") except OSError: if not self.dry_run: raise f = None else: encoding, lines = tokenize.detect_encoding(f.readline) f.seek(0) first_line = f.readline() if not first_line: self.warn("%s is an empty file (skipping)" % script) continue match = first_line_re.match(first_line) if match: adjust = True post_interp = match.group(1) or b'' if adjust: log.info("copying and adjusting %s -> %s", script, self.build_dir) updated_files.append(outfile) if not self.dry_run: if not sysconfig.python_build: executable = self.executable else: executable = os.path.join( sysconfig.get_config_var("BINDIR"), "python%s%s" % (sysconfig.get_config_var("VERSION"), sysconfig.get_config_var("EXE"))) executable = os.fsencode(executable) shebang = b"#!" + executable + post_interp + b"\n" # Python parser starts to read a script using UTF-8 until # it gets a #coding:xxx cookie. The shebang has to be the # first line of a file, the #coding:xxx cookie cannot be # written before. So the shebang has to be decodable from # UTF-8. try: shebang.decode('utf-8') except UnicodeDecodeError: raise ValueError( "The shebang ({!r}) is not decodable " "from utf-8".format(shebang)) # If the script is encoded to a custom encoding (use a # #coding:xxx cookie), the shebang has to be decodable from # the script encoding too. try: shebang.decode(encoding) except UnicodeDecodeError: raise ValueError( "The shebang ({!r}) is not decodable " "from the script encoding ({})" .format(shebang, encoding)) with open(outfile, "wb") as outf: outf.write(shebang) outf.writelines(f.readlines()) if f: f.close() else: if f: f.close() updated_files.append(outfile) self.copy_file(script, outfile) if os.name == 'posix': for file in outfiles: if self.dry_run: log.info("changing mode of %s", file) else: oldmode = os.stat(file)[ST_MODE] & 0o7777 newmode = (oldmode | 0o555) & 0o7777 if newmode != oldmode: log.info("changing mode of %s from %o to %o", file, oldmode, newmode) os.chmod(file, newmode) # XXX should we modify self.outfiles? return outfiles, updated_files class build_scripts_2to3(build_scripts, Mixin2to3): def copy_scripts(self): outfiles, updated_files = build_scripts.copy_scripts(self) if not self.dry_run: self.run_2to3(updated_files) return outfiles, updated_files


Examples: Securing Web Applications
Some basic setup is required before any of the example applications will run correctly. The examples use annotations, programmatic security, and/or declarative security to demonstrate adding security to existing web applications.
Here are some other locations where you will find examples of securing various types of applications:
Example: Securing an Enterprise Bean with Declarative Security
Example: Securing an Enterprise Bean with Programmatic Security
GlassFish samples: http://glassfish-samples.java.net/
To Set Up Your System for Running the Security Examples
To set up your system for running the security examples, you need to configure a user database that the application can use for authenticating users. Before continuing, follow these steps.
- Add an authorized user to the GlassFish Server. For the examples in this
chapter and in Chapter 41, Getting Started Securing Enterprise Applications, add a user to the file realm of the
GlassFish Server, and assign the user to the group TutorialUser:
- From the Administration Console, expand the Configurations node, then expand the server-config node.
- Expand the Security node.
- Expand the Realms node.
- Select the File node.
- On the Edit Realm page, click Manage Users.
- On the File Users page, click New.
- In the User ID field, type a User ID.
- In the Group List field, type TutorialUser.
- In the New Password and Confirm New Password fields, type a password.
- Click OK.
Be sure to write down the user name and password for the user you create so that you can use it for testing the example applications. Authentication is case sensitive for both the user name and password, so write down the user name and password exactly. This topic is discussed more in Managing Users and Groups on the GlassFish Server.
- Set up Default Principal to Role Mapping on the GlassFish Server:
- From the Administration Console, expand the Configurations node, then expand the server-config node.
- Select the Security node.
- Select the Default Principal to Role Mapping Enabled check box.
- Click Save.
Example: Basic Authentication with a Servlet
This example explains how to use basic authentication with a servlet. With basic authentication of a servlet, the web browser presents a standard login dialog that is not customizable. When a user submits his or her name and password, the server determines whether the user name and password are those of an authorized user and sends the requested web resource if the user is authorized to view it.
In general, the following steps are necessary for adding basic authentication to an unsecured servlet, such as the ones described in Chapter 3, Getting Started with Web Applications. In the example application included with this tutorial, many of these steps have been completed for you and are listed here simply to show what needs to be done should you wish to create a similar application. The completed version of this example application can be found in the tut-install/examples/security/hello2_basicauth/ directory.
Follow the steps in To Set Up Your System for Running the Security Examples.
Create a web module as described in Chapter 3, Getting Started with Web Applications for the servlet example, hello2.
Add the appropriate security annotations to the servlet. The security annotations are described in Specifying Security for Basic Authentication Using Annotations.
Build, package, and deploy the web application by following the steps in To Build, Package, and Deploy the Servlet Basic Authentication Example Using NetBeans IDE or To Build, Package, and Deploy the Servlet Basic Authentication Example Using Ant.
Run the web application by following the steps described in To Run the Basic Authentication Servlet.
Specifying Security for Basic Authentication Using Annotations
The default authentication mechanism used by the GlassFish Server is basic authentication. With basic authentication, the GlassFish Server spawns a standard login dialog to collect user name and password data for a protected resource. Once the user is authenticated, access to the protected resource is permitted.
To specify security for a servlet, use the @ServletSecurity annotation. This annotation allows you to specify both specific constraints on HTTP methods and more general constraints that apply to all HTTP methods for which no specific constraint is specified. Within the @ServletSecurity annotation, you can specify the following annotations:
The @HttpMethodConstraint annotation, which applies to a specific HTTP method
The more general @HttpConstraint annotation, which applies to all HTTP methods for which there is no corresponding @HttpMethodConstraint annotation
Both the @HttpMethodConstraint and @HttpConstraint annotations within the @ServletSecurity annotation can specify the following:
A transportGuarantee element that specifies the data protection requirements (that is, whether or not SSL/TLS is required) that must be satisfied by the connections on which requests arrive. Valid values for this element are NONE and CONFIDENTIAL.
A rolesAllowed element that specifies the names of the authorized roles.
For the hello2_basicauth application, the GreetingServlet has the following annotations:
@WebServlet(name = "GreetingServlet", urlPatterns = {"/greeting"}) @ServletSecurity( @HttpConstraint(transportGuarantee = TransportGuarantee.CONFIDENTIAL, rolesAllowed = {"TutorialUser"}))
These annotations specify that the request URI /greeting can be accessed only by users who have been authorized to access this URL because they have been verified to be in the role TutorialUser. The data will be sent over a protected transport in order to keep the user name and password data from being read in transit.
If you use the @ServletSecurity annotation, you do not need to specify security settings in the deployment descriptor. Use the deployment descriptor to specify settings for nondefault authentication mechanisms, for which you cannot use the @ServletSecurity annotation.
To Build, Package, and Deploy the Servlet Basic Authentication Example Using NetBeans IDE
- Follow the steps in To Set Up Your System for Running the Security Examples.
- In NetBeans IDE, from the File menu, choose Open Project.
- In the Open Project dialog, navigate to:
tut-install/examples/security/
- Select the hello2_basicauth folder.
- Select the Open as Main Project check box.
- Click Open Project.
- Right-click hello2_basicauth in the Projects pane and select Deploy.
This option builds and deploys the example application to your GlassFish Server instance.
To Build, Package, and Deploy the Servlet Basic Authentication Example Using Ant
- Follow the steps in To Set Up Your System for Running the Security Examples.
- In a terminal window, go to:
tut-install/examples/security/hello2_basicauth/
- Type the following command:
ant
This command calls the default target, which builds and packages the application into a WAR file, hello2_basicauth.war, that is located in the dist directory.
- Make sure that the GlassFish Server is started.
- To deploy the application, type the following command:
ant deploy
To Run the Basic Authentication Servlet
- In a web browser, navigate to the following URL:
https://localhost:8181/hello2_basicauth/greeting
You may be prompted to accept the security certificate for the server. If so, accept the security certificate. If the browser warns that the certificate is invalid because it is self-signed, add a security exception for the application.
An Authentication Required dialog box appears. Its appearance varies, depending on the browser you use.
- Type a user name and password combination that corresponds to a user who
has already been created in the file realm of the GlassFish Server and
has been assigned to the group of TutorialUser; then click OK.
Basic authentication is case sensitive for both the user name and password, so type the user name and password exactly as defined for the GlassFish Server.
The server returns the requested resource if all the following conditions are met.
A user with the user name you entered is defined for the GlassFish Server.
The user with the user name you entered has the password you entered.
The user name and password combination you entered is assigned to the group TutorialUser on the GlassFish Server.
The role of TutorialUser, as defined for the application, is mapped to the group TutorialUser, as defined for the GlassFish Server.
- Type a name in the text field and click the Submit button.
Because you have already been authorized, the name you enter in this step does not have any limitations. You have unlimited access to the application now.
The application responds by saying “Hello” to the name you typed.
Next Steps
For repetitive testing of this example, you may need to close and reopen your browser. You should also run the ant undeploy and ant clean targets or the NetBeans IDE Clean and Build option to get a fresh start.
Example: Form-Based Authentication with a JavaServer Faces Application
This example explains how to use form-based authentication with a JavaServer Faces application. With form-based authentication, you can customize the login screen and error pages that are presented to the web client for authentication of the user name and password. When a user submits his or her name and password, the server determines whether the user name and password are those of an authorized user and, if authorized, sends the requested web resource.
This example, hello1_formauth, adds security to the basic JavaServer Faces application shown in Web Modules: The hello1 Example.
In general, the steps necessary for adding form-based authentication to an unsecured JavaServer Faces application are similar to those described in Example: Basic Authentication with a Servlet. The major difference is that you must use a deployment descriptor to specify the use of form-based authentication, as described in Specifying Security for the Form-Based Authentication Example. In addition, you must create a login form page and a login error page, as described in Creating the Login Form and the Error Page.
The completed version of this example application can be found in the tut-install/examples/security/hello1_formauth/ directory.
Creating the Login Form and the Error Page
When using form-based login mechanisms, you must specify a page that contains the form you want to use to obtain the user name and password, as well as a page to display if login authentication fails. This section discusses the login form and the error page used in this example. Specifying Security for the Form-Based Authentication Example shows how you specify these pages in the deployment descriptor.
The login page can be an HTML page or a servlet, and it must return an HTML page containing a form that conforms to specific naming conventions (see the Java Servlet 3.0 specification for more information on these requirements). To do this, include the elements that accept user name and password information between <form></form> tags in your login page. The content of an HTML page or servlet for a login page should be coded as follows:
<form method="post" action="j_security_check"> <input type="text" name="j_username"> <input type="password" name= "j_password"> </form>
The full code for the login page used in this example can be found at tut-install/examples/security/hello1_formauth/web/login.xhtml. Here is the code for this page:
 <h:head> <title>Login Form</title> </h:head> <h:body>
Hello, please log in:
 <form method="post" action="j_security_check"> <h:panelGrid columns="2"> <h:outputLabel for="j_username" value="Please type your user name:"/> <input type="text" id="username" name="j_username" size="25"/> <h:outputLabel for="j_password" value="Please type your password:"/> <input type="password" id="password" name="j_password" size="15"/> <h:outputText value=" " /> <input type="submit" name="submit" value="Submit"/> <input type="reset" name="reset" value="Reset"/> </h:panelGrid> </form> </h:body>
The login error page is displayed if the user enters a user name and password combination that is not authorized to access the protected URI. For this example, the login error page can be found at tut-install/examples/security/hello1_formauth/web/error.xhtml. For this example, the login error page explains the reason for receiving the error page and provides a link that will allow the user to try again. Here is the code for this page:
  <title>Login Error</title>
Invalid user name or password.

Please enter a user name or password that is authorized to access this application. For this application, this means a user that has been created in the file realm and has been assigned to the group of TutorialUser.

Return to login page

Specifying Security for the Form-Based Authentication Example
This example takes a very simple servlet-based web application and adds form-based security. To specify form-based instead of basic authentication for a JavaServer Faces example, you must use the deployment descriptor.
The following sample code shows the security elements added to the deployment descriptor for this example, which can be found in tut-install/examples/security/hello1_formauth/web/WEB-INF/web.xml.
<security-constraint> <display-name>Constraint1</display-name> <web-resource-collection> <web-resource-name>wrcoll</web-resource-name> <description/> <url-pattern>/*</url-pattern> </web-resource-collection> <auth-constraint> <description/> <role-name>TutorialUser</role-name> </auth-constraint> </security-constraint> <login-config> <auth-method>FORM</auth-method> <realm-name>file</realm-name> <form-login-config> <form-login-page>/faces/login.xhtml</form-login-page> <form-error-page>/faces/error.xhtml</form-error-page> </form-login-config> </login-config> <security-role> <description/> <role-name>TutorialUser</role-name> </security-role>
To Build, Package, and Deploy the Form-Based Authentication Example Using NetBeans IDE
- Follow the steps in To Set Up Your System for Running the Security Examples.
- In NetBeans IDE, from the File menu, choose Open Project.
- In the Open Project dialog, navigate to:
tut-install/examples/security/
- Select the hello1_formauth folder.
- Select the Open as Main Project check box.
- Click Open Project.
- Right-click hello1_formauth in the Projects pane and select Deploy.
To Build, Package, and Deploy the Form-Based Authentication Example Using Ant
- Follow the steps in To Set Up Your System for Running the Security Examples.
- In a terminal window, go to:
tut-install/examples/security/hello1_formauth/
- Type the following command at the terminal window or command prompt:
ant
This target will spawn any necessary compilations, copy files to the tut-install/examples/security/hello1_formauth/build/ directory, create the WAR file, and copy it to the tut-install/examples/security/hello1_formauth/dist/ directory.
- To deploy hello1_formauth.war to the GlassFish Server, type the following command:
ant deploy
To Run the Form-Based Authentication Example
To run the web client for hello1_formauth, follow these steps.
- Open a web browser to the following URL:
https://localhost:8181/hello1_formauth/
The login form opens in the browser.
- Type a user name and password combination that corresponds to a user who
has already been created in the file realm of the GlassFish Server and
has been assigned to the group of TutorialUser.
Form-based authentication is case sensitive for both the user name and password, so type the user name and password exactly as defined for the GlassFish Server.
- Click the Submit button.
If you entered My_Name as the name and My_Pwd for the password, the server returns the requested resource if all the following conditions are met.
A user with the user name My_Name is defined for the GlassFish Server.
The user with the user name My_Name has a password My_Pwd defined for the GlassFish Server.
The user My_Name with the password My_Pwd is assigned to the group TutorialUser on the GlassFish Server.
The role TutorialUser, as defined for the application, is mapped to the group TutorialUser, as defined for the GlassFish Server.
When these conditions are met and the server has authenticated the user, the application appears.
- Type your name and click the Submit button.
Because you have already been authorized, the name you enter in this step does not have any limitations. You have unlimited access to the application now.
The application responds by saying “Hello” to you.
Next Steps
For additional testing and to see the login error page generated, close and reopen your browser, type the application URL, and type a user name and password that are not authorized.
Note - For repetitive testing of this example, you may need to close and reopen your browser. You should also run the ant clean and ant undeploy commands to ensure a fresh build if using the Ant tool, or select Clean and Build then Deploy if using NetBeans IDE.


Truck and car crash halts M1 traffic, Gold Coast
by Monique StClairThere are already massive delays on the M1 southbound this afternoon, after a truck and car crash.
It happened in the left lane between exit 89 and exit 92 (through Burleigh Heads/Palm Beach) just before 2.00pm.

The Department of Transport and Main Roads has warned of delays throughout the afternoon.
At the time of writing, the traffic was banked back up to Robina.
Northbound traffic through that area is also experiencing delays.
It’s understood paramedics haven’t been called to the scene.


Delay the evaluation of a paramless async function and cache the result (see thunk).
npm install thunky
Let’s make a simple function that returns a random number 1 second after it is called for the first time
``` js var thunky = require(‘thunky’);
var test = thunky(function(callback) { // the inner function should only accept a callback console.log(‘waiting 1s and returning random number’); setTimeout(function() { callback(Math.random()); }, 1000); });
test(function(num) { // inner function is called the first time we call test console.log(num); // prints random number });
test(function(num) { // subsequent calls waits for the first call to finish and return the same value console.log(num); // prints the same random number as above }); ```
Thunky makes it easy to implement a lazy evaluation pattern.
``` js var getDb = thunky(function(callback) { db.open(myConnectionString, callback); });
var queryDb = function(query, callback) { getDb(function(err, db) { if (err) return callback(err); db.query(query, callback); }); };
queryDb(‘some query’, function(err, result) { … } );
queryDb(‘some other query’, function(err, result) { … } ); ```
The first time getDb
is called it will try do open a connection to the database.
Any subsequent calls will just wait for the first call to complete and then call your callback.
A nice property of this pattern is that it easily allows us to pass any error caused by getDb
to the queryDb
callback.
If the thunk callback is called with an Error
object as the first argument it will not cache the result
``` js var fails = thunky(function(callback) { console.log(‘returning an error’); callback(new Error(‘bad stuff’)); });
fails(function(err) { // inner function is called console.log(err); });
fails(function(err) { // inner function is called again as it returned an error before console.log(err); }); ```
MIT


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WuhooHandle | platform_window |
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WuhooConvertRGBA | convert_rgba |
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int | window_flags |
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WuhooFlags | flags |
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int | global_mods |
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int | width |
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int | height |
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int | cwidth |
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int | cheight |
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int | x |
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int | y |
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WuhooBoolean | is_initialized |
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WuhooBoolean | is_alive |
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WuhooByte | memory [256] |
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The documentation for this struct was generated from the following file: