International Obfuscated C Code Tattoo

chongo writes "Some people eschew obfuscation while other live for it. Thomas Scovell has taken obfuscation to a completely new personal level by obtaining the very first International Obfuscated C Code Tattoo. We (the IOCCC Judges) are pleased that Thomas has honored the 1984 anonymous IOCCC winning entry by placing the source code on his arm: the very first IOCCC winner to receive this distinction. The anonymous winner (a person who known for various things on the Internet and has been programming in and associated with C for decades) feels honored by the tattoo as well."

Comments

What I want to know is...

[CommanderData]

Could I be arrested for tattooing this on my arm:

#!/usr/bin/perl -w
# 531-byte qrpff-fast, Keith Winstein and Marc Horowitz <sipb-iap-dvd@mit.edu>
# MPEG 2 PS VOB file on stdin -> descrambled output on stdout
# arguments: title key bytes in least to most-significant order
$_='while(read+STDIN,$_,2048){$a=29;$b=73;$ c=142;$t=255;@t=map{$_%16or$t^=$c^=(
$m=(11,10,11 6,100,11,122,20,100)[$_/16%8])&110;$t^=(72,@z=(64, 72,$a^=12*($_%16
-2?0:$m&17)),$b^=$_%64?12:0,@z)[ $_%8]}(16..271);if((@a=unx"C*",$_)[20]&48){$h
=5; $_=unxb24,join"",@b=map{xB8,unxb8,chr($_^$a[--$h+8 4])}@ARGV;s/...$/1$&/;$
d=unxV,xb25,$_;$e=256|(or d$b[4])<<9|ord$b[3];$d=$d>>8^($f=$t&($d>>12^$d>>4^
$d^$d/8))<<17,$e=$e>>8^($t&($g=($q=$e>>14&7^$e)^ $q*8^$q<<6))<<9,$_=$t[$_]^
(($h>>=8)+=$f+(~$g&$t) )for@a[128..$#a]}print+x"C*",@a}';s/x/pack+/g;eval

I'd be a walking violation of the DMCA :)

entry explained

[voisine]

int i;main(){for(;i["]i;++i){--i;}"];read('-'-'-',i+++ "hell\
o, world!\n",'/'/'/'));}read(j,i,p){write(j/p+p,i---j ,i/i);}
==== add some whitespace ==========
int i;
main()
{
for (; i["]i;++i){--i;}"]; read('-' - '-', i++ + "hello, world!\n", '/' / '/'));
}

read(j, i, p)
{
write(j / p + p, i-- - j, i / i);
}
===== and char subtraced from itself is 0, and char or pointer divided by itself is 1 =====
int i;
main()
{
for (; i["]i;++i){--i;}"]; read(0, i++ + "hello, world!\n", 1));
}

read(j, i, p)
{
write(j / p + p, i-- - j, 1);
}
======= j is always 0, p is always 1, lets remove them ======
int i;
main()
{
for (; i["]i;++i){--i;}"]; read(i++ + "hello, world!\n"));
}

read(i)
{
write(0 / 1 + 1, i-- - 0, 1);
}
======= 0 / 1 + 1 is 1, subtracting 0 does nothing, decrementing a local variable this is never used afterward also does nothing =======
int i;
main()
{
for (; i["]i;++i){--i;}"]; read(i++ + "hello, world!\n"));
}

read(i)
{
write(1, i, 1);
}
======== replace read(i) with write(1, i, 1) =====
int i;
main()
{
for (; i["]i;++i){--i;}"]; write(1, i++ + "hello, world!\n", 1));
}
====== i[n] can be rewritten *(i + n) or *(n + i) ======
int i;
main()
{
for (; *("]i;++i){--i;}" + i); write(1, "hello, world!\n" + i++, 1));
}
=== as i gets incrimented, we dereference the next char of the string which is always non-zero till we hit the null terminator, all the matters is that the string is the same length as "hello, world!\n" =====
int i;
main()
{
for (; *("hello, world!\n" + i); write(1, "hello, world!\n" + i++, 1));
}
===== so now we can see we incriment i, printing out the next character of hello world till we hit the null terminator ====

Re:entry explained

[voisine]

global variables are initialized to zero, local stack variables are undefined

Re:entry explained

[voisine]

Acording to my K&R book here (second edition which was updated long after 1984) static variables are initialied by default to 0, and it also states that variables defined outside of functions are implicitly static.

Re:So... who was it?

[chongo]

FYI: I asked the anonymous author a while back about Perl. They said:

" Sometimes write in Perl when the wind is right. :-) (my smiley added) No seriously, Perl does have its place. I do use it for short scripting sometimes although in those years (back in 1984) it was sed, awk, grep and sh."