国际C语言混乱代码大赛（IOCCC, The International Obfuscated C Code Contest）是一项国际编程赛事，从1984年开始，每年举办一次（1997年、1999年、2002年、2003年和2006年例外）。目的是写出最有创意的最让人难以理解的C语言代码。
我自认为C语言学的还可以接受，对于C语言编程有一定了解，但是看到别人的代买还是完全自愧不如，还是要多学习，以后的路还很长。
以下针对我们第一个程序的各种版本
version 0
这个版本是我们刚开始学编程的时候编写的，很容易理解咯
#include <stdio.h>
int main(void)
{
printf("Hello World !");
return 0;
}
version 1
这个版本就没有那么容易理解了，主要是里面有define宏，其实仔细分析，还是可以解决的
/*hello world version 1*/
#define _________ }
#define ________ putchar
#define _______ main
#define _(a) ________(a);
#define ______ _______(){
#define __ ______ _(0x48)_(0x65)_(0x6C)_(0x6C)
#define ___ _(0x6F)_(0x2C)_(0x20)_(0x77)_(0x6F)
#define ____ _(0x72)_(0x6C)_(0x64)_(0x21)
#define _____ __ ___ ____ _________
#include<stdio.h>
_____<pre name="code" class="cpp">main(){
putchar(0x48);putchar(0x65);putchar(0x6C);
putchar(0x6C); putchar(0x6F);putchar(0x2C);
putchar(0x20);putchar(0x77);putchar(0x6F);
putchar(0x72);putchar(0x6C);putchar(0x64);
putchar(0x21); }
version 2
此版本也主要是对数组y[]操作，以及指针Z进行的++操作，最后打印出来
/*hello world version 2*/
#include<stdio.h>
main()
{
int x=0,y[14],*z=&y;
*(z++)=0x48;
*(z++)=y[x++]+0x1D;
*(z++)=y[x++]+0x07;
*(z++)=y[x++]+0x00;
*(z++)=y[x++]+0x03;
*(z++)=y[x++]-0x43;
*(z++)=y[x++]-0x0C;
*(z++)=y[x++]+0x57;
*(z++)=y[x++]-0x08;
*(z++)=y[x++]+0x03;
*(z++)=y[x++]-0x06;
*(z++)=y[x++]-0x08;
*(z++)=y[x++]-0x43;
*(z++)=y[x]-0x21;
x=*(--z);
while(y[x]!=NULL)
putchar(y[x++]);
}
此版本依然是简单的宏替换以及ansi字符的运算
/*hello world version 3*/
#include<stdio.h>
#define __(a) goto a;
#define ___(a) putchar(a);
#define _(a,b) ___(a) __(b);
main()
{
_:
__(t)a:
_('r',g)b:
_('$',p)
c:
_('l',f)d:
_(' ',s)e:
_('a',s)
f:
_('o',q)g:
_('l',h)h:
_('d',n)
i:
_('e',w)j:
_('e',x)k:
_('\n',z)
l:
_('H',l)m:
_('X',i)n:
_('!',k)
o:
_('z',q)p:
_('q',b)q:
_(',',d)
r:
_('i',l)s:
_('w',v)t:
_('H',j)
u:
_('a',a)v:
_('o',a)w:
_(')',k)
x:
_('l',c)y:
_('\t',g)z:
___(0x0)
}
main()
{
_:
goto t;a:
putchar('r'); goto g;;b:
putchar('$'); goto p;;
c:
putchar('l'); goto f;;d:
putchar(' '); goto s;;e:
putchar('a'); goto s;;
f:
putchar('o'); goto q;;g:
putchar('l'); goto h;;h:
putchar('d'); goto n;;
i:
putchar('e'); goto w;;j:
putchar('e'); goto x;;k:
putchar('\n'); goto z;;
l:
putchar('H'); goto l;;m:
putchar('X'); goto i;;n:
putchar('!'); goto k;;
o:
putchar('z'); goto q;;p:
putchar('q'); goto b;;q:
putchar(','); goto d;;
r:
putchar('i'); goto l;;s:
putchar('w'); goto v;;t:
putchar('H'); goto j;;
u:
putchar('a'); goto a;;v:
putchar('o'); goto a;;w:
putchar(')'); goto k;;
x:
putchar('l'); goto c;;y:
putchar('\t'); goto g;;z:
putchar(0x0);
}
这个版本主要是运用递归的思想打印程序
/*hello world version 4*/
int n[]=
{
0x48,0x65,0x6C,0x6C,0x6F,
0x2C,0x20,0x77,0x6F,0x72,
0x6C,0x64,0x21,0x0A,0x00
},*m=n;
main(n)
{
putchar
(*m)!='\0'?main
(m++):exit(n++);
}
这个版本程序主要是考察位运算，主要是左移运算，运算结果是将hello world 的ansi编码存储到数组中，最后打印
main()
{
int i,n[]= {
(((1<<1)<<(1<<1)<<(1<<
1)<<(1<<(1>>1)))+((1<<1)<<(1<<1))), (((1
<<1)<<(1<<1)<<(1<<1)<<(1<<1))-((1<<1)<<(
1<<1)<<(1<<1))+((1<<1)<<(1<<(1>>1)))+ (1
<<(1>>1))),(((1<<1)<<(1<<1)<<(1<<1)<< (1
<<1))-((1<<1)<<(1<<1)<<(1<<(1>>1)))- ((1
<<1)<<(1<<(1>>1)))),(((1<<1)<<(1<<1)<<(1
<<1)<<(1<<1))-((1<<1)<<(1<<1)<<(1<<(1>>1
)))-((1<<1)<<(1<<(1>>1)))),(((1<<1)<< (1
<<1)<<(1<<1)<<(1<<1))-((1<<1)<<(1<<1)<<(
1<<(1>>1)))-(1<<(1>>1))),(((1<<1)<<(1<<1
)<<(1<<1))+((1<<1)<<(1<<1)<<(1<<(1>>1)))
-((1<<1)<<(1<<(1>>1)))),((1<<1)<< (1<<1)
<<(1<<1)),(((1<<1)<<(1<<1)<<(1<<1)<<(1<<
1))-((1<<1)<<(1<<1))-(1<<(1>>1))),(((1<<
1)<<(1<<1)<<(1<<1)<<(1<<1))-((1<<1)<< (1
<<1)<<(1<<(1>>1)))-(1<<(1>>1))), (((1<<1
)<<(1<<1)<<(1<<1)<<(1<<1))- ((1<<1)<< (1
<<1)<<(1<<(1>>1)))+(1<<1)), (((1<<1)<< (
1<<1)<<(1<<1)<< (1<<1))-((1<<1)<< (1<<1)
<<(1<<(1>>1)))-((1<<1) <<(1<< (1>>1)))),
(((1<<1)<< (1<<1)<<(1<<1)<< (1<<1))- ((1
<<1)<<(1<<1)<<(1<<1))+((1<<1)<< (1<<(1>>
1)))), (((1<<1)<<(1<<1) <<(1<<1))+(1<<(1
>>1))),(((1<<1)<<(1<<1))+((1<<1)<< (1<<(
1>>1))) + (1<< (1>>1)))
};
for(i=(1>>1); i
<(((1<<1) <<(1<<1))+((1 <<1)<< (1<<(1>>1
))) + (1<<1)); i++) printf("%c",n[i]);
}
这个版本主要考察的是前缀单目运算符++和后缀运算符++以及双目运算符+的优先级问题，此版本需要保存为cpp后缀运算
#include <stdio.h>
#define _(_) putchar(_);
int main(void)
{
int i = 0;
_(
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++i
)
_(
++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++i
)
_(
++++++++++++++i
)
_(
--++i
)
_(
++++++i
)
_(
------
----------------------------
----------------------------
----------------------------
----------------------------
----------------i
)
_(
--------
----------------i
)
_(
++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++i
)
_(
----------------i
)
_(
++++++i
)
_(
------------i
)
_(
--------
--------i
)
_(
----------------
----------------------------
----------------------------
----------------------------
----------------------------
------i
)
_(
------------------
----------------------------i
)
return i;
}
结果如下所示，针对每个运算符的优先级进行结合运算
<pre name="code" class="cpp"><pre name="code" class="cpp">int main(void)
{
int i = 0;
putchar(++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++i);
putchar(++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++i);
putchar(++++++++++++++i);
putchar(--++i);
putchar(++++++i);
putchar(------
----------------------------
----------------------------
----------------------------
----------------------------
----------------i);
putchar(-------- ----------------i);
putchar(++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++++
++++++++++++++++++++++++++i);
putchar(----------------i);
putchar(++++++i);
putchar(------------i);
putchar(-------- --------i);
putchar(----------------
----------------------------
----------------------------
----------------------------
----------------------------
------i);
putchar(------------------
----------------------------i);
return i;
}原文：http://blog.csdn.net/u011889952/article/details/44596225