今天翻前两天刚送到的《C专家编程》，章节8.10的那个IOCCC 1987年的获奖作品让我觉得很好玩。

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2. main(){ printf(&unix["/021%six/012/0"], (unix)["have"] + "fun" - 0x60);}

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作者这里利用的第一个技巧并不算太晦涩："a = i[a]=  *(a+i)“ 即下标运算符的可交换性。

但是这个老天爷的unix和&unix是怎么回事？

我把自己脑子里有印象的C的最隐蔽的角落回想了一遍，难道是trigraph之类的特殊符号？查了查不是

正推不行，就反推吧。编译运行后，输出结果是“unix“。如此说来“(unix)["have"] + "fun" - 0x60)“这个表达式的结果应该是指向"un"的char * ；因此，“(unix)["have"] - 0x60"的值应该是整数1才对；因此“(unix)["have"]"的值应该是0x61；这就豁然开朗了，0x61='a'，是"have"中的第二个字符，所以unix的值必然为1。

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这个反推是否正确呢? 利用gcc看一下只进行preprocessor处理后的代码：

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2. whodare@whodare:~/programming/c++$ gcc -E 1.c
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3. main()
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4. {
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5.     printf(&1["/021%six/012/0"], (1)["have"] + "fun" - 0x60) ;
   
6. }

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正确。unix看来是属于预定义的marco，值为1。拿“gcc predefined marco"为关键字google了一下，找到可信资料了：

http://gcc.gnu.org/onlinedocs/cpp/System_002dspecific-Predefined-Macros.html#System_002dspecific-Predefined-Macros

原来unix这个宏是属于gcc提供的 System-specific Predefined Macros，因此这个奇妙的函数只能在*nix环境下正常工作。

% ~( j9 I" D! ?; g

可以查看更多的gcc提供的predefined marco

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2. gcc -dM -E 1.c

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顺便附上历史文献，哈哈

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2. Best One Liner:
   
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4. 
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5.     David Korn
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6.     AT&T Bell Labs
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7.     MH 3C-526B, AT&T Bell Labs
   
8.     Murray Hill, NJ
   
9.     07974
   
10.     USA
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12.    The Judges believe that this is the best one line entry ever received.
    
13. Compile on a UN*X system, or at least using a C implementation that
    
14. fakes it.  Very few people are able to determine what this program
    
15. does by visual inspection.  I suggest that you stop reading this
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16. section right now and see if you are one of the few people who can.
    
17. 
    
18. 
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19. Several points are important to understand in this program:
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22.     1) What is the symbol `unix' and what is its value in the program?
    
23.        Clearly `unix' is not a function, and since `unix' is not declared
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24.        to be a data type (such as int, char, struct foo, enum, ...)
    
25.        what must `unix' be?
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27. 
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28.     2) What is the value of the symbol "have"?  (hint: the value is
    
29.        NOT 4 characters, or 'h', or a string)  Consider the fact that:
    
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32.         char *x;
    
33. 
    
34. 
    
35.       defines a pointer to a character (i.e. an address), and that
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36.       the `=' assigns things is compatible types.  Since:
    
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38. 
    
39.             x = "have";
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42.       is legal C, what type of value is "have"?
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44. 
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45.     3) Note that the following expressions yield the same value:
    
46. (unix)["have"] + "fun" - 0x60)
    
47.         x[3]    *(x+3)      *(3+x)
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50.        since addition is communitive.  What can be said about the value:
    
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53.         3[x]

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