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。

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
   
3. main()
   
4. {
   
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。

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

1. 
   
2. Best One Liner:
   
3. 
   
4. 
   
5.     David Korn
   
6.     AT&T Bell Labs
   
7.     MH 3C-526B, AT&T Bell Labs
   
8.     Murray Hill, NJ
   
9.     07974
   
10.     USA
    
11. 
    
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
    
16. section right now and see if you are one of the few people who can.
    
17. 
    
18. 
    
19. Several points are important to understand in this program:
    
20. 
    
21. 
    
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
    
24.        to be a data type (such as int, char, struct foo, enum, ...)
    
25.        what must `unix' be?
    
26. 
    
27. 
    
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:
    
30. 
    
31. 
    
32.         char *x;
    
33. 
    
34. 
    
35.       defines a pointer to a character (i.e. an address), and that
    
36.       the `=' assigns things is compatible types.  Since:
    
37. 
    
38. 
    
39.             x = "have";
    
40. 
    
41. 
    
42.       is legal C, what type of value is "have"?
    
43. 
    
44. 
    
45.     3) Note that the following expressions yield the same value:
    
46. (unix)["have"] + "fun" - 0x60)
    
47.         x[3]    *(x+3)      *(3+x)
    
48. 
    
49. 
    
50.        since addition is communitive.  What can be said about the value:
    
51. 
    
52. 
    
53.         3[x]

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