C语言混乱大赛1987年获奖作品

时间：2020-05-06 10:18:06

关键字： C语言大赛 UNIX BSP

手机看文章

扫描二维码
随时随地手机看文章

[导读]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。 main(){ printf(&unix["/021%six/012/0"], (unix)["have"] + "fun" - 0x60);} 作者这里利用的第一个技巧并不算太晦涩："a[i] = i[a]= *(a+i)“ 即下标运算符的可

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。

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

作者这里利用的第一个技巧并不算太晦涩："a[i] = 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：

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

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

gcc -dM -E 1.c

顺便附上历史文献，哈哈

Best One Liner:

 David Korn AT&T Bell Labs MH 3C-526B, AT&T Bell Labs Murray Hill, NJ 07974 USA
 The Judges believe that this is the best one line entry ever received.Compile on a UN*X system, or at least using a C implementation thatfakes it. Very few people are able to determine what this programdoes by visual inspection. I suggest that you stop reading thissection right now and see if you are one of the few people who can.

Several points are important to understand in this program:

 1) What is the symbol `unix' and what is its value in the program? Clearly `unix' is not a function, and since `unix' is not declared to be a data type (such as int, char, struct foo, enum, ...) what must `unix' be?

 2) What is the value of the symbol "have"? (hint: the value is NOT 4 characters, or 'h', or a string) Consider the fact that:

 char *x;

 defines a pointer to a character (i.e. an address), and that the `=' assigns things is compatible types. Since:

 x = "have";

 is legal C, what type of value is "have"?

 3) Note that the following expressions yield the same value:(unix)["have"] + "fun" - 0x60) x[3] *(x+3) *(3+x)

 since addition is communitive. What can be said about the value:

 3[x]