C語言混亂大賽1987年獲獎作品

時間：2020-05-06 10:18:06

關(guān)鍵字： C語言大賽 UNIX BSP

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[導(dǎo)讀]今天翻前兩天剛送到的《C專家編程》，章節(jié)8.10的那個IOCCC 1987年的獲獎作品讓我覺得很好玩。 main(){?printf(&unix["/021%six/012/0"],?(unix)["have"]?+?"fun"?-?0x60);} 作者這里利用的第一個技巧并不算太晦澀："a[i] = i[a]=? *(a+i)“ 即下標運算符的可

今天翻前兩天剛送到的《C專家編程》，章節(jié)8.10的那個IOCCC 1987年的獲獎作品讓我覺得很好玩。

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

作者這里利用的第一個技巧并不算太晦澀："a[i] = i[a]= *(a+i)“ 即下標運算符的可交換性。

但是這個老天爺?shù)膗nix和&unix是怎么回事？

我把自己腦子里有印象的C的最隱蔽的角落回想了一遍，難道是trigraph之類的特殊符號？查了查不是

正推不行，就反推吧。編譯運行后，輸出結(jié)果是“unix“。如此說來“(unix)["have"] + "fun" - 0x60)“這個表達式的結(jié)果應(yīng)該是指向"un"的char * ；因此，“(unix)["have"] - 0x60"的值應(yīng)該是整數(shù)1才對；因此“(unix)["have"]"的值應(yīng)該是0x61；這就豁然開朗了，0x61='a'，是"have"中的第二個字符，所以unix的值必然為1。

這個反推是否正確呢? 利用gcc看一下只進行preprocessor處理后的代碼：

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

正確。unix看來是屬于預(yù)定義的marco，值為1。拿“gcc predefined marco"為關(guān)鍵字google了一下，找到可信資料了：

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

原來unix這個宏是屬于gcc提供的 System-specific Predefined Macros，因此這個奇妙的函數(shù)只能在*nix環(huán)境下正常工作。

可以查看更多的gcc提供的predefined marco

gcc -dM -E 1.c

順便附上歷史文獻，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]