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

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

gcc -dM -E 1.c

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]