/* ioccc.c */
/* IOCCC best one-liner winner 1987 by David Korn ---
main() { printf(&unix["\021%six\012\0"],(unix)["have"]+"fun"-0x60);}
from <http://www.ioccc.org/years.html#1987>
*/
/* A detailed set of samples to show how this works
by David Ireland, copyright (C) 2002.
Modified by William Cheung
for GCC Version 4.8.3 on CentOS 7 (x86_64)
See
http://www.di-mgt.com.au/src/korn_ioccc.txt
for original code
*/
#include <stdio.h>
int main()
{
// int unix;
// We do not need to declare 'unix', or we will get an error:
// expected identifier or ‘(’ before numeric constant
// because unix is a predefined macro that expands to 1 (only on
// unix-like systems perhaps)
printf("unix=%d\n", unix); /* =1 */
/* This prints the string "un",
i.e. "fun" starting at offset [1] */
printf("%s\n","fun"+1);
/* This prints 97 = the int value of the 2nd char 'a' */
printf("%d\n", "have"[1]);
/* just like this */
printf("%d\n", 'a');
/* ditto because x[1] = 1[x] */
printf("%d\n", (1)["have"]);
/* 97 - 96 = 0x61 - 0x60 = 1 */
printf("%d\n", (1)["have"] - 0x60);
/* So this is the same as "fun" + 1, printing "un" */
printf("%s\n", "fun" + ((1)["have"] - 0x60));
/* Rearrange and use unix variable instead of 1 */
printf("%s\n", (unix)["have"]+"fun"-0x60);
/* ...thus we have the first argument in the printf call. */
/* Both these print the string "bcde", ignoring the 'a' */
printf("%s\n", "abcde" + 1);
printf("%s\n", &"abcde"[1]);
/* so does this */
printf("%s\n", &(1)["abcde"]);
/* and so does this (NB [] binds closer than &) */
printf("%s\n", &unix["abcde"]);
/* This prints "%six" + newline */
printf("%s", &"?%six\n"[1]);
/* So does this: note that
\012 = 0x0a = \n,
the first \021 char is ignored,
and the \0 is superfluous, probably just for symmetry */
printf("%s", &"\021%six\012\0"[1]);
/* and so does this */
printf("%s", &unix["\021%six\012\0"]);
/* Using this as a format string, we can print "ABix" */
printf(&unix["\021%six\012\0"], "AB");
/* just like this does */
printf("%six\n", "AB");
/* So, we can print "unix" like this */
printf("%six\n", (unix)["have"]+"fun"-0x60);
/* or, finally, like this */
printf(&unix["\021%six\012\0"],(unix)["have"]+"fun"-0x60);
return 0;
}
