Signed integers have some weirdness attached. The number that's one followed by all zeroes in binary (INT_MIN in limits.h) is defined as negative, because the sign bit is set. But, the rules for 2's complement arithmetic predict that -INT_MIN == INT_MIN. So it's not a normal number.
#include <limits.h>
#include <stdio.h>
void main(void) {
int x = INT_MIN;
printf("INT_MIN = %d\n", x);
printf("INT_MIN * -1 = %d\n", x * -1);
printf("INT_MIN / -1 = %d\n", x / -1);
} $ gcc -std=c99 -S intmin.c -o intmin.s
$ clang -std=c99 -S intmin.c -o intmin.sc
$ grep -i div intmin.s*
intmin.sc: idivl %esi
$ clang --version
Debian clang version 3.5-1~exp1 (trunk) (based on LLVM 3.5)
Target: x86_64-pc-linux-gnu
Thread model: posix
$ clang -o intmin_c -std=c99 intmin.c
$ ./intmin_c
INT_MIN = -2147483648
INT_MIN * -1 = -2147483648
Floating point exception
$ gcc --version
gcc-4.7.real (Debian 4.7.2-5) 4.7.2
(...)
$ gcc -o intmin -std=c99 intmin.c
$ ./intmin
INT_MIN = -2147483648
INT_MIN * -1 = -2147483648
INT_MIN / -1 = -2147483648
$ tcc -v
tcc version 0.9.25
$ tcc -run intmin.c
INT_MIN = -2147483648
INT_MIN * -1 = -2147483648
Floating point exception
$ cat intmin.c
#include <limits.h>
#include <stdio.h>
int main(void) { // Change from void to int for c99
int x = INT_MIN;
printf(" INT_MIN = %d\n", x);
printf("INT_MIN * -1 = %d\n", x * -1);
printf("INT_MIN / -1 = %d\n", x / -1);
return 0; // return 0 - c99
}
It's also a good example of how undefined behavior allows for optimizations. The compiler is able to evaluate your expression at compile time rather than emitting a division instruction, even though this changes how the program behaves.
C sure is fun.
int abs_s(int x) {
assert(x != INT_MIN);
return (x >= 0) ? x : -x;
}(On that note: Are there even machines left to write code for that don't use two's complement? Or 8 bits per byte?)