3 ms·
I'd say: #include<stdio.h> void output(x) { int y; printf("%i\n", x); y = 1 / (1000 - x); output(x + 1); } int main() { output(1); } Wit
by orblivion 16y ago
I'd say:
#include<stdio.h>
void output(x)
{
int y;
printf("%i\n", x);
y = 1 / (1000 - x);
output(x + 1);
}
int main()
{
output(1);
}
With gcc on Linux it has the added bonus of "Floating Point Exception", don't know if that disqualifies it.
EDIT: Nevermind, I got it:
#include<stdio.h>
int output(x)
{
printf("%i\n", x);
(x >= 1000 ) || output(x + 1);
return 1;
}
int main()
{
output(1);
}