I ran this simple example:
#include stdlib.h
#include stdio.h
int fib(int AnArg) {
if (AnArg = 2) return (1);
return (fib(AnArg-1)+fib(AnArg-2));
}
int main(int argc, char* argv[]) {
int n = atoi(argv[1]);
printf(fib(%i)=%i\n, n, fib(n));
}
through system gcc and gcc built from sources.
Yuri y...@rawbw.com writes:
I ran this simple example:
#include stdlib.h
#include stdio.h
int fib(int AnArg) {
if (AnArg = 2) return (1);
return (fib(AnArg-1)+fib(AnArg-2));
}
int main(int argc, char* argv[]) {
int n = atoi(argv[1]);
printf(fib(%i)=%i\n, n, fib(n));
}
through
Leonidas Tsampros wrote:
I'm pretty sure that a small difference in execution time does not mean
that the produced code is different.
Actually, execution time of a process is very sensitive to the
environment of this process. See for instance:
