Why sqrt become much faster without -O2 in g++ on my computer?

Question

Consider the following code:

#include <cstdio>
#include <cmath>

const int COUNT = 1000000000;

int main()
{
    double sum = 0;
    for (int i = 1; i <= COUNT; ++i) {
        sum += sqrt(i);
    }
    printf("%f\n", sum);
    return 0;
}

Without -O2, it runs only 2.9s on my computer, while it runs 6.4s with -O2.

My computer is Fedora 23, with g++ 5.3.1.

I have tried the same thing on Ubuntu 14.04 (with g++ 4.8), it doesn't have the problem (all 6.4s).

Answer 1

Naive version uses call to glibc sqrt function.

Optimized version uses SSE sqrtsd instruction. But after instruction has completed, it checks that result value is not a NaN. If result value is NaN then it calls glibc sqrt function to setup proper error flags (see manual page for math_error(7)). See Why does compiler generate additional sqrts in the compiled assembly code for detailed explanation.

Why gcc thinks that this is faster? Nobody knows. If you are sure that your numbers don't generate NaNs, use -fno-math-errno compile option.