Question

The below code in unix takes ~9s reported by time command.

int main()
{
    double u = 0;
    double v = 0;
    double w = 0;
    int i;
    for (i = 0;i < 1000000000;++i) {
        v *= w;
        u += v;
    }
    printf("%lf\n",u);
}

I don’t understand why the execution times almost double when i change v *= w;withv *= u;

Answer 1

When you change v *= w to v *= u then there is an inter-dependency between the 2 statements. Hence, the first statement has to be completed before executing u += v which could be the reason for the increased performance as the compiler can’t parallelize the execution.