Question

I was just looking at some algorithms for prime numbers and came across this:

for(int i=2;i*i <= n;i++)
{/*assume no operations here*/} 

I was just wondering if the above loop will be faster than the following or not?

int x=sqrt(n);
for(int i=2;i<=x;i++)
{/*nop*/}

Answer 1

It depends on the value of n, of course. Anyway, sqrt() is not guaranteed to give you the right result: due to rounding reasons, you might end up with a value of x which is one less than expected and ruin the algorithm. Rather than going for a micro-optimisation, I would stick to correctness here and use the original version, which is guaranteed to give correct results.