Checking if a number is perfect square?

Question

I think there is a precision problem with the following code:

bool isPerfectSquare(long long n){
    long long squareRootN=(long long)(sqrt(n)+0.5);

    return          


        

Answer 1

You use round.

bool isPerfectSquare(long long n){
    long long squareRootN=(long long)round((sqrt(n)));

    if(squareRootN*squareRootN == n) {
        return true; 
    }
     else {
        return false; 
     }

round rounds the number to the nearest rounding. The function will be true only if n is a perfect square.