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 might use std::sqrt as a guess and test with multiplication:

#include 
#include 
#include 

bool isPerfectSquare(long long n){
    double guess = sqrt(n);
    long long r = std::floor(guess);
    if(r*r == n) return true;
    else {
        r = std::ceil(guess);
        return r*r == n;
    }
}

int main() {
    const long long Limit = std::pow(10, std::numeric_limits::digits10 / 2);
    std::cout << " Limit: " << Limit << '\n';
    for (long long i = 0; i < Limit; ++i) {
        if( ! isPerfectSquare(i*i)) {
            std::cout << "Failure: " << i << '\n';
            return 0;
        }
    }
    std::cout << "Success\n";
}