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

The code is fine!

I have taken the time to write a small test. As the input range is limited, we can simply validate the function for every input.

Not that I did have to add an explicit cast to double for the sqrt function for it to compile. (using MS VC++ 10)

#include 
#include 
#include 

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

    return squareRootN*squareRootN == n;
}


int main()
{
    // for the input range,
    // generate a set with numbers that are known to be perfect squares.
    // all the rest are not.
    std::set perfectSquares;
    for(long long i = 1ll; i <= 100000ll; ++i) {
        perfectSquares.insert(i*i);
    }
    std::cout << "Created lookup." << std::endl;

    // now test the function for all the numbers in the input range   
    int progress = -1;
    for(long long i = 1ll; i <= 10000000000ll; ++i) {
        bool expected = (perfectSquares.count(i) == 1);
        bool actual = isPerfectSquare(i);

        if(expected != actual) {
            std::cout << "Failed for " << i << std::endl;
        }

        int newprogress = i / 100000000ll;
        if(newprogress != progress) {
            progress = newprogress;
            std::cout << "Progress " << progress << "%" << std::endl;
        }
    }
    std::cout << "Test finished." << std::endl;
}

Results? Pass for all values! The problem must be in the code that uses the function. Maybe add input range validation to the function.