Swift 3 - Find if the Number is a perfect Square

Question

I want to know if there is a way of finding if a number is a perfect square in Swift. I have the user enter a number to check if it is a perfect square. Is there a statement or a function? Thank you in advance!

Answer 1

Try something like this:

var number = 9.0

let root = sqrt(number)
let isInteger = floor(root) == root
print("\(root) is \(isInteger ? "perfect" : "not perfect")")

That "isInteger" bit I found in this related question.

Answer 2

edit/update: Swift 5.2

extension BinaryInteger {
    var isPerfectSquare: Bool {
        guard self >= .zero else { return false }
        var sum: Self = .zero
        var count: Self = .zero
        var squareRoot: Self = .zero
        while sum < self {
            count += 2
            sum += count
            squareRoot += 1
        }
        return squareRoot * squareRoot == self
    }
}

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Playground

4.isPerfectSquare     // true
7.isPerfectSquare     // false
9.isPerfectSquare     // true
(-9).isPerfectSquare  // false