Challenge:

The challenge is this: write a function that accepts an integer from 1 through 10,000, and returns the integer square root of that number. That sounds easy, but there are some catches:

You can’t use Swift’s built-in sqrt() function or similar – you need to find the square root yourself. If the number is less than 1 or greater than 10,000 you should throw an “out of bounds” error. You should only consider integer square roots – don’t worry about the square root of 3 being 1.732, for example. If you can’t find the square root, throw a “no root” error. As a reminder, if you have number X, the square root of X will be another number that, when multiplied by itself, gives X. So, the square root of 9 is 3, because 3x3 is 9, and the square root of 25 is 5, because 5x5 is 25.

import Cocoa

enum possibleErrors: Error{
    case outOfBounds, notPerfectSquare
}
var numberToCheck = 3
func squareRootOf(_ number: Int)throws{
    var value = 0
    if number < 1 || number > 10000{
        throw possibleErrors.outOfBounds
    }
    for _ in 1...100{
        value += 1
        if value * value == number{
            print("The square root of \(number) is \(value).")
            return
        }
    }
    throw possibleErrors.notPerfectSquare
}

do{
    try squareRootOf(numberToCheck)
}
catch possibleErrors.outOfBounds{
    print("The number has to be more than 0 and less than 10,000.")
}
catch possibleErrors.notPerfectSquare{
    print("The number has to be a perfect square.")
}

For this:

var value = 0
for _ in 1...100{
    value += 1
    if value * value == number{
        print("The square root of \(number) is \(value).")
        return
    }
}

you can just do this and save yourself having to declare that extra variable:

for value in 1...100 {
    if value * value == number {
        print("The square root of \(number) is \(value).")
        return
    }
}