In this checkpoint I have to create a function that finds the square root of a number and returns it without using sqrt() or similar methods. As per the hint in the course, I brute forced all the possible square roots with a for loop that checks if the input is equal to i * i and then breaks if it gets a match. But I can't figure out how to return the value of 'i' as an integer in the function. Would appreciate any help.

enum numberError: Error {
    case outOfBounds, noRoot
}

func checkSquareRoot(_ number: Int) throws -> Int {
    if number < 1 || number > 10_000 {
        throw numberError.outOfBounds
    }

    for i in 1...100 {
        if number == i * i {
            let root = i
            break
        } else {
            throw numberError.noRoot
        }
    }
    return root
    // How do I return the value of root / i ?
}

let integer = 25

do {
    let result = try checkSquareRoot(integer)
    print("Square root of \(integer) is \(result).")
} catch numberError.outOfBounds {
    print("\(integer) is out of bounds")
} catch numberError.noRoot {
    print("\(integer) is not a perfect square root.")
} catch {
    print("Error.")
}

Found a workaround! It's not perfect but the code runs correctly.

import Cocoa

enum numberError: Error {
    case outOfBounds, noRoot
}

func checkSquareRoot(_ number: Int) throws -> Int {
    if number < 1 || number > 10_000 {
        throw numberError.outOfBounds
    }

    var root = 0 // initializes a variable to store and return the result of the loop.

    for i in 1...100 {
        if number == i * i {
            root = i // assignes the value of i to root
            break
        }
    }

    if root == 0 { // if root hasn't received a new value, throws an error.
        throw numberError.noRoot
    }

    return root
}

let input = 100

do {
    let result = try checkSquareRoot(input)
    print("Square root of \(input) is \(result).")
} catch numberError.outOfBounds {
    print("\(input) is out of bounds")
} catch numberError.noRoot {
    print("\(input) is not a perfect square root.")
} catch {
    print("Error.")
}

The only problem I see is throwing the numberError.noRoot. Is there a more elegant way to do this? If I put it in the loop as an 'else' it would just throw an error after the first iteration if the square root isn't 1.

You can just return i from within your loop.

As for throwing noRoot, do that after running through the entire loop without a hit.

Wow! It was so easy. For some reason when I tried returning i inside the loop Xcode complained so I just gave up on it. Thanks for the help!

Wow thank you both!

for i in 1...100 {
        if number == i * i {
            return i
        }

the return i doesnt make the program out of the loop, it still doing the loop until reach 100. For me I prefer with break inside the loop so the program dont waste any more time.

Change the for loop in the above code to this:

    for i in 1...100 {
        print(i)
        if number == i * i {
            return i
        }
    }

and you will see that the loop only runs 10 times before hitting a match and causing the function to return.

return will cause the loop to exit the function early.