SOLVED: Help with Checkpoint 4

@mliapich

Nov '21

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.")
}

5

@mliapich

Nov '21

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.

3

@roosterboy HWS+

Nov '21

You can just return i from within your loop.

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

2

@NigelGee HWS+

Nov '21

Instead of break return i.

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

and delete return root

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 {
            return i
        }
    }

    throw numberError.noRoot
}

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.")
}

6

@mliapich

Nov '21

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!

3

@michaelpgalen

Dec '21

Wow thank you both!

2

@Rooster_Swift

Dec '21

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.

2

@roosterboy HWS+

Dec '21

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.

playground loop output

return will cause the loop to exit the function early.

4

	@mliapich Nov '21 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.") } 5
	@mliapich Nov '21 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. 3
	@roosterboy HWS+ Nov '21 You can just return `i` from within your loop. As for throwing `noRoot`, do that after running through the entire loop without a hit. 2
	@NigelGee HWS+ Nov '21 Instead of break `return i`. `for i in 1...100 { if number == i * i { return i } }` and delete `return root` 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 { return i } } throw numberError.noRoot } 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.") } 6
	@mliapich Nov '21 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! 3
	@michaelpgalen Dec '21 Wow thank you both! 2
	@Rooster_Swift Dec '21 `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. 2
	@roosterboy HWS+ Dec '21 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. 4

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