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iOS

Jordan Gorrell
Jordan Gorrell
6,627 Points
Posted by Jordan Gorrell
Jordan Gorrell
6,627 Points

FizzBuzz Function Error

Hi everyone, I wanted to take the FizzBuzz exercise in Swift a step further and make it into a function where you can select the range of numbers that the are getting "FizzBuzzed." I'm having a problem with it. I'm getting an error that says:

"ERROR at line 25, col 11: missing argument label 'high:' in call fizzBuzzer(100,150) ^ high: "

Here is my code. I'd really like to know what is going wrong with it so any help is appreciated. I'm assuming it's a simple fix and I've just missed something. Thanks

let fizz = "Fizz"
let buzz = "Buzz"

func fizzBuzzer(low: Int, high: Int) {
    let counter = low...high

    for count in counter {
        switch count {
            case count where count % 3 == 0 && count % 5 == 0:
                print("\(count) makes \(fizz+buzz)")
            case count where count % 3 == 0:
                print("\(count) makes \(fizz)")
            case count where count % 5 == 0:
                print("\(count) makes \(buzz)")
            default:
                print("\(count)")
        }
    }
}

fizzBuzzer(100,150)

1 Answer

Jason Anders
MOD
Jason Anders
Treehouse Moderator 145,692 Points
Jason Anders
Jason Anders
Treehouse Moderator 145,692 Points

Hey Jordan,

You are missing the second external variable name in your function call. If you put 

fizzBuzzer(100, high: 150)

your code will execute with no errors. I could go into a very long explanation, but I would mostly be quoting the Apple Dev Docs. So, if you want, you can read up on functions and function calls at your leisure.

But, other than that... nice code. :)

:dizzy: