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iOS Functions in Swift Adding Power to Functions Function Parameters

Martel Storm
Martel Storm
4,157 Points

Stumped. Can someone explain to me why this is not correct?

XXX Bummer! You need to assign the result of the function evaluation to the constant named result. Thanks in advance!

Shout out to the fellas who have already been answering my questions. You're all very helpful and are much appreciated.

functions.swift
// Enter your code below
func getRemainder(value a: Int, divisor b: Int) -> (Int){
    let value = a % b
    return value
}
let result = getRemainder(value: 10, divisor: 3)

2 Answers

David Papandrew
David Papandrew
8,386 Points

I think the code challenge is a little finicky. Remove the parens from your function signature return type and it will pass.

Specifically:

func getRemainder(value a: Int, divisor b: Int) -> Int {
    return a % b
}

let result = getRemainder(value: 10, divisor: 4)

Playgrounds has no problem with -> (Int) but the Treehouse code evaluator doesn't like it.

Good luck.

Martel Storm
Martel Storm
4,157 Points

Hey hey that worked wonderfully! Thank you