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iOS Functions in Swift Adding Power to Functions Function Parameters

Code Challenge: Function with argument labels not executing. What am I missing?

In this task we're going to write a simple function that takes two numbers and returns the remainder of dividing one number by the other.

Step 1: Declare a function named getRemainder that takes two parameters, aand b, both of type Int, and returns the value, also of type Int, obtained by carrying out the operation a modulo b. In case you've forgotten, the modulo operator is also called the remainder operator.

Step 2: The local names of the parameters are convenient but they make it hard to figure out the meaning of the function when we call it. Add two external names - value, for the first parameter and divisor for the second.

functions.swift
func getRemainder(a value: Int, b divisor: Int) -> Int {
  let aModuloB = a % b

  return aModuloB
}

You almost got it,

func getRemainder(a value: Int, b divisor: Int) -> Int {
  let aModuloB = value % divisor

  return aModuloB
}

Actually now that I'm reading this it may be this solution, I haven't taken the course so I was just assuming,

func getRemainder(value a: Int, divisor b: Int) -> Int {
    let aModuloB = a % b

    return aModuloB
}

Either way you are just mixing up your local and external params. Here 'value' and 'divisor' are just external names to help you understand what is going into the function, you don't actually use them in the method.

2 Answers

Jeff McDivitt
Jeff McDivitt
23,970 Points

This is the cleanest way to write this function

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

let result = getRemainder(value: 10, divisor: 3)

Hey man looks like you already got it though there is a cleaner way to write this :)

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