iOS Functions in Swift Adding Power to Functions Function Parameters

Gerald Ford
Gerald Ford
6,834 Points

Make sure your function accepts two parameters, with the correct external and local names

Not working please some help

functions.swift
// Enter your code below 
func getRemainder(a value: Int, b divisor: Int) -> Int {
let operation = a % b
return operation
}
Gerald Ford
Gerald Ford
6,834 Points

//Have tried // Enter your code below func getRemainder(a value: 10 , b divisor: 5) -> 2 { let operation = 10 % 5 return operation } //Then this // Enter your code below func getRemainder(a value: Int, b divisor: Int) -> Int{ let operation = value % divisor return operation } //Before this // Enter your code below func getRemainder(a value: Int, b divisor: Int) -> Int{ let operation = a % b return operation }

2 Answers

James Vlok
James Vlok
14,535 Points

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

a and b are the external names for the variable. value and divisor are the names for the variable in the function. eg.. when you call the function you would do the following:

getRemainder(a: 5, b: 10)

hence a and b here are used externally from the function

In the function: let operation = value % divisor

value and divisor are used.

So in the function you will use value instead of a and divisor instead of b.

value and divisor are just names for the same variable to make the code more readable.

Gerald Ford
Gerald Ford
6,834 Points

// not working func getRemainder(a value: Int, b divisor: Int) -> Int { let operation = value % divisor return operation }

James Vlok
James Vlok
14,535 Points

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

This should work. the question states that the external variable names need to be value and divisor. In the function definition just swap value and a , divisor and b

Sorry didn't see that it was apart of a code challenge.

All the best