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iOS Functions in Swift Adding Power to Functions Function Parameters

Nigel Matheson
PLUS
Nigel Matheson
Courses Plus Student 1,166 Points
Posted by Nigel Matheson
Nigel Matheson
Courses Plus Student 1,166 Points

I don't know why this is not working

I believe I had got this right but it still doesn't work?

Help!

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


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

2 Answers

Jeff McDivitt
Jeff McDivitt
23,970 Points
Jeff McDivitt
Jeff McDivitt
23,970 Points

In your function the values are of type Int. You do not input numbers until you call the function

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

return a % b

}