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iOS Functions in Swift Adding Power to Functions Function Parameters

what is wrong??

?

functions.swift
// Enter your code below
func getRemainder (a value: Int , b divisor: Int) -> (Int) {
    return value % divisor
}
getRemainder(a: 10, b: 4)

1 Answer

John Marley
John Marley
8,740 Points

Hi there,

The issue is that the internal and external parameters need to be switched over. I was stuck on this for ages - the passing code reads:

// Switching over value with a and divisor with b in the brackets
// Also switching out value and divisor in the body of the function
// with a and b (the now local parameters)

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

getRemainder(value: 10, divisor: 4)

Using 'a' and 'b' inside the function is fine as we know what they represent, in calling the function though we'd rather it say "value" and "divisor" as then we know what values we're entering and why.