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iOS Swift Functions and Optionals Optionals Exercise: isDivisible function

Posted by František Bakyta
František Bakyta
1,452 Points

How can I access the named input parameters "dividend" and "divisor" in the print out statement?

// isDivisible

func isDivisible(#dividend: Int, #divisor: Int) ->Bool? {
    if dividend % divisor == 0 {
        return true
    } else {
        return nil
    }
}

if let result = isDivisible(dividend: 15, divisor: 4) {
    println("\(dividend):\(divisor) is divisible.")
} else {
    println("\(dividend):\(divisor) is not divisible.")
}

1 Answer

Stone Preston
Stone Preston
42,016 Points
Stone Preston
Stone Preston
42,016 Points

How can I access the named input parameters "dividend" and "divisor" in the print out statement?

I dont think you can really accomplish that outside of the function body. if this println was occuring inside the function you could

you can assign them to constants before making the funtion call and pass the constants in as arugments:

func isDivisible(#dividend: Int, #divisor: Int) ->Bool? {
    if dividend % divisor == 0 {
        return true
    } else {
        return nil
    }
}

let dividend = 15
let divisor = 4
if let result = isDivisible(dividend: dividend, divisor: divisor) {
    println("\(dividend):\(divisor) is divisible.")
} else {
    println("\(dividend):\(divisor) is not divisible.")
}
František Bakyta
František Bakyta
1,452 Points

Great, thank you Stone for your quick answer and help. If I don't assign them to constants before the function call, there is no way of accessing them in the code like I tried above in my answer? What is the purpose of the "named input parameter" for "external use" if I cannot use it in the code for example for printing out the inputs in the function?

Stone Preston
Stone Preston
42,016 Points
Stone Preston
Stone Preston
42,016 Points

the external parameter name allows you to understand what the parameter actually is when you call the function since you have to provide the parameter name when calling a function with external parameters.

if you did not provide external parameters, the function call would look like this:

isDivisible(15, 4)

what is 15? what is 4? which one is the dividend? see the reason external parameters are helpful?

isDivisible(dividend: 15, divisor: 4)

in the above call, its clear what each argument is.

now, if you wanted to add the println inside the function you would have access the the parameters:

func isDivisible(#dividend: Int, #divisor: Int) ->Bool? {
    if dividend % divisor == 0 {
        println("\(dividend):\(divisor) is divisible.")
        return true
    } else {
        println("\(dividend):\(divisor) is not divisible.")
        return nil
    }
}
František Bakyta
František Bakyta
1,452 Points

Thank you very much Stone, now I fully understand what you tried to explain me :-)