Swift Closure why does calling function return error?

Question

just learning about closures and nesting functions. Given the nested function below:

func printerFunction() -> (Int) -> () {
    var runningTotal = 0
    func printInteger(number: Int) {
        runningTotal += 10
        println("The running total is: \(runningTotal)")
    }
    return printInteger
}

Why does calling the func itself have an error, but when I assign the func to a constant have no error? Where is printAndReturnIntegerFunc(2) passing the 2 Int as a parameter to have a return value?

printerFunction(2) // error
let printAndReturnIntegerFunc = printerFunction() 
printAndReturnIntegerFunc(2) // no error. where is this 2 going??

cannot invoke 'printerFunction' with an argument list of type '(Int)' - is the error message. I suppose i'm just basically confused why when I assign printerFunction() to a constant, I can pass variables through the constant but I have no idea where that variable actually gets used in the function itself. — Chris
– Chris, Commented Jun 25, 2015 at 5:31
The one-line equivalent of your last two lines is printerFunction()(2). — ahruss
– ahruss, Commented Jun 25, 2015 at 5:37

Dharmesh Kheni · Accepted Answer · 2015-06-25 05:34:47Z

6

First of all you are getting error here printerFunction(2) because printerFunction can not take any argument and If you want to give an argument then you can do it like:

func printerFunction(abc: Int) -> (Int) -> (){


}

And this will work fine:

printerFunction(2)

After that you are giving reference of that function to another variable like this:

let printAndReturnIntegerFunc = printerFunction()

which means the type of printAndReturnIntegerFunc is like this:

enter image description here

that means It accept one Int and it will return void so this will work:

printAndReturnIntegerFunc(2)

answered Jun 25, 2015 at 5:34

Dharmesh Kheni

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Comments

Eric Qian · Accepted Answer · 2015-06-25 05:33:36Z

5

(1) The function signature of printerFunction is () -> (Int) -> () which means it takes no parameter and returns another function, thats why when you try to call printerFunction(2) with a parameter gives you an Error.
(2) And the signature of the returned function is (Int) -> () which means it takes one parameter of Int and returns Void. So printAndReturnIntegerFunc(2) works

answered Jun 25, 2015 at 5:33

Eric Qian

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