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iOS Closures in Swift 2 First Class Functions Functions as Data

Kevin Gutowski
Kevin Gutowski
4,082 Points
Posted by Kevin Gutowski
Kevin Gutowski
4,082 Points

Why didn't we specify what the optional parameter was for someFunction?

When we defined someFunction's optional parameter as String, why wasn't it defined as (aString: String) to match the original printString function?

let someFunction: String -> Void = printString
someFunction("Another string")

vs

let someFunction: (aString: String) -> Void = printString
someFunction(aString: "Another string")

1 Answer

Jeroen de Vrind
Jeroen de Vrind
29,772 Points
Jeroen de Vrind
Jeroen de Vrind
29,772 Points

If you declare the type of the constant 'someFunction' you only specify it's type: String -> Void. Your second example is not a type declaration and I'm sure the compiler will give an error. Here you've put an parameter as if you declare a concrete method. So things are mixed up.

The type of someFunction is String -> Void, so you can assign a method of this type to this constant. 'printString' is a method of this type:

func printString(aString: String)

is of type String -> Void. And when you have assigned this function to the constant someFunction, then you can pass a string as the value for the argument of the printString method as well to the someFunction variable.

someFunction has not an optional parameter String. There are no questionmarks. It's not a function, but a constant which can hold a function of type String->Void