A func returning a func in Swift

Question

This gives me an error in the playground.

func returnAFunc() -> ()
{
    func f(){ println("hello") }

    return f
}

Now I read that as a func named "returnAFunc" which returns another func which doesn't return a value. Correct? But it doesn't work. I have to do this:

func returnAFunc() -> () -> ()     // or this (() -> ())
{
    func f(){ println("hello") }

    return f
}

Ok that doesn't seem right. Can someone explain?

Answer 1

Lets break down your definition:

func returnAFunc() -> ()

• func keyword
• returnAFunc function name
• () parameters expected (void/none)
• -> read as "returns"
• () void return type

So this function takes no parameters, and returns nothing.

The working version breaks down like this:

func returnAFunc() -> () -> ()

• func keyword
• returnAFunc function name
• () parameters expected (void/none)
• -> read as "returns"
• () -> () return type is another function specification:
  • () returned function takes no parameters
  • -> read as "returns"
  • () returned function returns void

So this function takes no parameters, and returns (a function which takes no parameters and returns nothing). (Eesh, even in English I feel the need to add extra parentheses to that...)

Answer 2

The type signature of your first returnAFunc is () -> () , which means a function that takes no arguments and returns nothing. The type for your second function is () -> () -> () , which means a function that takes no arguments and returns a function with type () -> () (ie a function that takes no arguments and returns nothing).

I think your basic misunderstanding is that () represents a function type. Instead, it's the same thing as Void. Once you understand this, the only other potential "gotcha" is to remember that the arrow is right-associative.

EDIT: Since I got the checkmark, I think it's worth calling attention to IMSoP's answer , which does a great job of explaining how to read the type in more detail.

Answer 3

() is an empty tuple. Swift uses it as the return type of a void function, so

  () -> ()

Is the signature of a function that takes no parameters and returns nothing (is void)

So,

func myFunc() -> ()

declares a function that returns nothing (is void), and

func myFunc() -> () -> ()

declares a function that returns a function, which takes no arguments and returns nothing.

Answer 4

Your first example reads as "Function returnAFunc takes no arguments, and returns no values." Your second example correctly defines the return type as "Function returnAFunc takes no arguments, and returns a function that takes no arguments and returns no arguments."

The syntax is a bit wonky, to be sure, but once you get used to it, it's pretty straightforward to see what's going on.

In addition, the Swift type void is represented as " () ", so you could also interpret this as "Function returnAFunc takes a void argument (which means no argument) and returns a void type (which means no argument)." I'm not sure that's actually what's happening behind the scenes, though.