Swift：如何将函数类型值设置为可选变量？

Question

I want to use function as a completion for another function, when it's needed. But I have a trouble that best described with code:

...{
   var myCompletion : (() -> Void)? = nil
   if condition {
       myCompletion = myCompletionFunction() // <==========problem here
   }
   someMethodThatCouldHaveCompletion(myCompletion)
   ...
}

func someMethodThatCouldHaveCompletion(completion: (() -> Void)?){
...
}

func myCompletionFunction() {
...
}

I have an error: Cannot assign a value of type '()' to a value of type '(()->Void)?'

I tried to add as! (() -> Void)? as! (() -> Void)? after setting function to myCompletion variable, but receive another error Cannot downcast from '()' to a more optional type '(()->Void)?'

So, is there a way to solve it? I have a sense that I miss something with completion: (() -> Void)? type.

Answer 1

The type of myCompletion is fine, the problem is here:

myCompletion = myCompletionFunction()

You aren't assigning myCompletion to myCompletionFunction ; you're calling myCompletionFunction and assigning it's return value to myCompletion , causing the type mismatch.

Change that line to:

myCompletion = myCompletionFunction