early return from a void-func in Swift?

Question

Can someone explain me the following behaviour in Swift?

func test() -> Bool {
    print(\"1 before return\")
    return false
    print(\"1 after return\         


        

Answer 1

func noReturn() {...}

is the same as

func noReturn() -> (Void) {...} //or func noReturn() -> ()

and since print(...) has the same signature it is ok to call return print(...) in a void function