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")
}


func test2() {
    print("2 before return")
    return
    print("2 after return")
}


test()
test2()

returns:

1 before return
2 before return
2 after return

I would expect that print("2 after return") would never be executed since it is after a return statement.

Is there something I am missing?

(tested with Swift 4 / 4.1 & Xcode 9.2 / Xcode 9.3 beta 2)

Answer 1

It is a tricky thing, Swift doesn't require semi-colons (they are optionally used), this makes Swift compiler automatically deduce whether is the next line should be a new line, or a completion for the old one. print() is a function that returns void. So the word return print("something"), is valid. So

return
print("Something")

could be deduced as return print("Something")

Your solution is to write

return;
print("Something")