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

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")