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 test2() is similar to func test2() -> Void

So your code gets treated as,

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

Adding semicolon after print should fix it.

func test2() -> Void {
  print("2 before return")
  return; print("2 after return")
}

You can also see the error if you place value type after return line and you will understand more,

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

error: unexpected non-void return value in void function 2 ^