Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

When did `guard let foo = foo` become legal?

Tags:

swift

guard

Back in November of 2016 I posted a question asking why I couldn't use guard to create an unwrapped version of a variable using the same name as the optional, like you can with if let:

Link: Why isn't guard let foo = foo valid?

When I wrote that question, the code below would fail to compile with an error that "Definition conflicts with previous value":

//Test of using guard to create an unwrapped version of a var, like if let
func guardTest(_ viewController: UIViewController?) -> UIViewController? {
  // Check if the current viewController exists
  print(String(describing: viewController))
  guard let viewController = viewController else {
    return nil
  }
  print(String(describing: viewController))

  return viewController
}

However, I just found some code at work that does this, and it now compiles without complaint and does what I want it to do! When run, the print statements show that foo is an optional before the guard, and an unwrapped optional after:

viewController = Optional(<TrochoidDemo.ViewController: 0x7ff16a039a00>)
viewController = <TrochoidDemo.ViewController: 0x7ff16a039a00>

(I added the test function guardTest(_:) to my latest open source project if you want to try it out. It's available on Github at https://github.com/DuncanMC/TrochoidDemo)

I'm happy that this construct now works as I want it to, but confused as to why it's now legal, and when the change occurred.

Is anybody aware of a recent change to the language definition that makes this construct work where it didn't before?

like image 671
Duncan C Avatar asked Jan 25 '17 14:01

Duncan C


1 Answers

TL;DR

guard let foo = foo is legal if foo was defined in another scope.


The example from your linked question:

func test()
{
  let a: Int? = 1

  guard let a = a else{
    return
  }
  print("a = \(a)")
}

still doesn't work because the guard statement is trying to create another variable a in the same scope.

This example:

//Test of using guard to create an unwrapped version of a var, like if let
func guardTest(_ viewController: UIViewController?) -> UIViewController? {
  // Check if the current viewController exists
  print(String(describing: viewController))
  guard let viewController = viewController else {
    return nil
  }
  print(String(describing: viewController))

  return viewController
}

works for the same reason that this does:

func test(a: Int)
{
    print(type(of: a))  // Int

    let a = 3.14

    print(type(of: a))  // Double
}

The parameter to the function is defined in a different scope, so Swift allows you to create a local variable with the same name.

like image 151
vacawama Avatar answered Nov 14 '22 18:11

vacawama