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?
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.
来源:https://stackoverflow.com/questions/41854162/when-did-guard-let-foo-foo-become-legal