Unwrapping Swift optional without variable reassignment

Question

When using optional binding to unwrap a single method call (or optional chaining for a long method call chain), the syntax is clear and understandable:

if le         


        

Answer 1

No. You should unwrap your optionals just redefining it with the same name as you mentioned. This way you don't need to create a second var.

func someFunction(childTitle: String?) {
    if let childTitle = childTitle {
        ...
    }
}

update: Xcode 7.1.1 • Swift 2.1

You can also use guard as follow:

func someFunction(childTitle: String?) {
    guard let childTitle = childTitle else {
        return
    }

    // childTitle it is not nil after the guard statement
    print(childTitle)
}

Answer 2

Here's the only alternative I'm aware of.

func someFunction(childTitle: String?) {
    if childTitle != nil {
        ...
    }
}

But then childTitle is still an optional, so you would have to unwrap it every time you use it: childTitle!.doSomething(). It shouldn't affect performance, though.