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 let childTitle = theItem.getChildItem()?.getTitle() {
    ...
}

But, when provided the variable as a parameter, I find myself either using:

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

or even just redefining it with the same name:

if let childTitle = childTitle { ... }

And I've started wondering if there is a shortcut or more efficient of performing a nil check for the sole purpose of using an existing variable. I've imagined something like:

if let childTitle { ... }

Does something like this exist, or at least an alternative to my above two interim solutions?

Answer 1

Edit/update: Xcode 14 • Swift 5.7

swift-evolution proposal SE-0345 if let shorthand for shadowing an existing optional variable:

if let childTitle {

}

---

guard let childTitle else {
    return
}

---

Original answer

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)
}