Reason for assigning optional to new variable in conditional statement in Swift

Question

I\'m going through the swift docs, and in the optional segment, it talks about using the question mark -- ? -- to signify variables that might be nil. This can

Answer 1

Take a look at the section on Optional Chaining in the docs. In the example you cite, there's not much difference. But in other cases, an if-let construction lets you get at an unwrapped value that comes from a series of optional references and method calls, without using implicit unwraps that can crash your app if you haven't considered all the possible bindings for a value in a chain.

It's also useful if you want to avoid recomputing a value. You can use it in a lot of the same ways you'd use an assignment in a conditional in (Obj)C (remember if (self = [super init])).

For example, if the optional being tested comes from a computed property:

var optionalName: String? {
get {
    if checkTouchID() {
        return "John Appleseed"
    } else {
        return nil
    }
}
}
var greeting = "Hello!"
if optionalName != nil {
    greeting = "Hello, \(optionalName)"
}

Paste that into a playground, along with a stub implementation of checkTouchID() that returns true, and you'll immediately see in the results area that the optionalName getter is executing twice. (This would be a problem in a more realistic scenario, because you probably don't want code like this to implicitly checkTouchID() or downloadFromServer() or billApplePay() twice.) If you use an if-let construction instead, you'll only execute the getter once.

In a series of chained optionals (like if let johnsStreet = john.residence?.address?.street in the docs linked above), you don't want to rewrite the whole chain in the body of the if statement, much less recompute it.