Providing a default value for an Optional in Swift?

Question

The idiom for dealing with optionals in Swift seems excessively verbose, if all you want to do is provide a default value in the case where it's nil:

if let value = optionalValue {     // do something with 'value' } else {     // do the same thing with your default value } 

which involves needlessly duplicating code, or

var unwrappedValue if let value = optionalValue {     unwrappedValue = value } else {     unwrappedValue = defaultValue } 

which requires unwrappedValue not be a constant.

Scala's Option monad (which is basically the same idea as Swift's Optional) has the method getOrElse for this purpose:

val myValue = optionalValue.getOrElse(defaultValue) 

Am I missing something? Does Swift have a compact way of doing that already? Or, failing that, is it possible to define getOrElse in an extension for Optional?

Answer 1

Update

Apple has now added a coalescing operator:

var unwrappedValue = optionalValue ?? defaultValue 

---

The ternary operator is your friend in this case

var unwrappedValue = optionalValue ? optionalValue! : defaultValue 

You could also provide your own extension for the Optional enum:

extension Optional {     func or(defaultValue: T) -> T {         switch(self) {             case .None:                 return defaultValue             case .Some(let value):                 return value         }     } } 

Then you can just do:

optionalValue.or(defaultValue) 

However, I recommend sticking to the ternary operator as other developers will understand that much more quickly without having to investigate the or method

Note: I started a module to add common helpers like this or on Optional to swift.

Answer 2

As of Aug 2014 Swift has coalescing operator (??) that allows that. For example, for an optional String myOptional you could write:

result = myOptional ?? "n/a"