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"

Answer 3

if you wrote:

let result = optionalValue ?? 50

and optionalValue != nil then result will be optional too and you will need unwrap it in future

But you can write operator

infix operator ??? { associativity left precedence 140 }

func ???<T>(optLeft:T?, right:T!) -> T!
{
    if let left = optLeft
    {
        return left
    }
    else { return right}
}

Now you can:

 let result = optionalValue ??? 50

And when optionalValue != nil then result will be unwraped

Answer 4

The following seems to work

extension Optional {
    func getOrElse<T>(defaultValue: T) -> T {
        if let value = self? {
            return value as T
        } else {
            return defaultValue
        }
    }
}

however the need to cast value as T is an ugly hack. Ideally, there should be a way to assert that T is the same as the type contained in the Optional. As it stands, type inferencing sets T based on the parameter given to getOrElse, and then fails at runtime if this does not match the Optional and the Optional is non-nil:

let x: Int?

let y = x.getOrElse(1.414) // y inferred as Double, assigned 1.414

let a: Int? = 5

let b: Double = a.getOrElse(3.14) // Runtime failure casting 5 to Double

Answer 5

If you are trying to do this with a String you can do this ..

string1 = string2.isEmpty ? "Default Value":string2