How to add an optional string extension?

Question

You can create a String extension like so:

extension String {
   func someFunc -> Bool { ... }
}

but what if you want it to apply to optional string?

var optionalString :String? = ""
optionalString!.someFunc() /* String? does not have a member someFunc */

Answer 1

In Swift 3.1 you can add an extension to optional values as well:

extension Optional where Wrapped == String {
  var isBlank: Bool {
    return self?.isBlank ?? true
  }
}

Answer 2

You can do it like this:

protocol OptionalType { typealias A; var opt: A? { get } }
extension Optional: OptionalType { var opt: A? { return self } }

protocol StringType { var get: String { get } }
extension String: StringType { var get: String { return self } }

extension Optional where Wrapped: StringType {
  func getOrElse(s: String) -> String {
    return self.opt?.get ?? s
  }
}

And:

let optStr: String? = nil
optStr.getOrElse("hello world")

The reason that you cannot constrain Optional or String for that matter, is because they are struct. By making pseudo-protocol for each, now we can constrain as we like.

I feel like swift has given up a lot of things just to make it easier for beginners to learn or maybe the language hasn't matured enough yet.

Answer 3

Extensions on Optional that return a String

As of Swift 3, you cannot directly constrain an extension method to an optional String. You can achieve the equivalent result with protocols as Daniel Shin's answer explains.

You can however, create an extension method on an Optional of any type and I've found some useful methods that have a String return value. These extensions are helpful for logging values to the console. I've used asStringOrEmpty() on a String optional when I want to replace a possible nil with empty string.

extension Optional {
    func asStringOrEmpty() -> String {
        switch self {
            case .some(let value):
                return String(describing: value)
            case _:
                return ""
        }
    }

    func asStringOrNilText() -> String {
        switch self {
            case .some(let value):
                return String(describing: value)
            case _:
                return "(nil)"
        }
    }
}

Example Use:

var booleanValue: Bool?
var stringValue: String?
var intValue: Int?

print("booleanValue: \(booleanValue.asStringOrNilText())")
print("stringValue: \(stringValue.asStringOrNilText())")
print("intValue: \(intValue.asStringOrNilText())")

booleanValue = true
stringValue = "text!"
intValue = 41

print("booleanValue: \(booleanValue.asStringOrNilText())")
print("stringValue: \(stringValue.asStringOrNilText())")
print("intValue: \(intValue.asStringOrNilText())")

Console Output:

booleanValue: (nil)
stringValue: (nil)
intValue: (nil)

booleanValue: true
stringValue: text!
intValue: 41

 

Optional different than nil pointer

These extensions illustrate that an Optional is different that a nil pointer. An Optional is a enum of a specified type (Wrapped) which indicates that it does or does not contain a value. You can write an extension on the Optional "container" even though it may not contain a value.

Excerpt from Swift Optional Declaration

enum Optional<Wrapped> : ExpressibleByNilLiteral {

    /// The absence of a value.
    case none

    /// The presence of a value, stored as `Wrapped`.
    case some(Wrapped)

    ...
}

In code, the absence of a value is typically written using the nil literal rather than the explicit .none enumeration case.

Answer 4

extension Optional where Wrapped == String {
var isNil: Bool {
    return self == nil
}

The above answer(written by @Vlad Hatko) works fine but in swift 4 there are some issues, so I changed it to this.

Answer 5

In Swift 4.1 I was getting a Optional is ambiguous for type lookup in this context build error. To fix, you have to explicitly add the Swift namespace to the type:

extension Swift.Optional where Wrapped == String {
    var isBlank: Bool {
        return self?.isBlank ?? true
    }
}

Answer 6

Update: For a workaround that works with Swift 2 and above, see Daniel Shin’s answer

---

An optional String isn't in and of itself a type, and so you cannot create an extension on an optional type. In Swift, an Optional is just an enum (plus a bit of syntactic sugar) which can either be None, or Some that wraps a value. To use your String method, you need to unwrap your optionalString. You can easily use optional chaining to achieve this:

optionalString?.someFunc()

If optionalString is not nil, someFunc will be called on it. An alternative (less concise) way of doing this is to use optional binding to establish whether or not optionalString has a value before trying to call the method:

if let string = optionalString {
    string.someFunc()    // `string` is now of type `String` (not `String?`)
}

---

In your example from the comments below, you needn't nest multiple if statements, you can check if the optional string is an empty string in a single if:

if optionalString?.isEmpty == true {
    doSomething()
}

This works because the expression optionalString?.isEmpty returns an optional Bool (i.e. true, false or nil). So doSomething() will only be called if optionalString is not nil, and if that string is empty.

Another alternative would be:

if let string = optionalString where string.isEmpty {
    doSomethingWithEmptyString(string)
}

Answer 7

Since Xcode 9.3, you can use this slight modification of @Vladyslav's answer:

extension Optional where Wrapped == String {

    var isEmpty: Bool {
        return self?.isEmpty ?? true
    }

}

Answer 8

found some trick swift 3

class A{
    var name:String!;
    init(_ name:String?){
        self.name = name;
    }
}

extension Optional where Wrapped == String {
    func compareText(_ other:String?)->Bool{
        switch (self,other){
        case let(a?,b?):
            return a < b;
        case (nil,_):
            return true;
        default:
            return false;
        }
    }
}

let words:[A] = [A("a"),A(nil),A("b"),A("c"),A(nil)];

// let sorted = words.sorted{ 0.name.compareText($1.name) }
// trick
let sorted = words.sorted{ ($0.name as String?).compareText($1.name) }

print(sorted.map{$0.name});