How to Add Documentation to Enum Associated Values in Swift

How to document an enum case with parameters?

/// Various coffee types.
enum Coffee {

    /// A Cappuccino.
    /// - cream: Is true if cream is added.
    case cappuccino(cream: Bool)
}

It'll show up as a bullet point when you click on cappuccino but won't show up if you click on the cream parameter.. But it's good enough I guess..

How to test equality of Swift enums with associated values

Swift 4.1+

As @jedwidz has helpfully pointed out, from Swift 4.1 (due to SE-0185, Swift also supports synthesizing Equatable and Hashable for enums with associated values.

So if you're on Swift 4.1 or newer, the following will automatically synthesize the necessary methods such that XCTAssert(t1 == t2) works. The key is to add the Equatable protocol to your enum.

enum SimpleToken: Equatable {
    case Name(String)
    case Number(Int)
}
let t1 = SimpleToken.Number(123)
let t2 = SimpleToken.Number(123)

Before Swift 4.1

As others have noted, Swift doesn't synthesize the necessary equality operators automatically. Let me propose a cleaner (IMHO) implementation, though:

enum SimpleToken: Equatable {
    case Name(String)
    case Number(Int)
}

public func ==(lhs: SimpleToken, rhs: SimpleToken) -> Bool {
    switch (lhs, rhs) {
    case let (.Name(a),   .Name(b)),
         let (.Number(a), .Number(b)):
      return a == b
    default:
      return false
    }
}

It's far from ideal — there's a lot of repetition — but at least you don't need to do nested switches with if-statements inside.

Swift enum size when associated value is a reference type

From Type Layout: Single-Payload Enums:

If the data type's binary representation has extra inhabitants, that is, bit patterns with the size and alignment of the type but which do not form valid values of that type, they are used to represent the no-data cases, with extra inhabitants in order of ascending numeric value matching no-data cases in declaration order.

Your example with more cases:

enum Opt<T> {
    case a, b, c, d, e, f, g, h, i, j, k
    case l, m, n, o, p, q, r, s, t, u, v
    case some(T)
}

class Person {
    var name: String
    init(name: String) { self.name = name }
}

print(unsafeBitCast(Opt<Person>.a, to: UnsafeRawPointer.self))
// 0x0000000000000000

print(unsafeBitCast(Opt<Person>.b, to: UnsafeRawPointer.self))
// 0x0000000000000002

print(unsafeBitCast(Opt<Person>.v, to: UnsafeRawPointer.self))
// 0x000000000000002a

let p = Person(name: "Bob")
print(unsafeBitCast(Opt.some(p), to: UnsafeRawPointer.self))
// 0x00006030000435d0

Apparently, 0x0, 0x2, ..., 0x2a are invalid bit patterns for a pointer, and therefore used for the additional cases.

The precise algorithm seems to be undocumented, one probably would have to inspect the Swift compiler source code.

Swift: Unable to switch on enum associated value

The problem is that you’re passing an Int16 value to the switch. You’re setting entry.mood, an Int16, to the raw value 1, but the switch wants your Mood type. So you have a type mismatch.

You can solve it by turning the value into a Mood:

switch Mood(rawValue: entry.mood)! {

How to compare enum with associated values by ignoring its associated value in Swift?

Edit: As Etan points out, you can omit the (_) wildcard match to use this more cleanly:

let number = CardRank.Number(5)
if case .Number = number {
    // Is a number
} else {
    // Something else
}

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Unfortunately, I don't believe that there's an easier way than your switch approach in Swift 1.2.

In Swift 2, however, you can use the new if-case pattern match:

let number = CardRank.Number(5)
if case .Number(_) = number {
    // Is a number
} else {
    // Something else
}

If you're looking to avoid verbosity, you might consider adding an isNumber computed property to your enum that implements your switch statement.

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