Enums, Equatable, and exhaustiveness

Say you have this Swift enum:

enum Expression {
    case number(Double)
    case string(String)
}

And you want it to conform to Equatable. Since the enum has associated values, Equatable conformance must be added manually. So you implement the == function:

extension Expression: Equatable {
    static func ==(lhs: Expression, rhs: Expression)
        -> Bool {
        switch (lhs, rhs) {
        case let (.number(l), .number(r)): return l == r
        case let (.string(l), .string(r)): return l == r
        default: return false
        }
    }
}

You handle the two cases where the two arguments are the same case and add a default pattern to return false otherwise. This is straightforward, short, and correct:

Expression.number(1) == .number(1) // → true
Expression.number(1) == .string("a") // → false

Default case makes exhaustiveness check uneffective

However, your implementation has a serious flaw: if you ever add another case to your enum, the compiler won’t alert you that your == implementation is now incomplete. Let’s add a third case to the enum:

enum Expression {
    case number(Double)
    case string(String)
    case bool(Bool)
}

This is totally fine as far as the compiler is concerned. Your code will now return wrong results, though:

Expression.bool(true) == .bool(true) // → false!

The default clause in the switch statement make the compiler’s exhaustiveness checks uneffective. For this reason, it’s generally a good idea to not use default in switch statements if you can avoid it.

Don’t use default in switch statements if you can avoid it.

Quadratic explosion of patterns

The downside of not having a default case is, of course, more boilerplate to write. Here’s a version of == that switches over the three-case enum exhaustively:

extension Expression: Equatable {
    static func ==(lhs: Expression, rhs: Expression)
        -> Bool {
        switch (lhs, rhs) {
        case let (.number(l), .number(r)): return l == r
        case let (.string(l), .string(r)): return l == r
        case let (.bool(l), .bool(r)): return l == r
        case (.number, .string),
             (.number, .bool),
             (.string, .number),
             (.string, .bool),
             (.bool, .number),
             (.bool, .string): return false
        }
    }
}

Phew! This isn’t fun to write, and it would get even worse with more cases. The number of states the switch statement must distinguish grows quadratically with the number of cases in the enum.

From quadratic to linear growth

You can make this considerably more manageable with some intelligent application of the _ placeholder pattern. While we saw above that a single default clause is not enough, one pattern per case is. The last six patterns in the switch statement become three:

extension Expression: Equatable {
    static func ==(lhs: Expression, rhs: Expression)
        -> Bool {
        switch (lhs, rhs) {
        case let (.number(l), .number(r)): return l == r
        case let (.string(l), .string(r)): return l == r
        case let (.bool(l), .bool(r)): return l == r
        case (.number, _),
             (.string, _),
             (.bool, _): return false
        }
    }
}

Even better, each additional case in the enum only adds two more lines to the switch statement — it doesn’t grow quadratically anymore. And you retain the benefit from the compiler’s exhaustiveness checks: adding a new case will raise an error in ==.

Sourcery

If this is still too much boilerplate for you, take a look at Krzysztof Zabłocki’s code generation tool Sourcery. Among many other things, it can auto-generate Equatable conformances for enums and other types (and keep them up to date).