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Why switch does not accept case nil in optional?

Time:06-25

I was working with optional in switch which I noticed switch cannot accept nil as a case!

func test(value: Bool?) {
    switch value {
    case true: print("true")
    case false: print("false")
    case nil: print("nil")
    }
}

Error: Switch must be exhaustive

then I used default for solving the issue, but I am looking to learn why nil cannot work in first code?

CodePudding user response:

I think this is just an incorrect diagnostic. You can add a default case, and this code will work, but I'm actually surprised that works.

You see, the value you're switching on is a Bool?, but 2 of the values you're comparing it against are true and false. The typical approach for this is to use pattern matching:

func test(value: Bool?) {
    switch value {
    case .some(true): print("true")
    case .some(false): print("false")
    case .none: print("nil")
    }
}

This works correctly, but it's a bit wordy. Luckily, Swift has a syntactic sugar for pattern matching against optionals, ? and nil:

func test(value: Bool?) {
    switch value {
    case true?: print("true")
    case false?: print("false")
    case nil: print("nil")
    }
}

Both of these snippets compile and work fine. It does open a new question as to why your original cases worked, if only you added the default.

I suggest you post about this on Swift forums, to get more input from the compiler nerds :)

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