if and switch expressions

Paul Hudson @twostraws

SE-0380 adds the ability for us to use if and switch as expressions in several situations. This produces syntax that will be a little surprising at first, but overall it does help reduce a little extra syntax in the language.

As a simple example, we could set a variable to either “Pass” or “Fail” depending on a condition like this:

let score = 800
let simpleResult = if score > 500 { "Pass" } else { "Fail" }
print(simpleResult)

Or we could use a switch expression to get a wider range of values like this:

let complexResult = switch score {
    case 0...300: "Fail"
    case 301...500: "Pass"
    case 501...800: "Merit"
    default: "Distinction"
}

print(complexResult)

You don’t need to assign the result somewhere in order to use this new expression syntax, and in fact it combines beautifully with SE-0255 from Swift 5.1 that allows us to omit the return keyword in single expression functions that return a value.

So, because both if and switch can now both be used as expressions, we can write a function like this one without using return in all four possible cases:

func rating(for score: Int) -> String {
    switch score {
    case 0...300: "Fail"
    case 301...500: "Pass"
    case 501...800: "Merit"
    default: "Distinction"
    }
}

print(rating(for: score))

You might be thinking this feature makes if work more like the ternary conditional operator, and you’d be at least partly right. For example, we could have written our simple if condition from earlier like this:

let ternaryResult = score > 500 ? "Pass" : "Fail"
print(ternaryResult)

However, the two are not identical, and there is one place in particular that might catch you out – you can see it in this code:

let customerRating = 4
let bonusMultiplier1 = customerRating > 3 ? 1.5 : 1
let bonusMultiplier2 = if customerRating > 3 { 1.5 } else { 1.0 }

Both those calculations produce a Double with the value of 1.5, but pay attention to the alternative value for each of them: for the ternary option I’ve written 1, and for the if expression I’ve written 1.0.

This is intentional: when using the ternary Swift checks the types of both values at the same time and so automatically considers 1 to be 1.0, whereas with the if expression the two options are type checked independently: if we use 1.5 for one case and 1 for the other then we’ll be sending back a Double and an Int, which isn’t allowed.

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