Why do closures require an explicit `self` when they're all non-escaping by default in Swift 3?

Question

I\'ve noticed that in Swift 2.2, closures marked as non-escaping with @noescape do not require an explicit self. In Swift 3, all closures are non-e

Answer 1

In Swift 3, all closures are non-escaping by default

No, in Swift 3, only closure function arguments (i.e function inputs that are functions themselves) are non-escaping by default (as per SE-0103). For example:

class A {

    let n = 5
    var bar : () -> Void = {}

    func foo(_ closure: () -> Void) {
        bar = closure // As closure is non-escaping, it is illegal to store it.
    }

    func baz() {
        foo {
            // no explict 'self.' required in order to capture n,
            // as foo's closure argument is non-escaping,
            // therefore n is guaranteed to only be captured for the lifetime of foo(_:)
            print(n)
        }
    }
}

As closure in the above example is non-escaping, it is prohibited from being stored or captured, thus limiting its lifetime to the lifetime of the function foo(_:). This therefore means that any values it captures are guaranteed to not remain captured after the function exits – meaning that you don’t need to worry about problems that can occur with capturing, such as retain cycles.

However, a closure stored property (such as bar in the above example) is by definition escaping (it would be nonsensical to mark it with @noescape) as its lifetime not limited to a given function – it (and therefore all its captured variables) will remain in memory as long as the given instance remains in memory. This can therefore easily lead to problems such as retain cycles, which is why you need to use an explicit self. in order to make the capturing semantics explicit.

In fact, case in point, your example code will create a retain cycle upon viewDidLoad() being called, as someClosure strongly captures self, and self strongly references someClosure, as it's a stored property.

It's worth noting that as an extension of the "stored function properties are always escaping" rule, functions stored in aggregates (i.e structures and enumerations with associated values) are also always escaping, as there's no restrictions on what you do with such aggregates. As pointed out by pandaren codemaster, this currently includes Optional – meaning that Optional<() -> Void> (aka. (() -> Void)?) is always escaping. The compiler might eventually make this a special case for function parameters though, given that optional is already built on a lot of compiler magic.

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Of course, one place where you would expect to be able to use the @noescape attribute is on a closure that’s a local variable in a function. Such a closure would have a predictable lifetime, as long as it isn’t stored outside of the function, or captured. For example:

class A {

    let n = 5

    func foo() {

        let f : @noescape () -> Void = {
            print(n)
        }

        f()
    }
}

Unfortunately, as @noescape is being removed in Swift 3, this won't be possible (What's interesting is that in Xcode 8 GM, it is possible, but yields a deprecation warning). As Jon Shier says, we’ll have to wait for it to be re-added to the language, which may or may not happen.

Answer 2

When you make a parameter with the @autoclosure attribute, the expression you pass as an argument is automatically wrapped into a closure for you. It makes your code cleaner. The @noescape keyword is more complex and interesting. It can be applied to a function parameter with any function type. The @noescape attribute indicates that a closure will be used inside a function body, before the function return is called. It means it won’t escape the function body.

Answer 3

Stored closures are considered escaping by default, even when they aren't really. There's no way to mark them non-escaping so we're stuck like this until they add @noescape back to the language, which they may or may not do. See this discussion on the swift-evolution mailing list.