Swift Closures: Must Capture Lists Be Exhaustive? - swift

Question

Suppose I have a Swift class like this:
#objc final MyClass : NSObject
{
let classPropertyString = "A class property"
func doStuff()
{
let localString = "An object local to this function"
DispatchQueue.global(qos: .userInitiated).async { [classPropertyString] in
// Do things with 'classPropertyString' and 'localString'
}
}
}
My question is: when I write a capture list, am I responsible for EXHAUSTIVELY listing all the things to which I want the closure to hold a strong reference?
In other words, if I omit localString from my capture list (as I've done here), will the closure still automatically capture a strong reference to it or am I in for a bad time?

Answer 1

There are several minor quirks with your question that make it tricky to answer clearly, but I think I understand the underlying concern, and the short answer is "no." But your example is impossible, so the answer is "it's impossible." And if it were possible, there'd be no strong reference (nor would there be a need for one), so the question still would be kind of "it's impossible." Even so, let's walk through what's going on here.
First, closure can't reference localString unless it's reassigned somehow in the comment inside doStuff(). closure is assigned at a level where localString is not in scope. Closures can only capture variables that are in scope when they are assigned, not when they're called. But let's go back to the original version of this question, before it was edited. That version did have the case you're describing:
#objc final myClass : NSObject
{
let classPropertyString = "A class property"
func doStuff()
{
let localString = "An object local to this function"
DispatchQueue.global(qos: .userInitiated).async { [classPropertyString] in // (1)
// Do things with 'classPropertyString' and 'localString'
}
// (2)
}
}
There's no problems here. classPropertyString is copied into the closure, avoiding any retain loops. localString is referenced by the closure, and so it's preserved as long as the closure exists.
Because you listed classPropertyString in the capture list, it is evaluated at point (1) and copied into the closure. Because you implicitly captured localString, it is treated as a reference. See Capture Lists in the Swift Programming Language Reference for some excellent examples of exactly how this works in different cases.
In no case (*) will Swift allow the underlying storage for something you're using in a closure to disappear behind your back. That's why the typical concern is excessive retains (memory leaks) rather than dangling references (crashes).
(*) "In no case" here is a lie. There are several ways that Swift will allow it, but almost all of them involve "Unsafe" which is your warning about that. The major exception is unowned, and of course anything involving ! types. And Swift is not typically thread-safe, so you need to be careful about that...
The last comment about thread-safety is a place where the subtle distinctions between implicit and explicit captures can really matter. Consider this case where you modify an implicitly captured value on two queues:
func doStuff() -> String
{
var localString = "An object local to this function"
DispatchQueue.global(qos: .userInitiated).async {
localString = "something else"
callFunction(localString)
}
localString = "even more changes"
return localString
}
What happens in that case? Good grief, never do that. I believe it's undefined behavior and that localString could be anything including corrupted memory, at least in the most general case (it might be defined behavior for calling .async; I'm not sure). But don't do it.
But for your normal cases, there is no reason to explicitly capture local variables. (I sometimes wish Swift had gone the C++ way and said it was required, but it isn't.)
Ok, one more way implicit and explicit are different that might drive home how they work. Consider a stateful closure like this (I build these pretty often):
func incrementor() -> () -> Int {
var n = 0
return {
n += 1
return n
}
}
let inc = incrementor()
inc() // 1
inc() // 2
inc() // 3
let inc2 = incrementor()
inc2() // 1
See how the local variable n is captured by the closure, and can be modified after it goes out of scope. And see how inc2 has its own version of that local variable. Now try that with explicit capture.
func incrementor() -> () -> Int {
var n = 0
return { [n] in // <---- add [n]
n += 1 // Left side of mutating operator isn't mutable: 'n' is an immutable capture
return n
}
}
Explicit captures are copies and they're immutable. Implicit captures are references, and so have the same mutability as the thing they reference.