[swift-users] inout params seem to have undefined behavior

David Sweeris davesweeris at mac.com
Sat Jun 11 16:06:26 CDT 2016

> On Jun 11, 2016, at 3:36 PM, Jens Alfke <jens at mooseyard.com> wrote:
>> On Jun 11, 2016, at 11:57 AM, David Sweeris via swift-users <swift-users at swift.org <mailto:swift-users at swift.org>> wrote:
>> You can’t pass a `let` as an `inout` argument. I’d guess that’s what’s happening is the `arr[2]` part is creating a temporary var to which the `&` part then provides a reference.
> But `arr` is a var, not a let.

I know. You’d said that you "can't believe a let variable gets changed”. I was just pointing out that you’re correct, in that the compiler will complain if you try to pass one as an in-out argument.

>> `b` is then dutifully modified in the function, but there’s no mechanism for copying it back into `arr` when `foo` returns
> No, it gets copied back using subscript assignment. Remember, `inout` isn’t really passing the address of the parameter (although the optimizer may reduce it to that.) It’s literally in-and-out: the caller passes the original value, the function returns the new value, the caller then stores the new value where the old value came from.

I don’t think it can… My recollection is that in Swift the subscript operator (`arr[2]` in this case) can refer to the setter xor the getter, but not both within the same statement. If that’s correct, for there to be a value to pass to the function, `arr[2]` must be referring to the getter version, which means that there’s no setter to update the value when `foo` returns.

> I am not a Swift guru, but I think the problem in this example is that there’s a sort of race condition in that last post-return stage: the function has returned new values for both `arr` and arr[2]`, both of which get stored back where they came from, but the ordering is significant because arr[2] will have a different value depending on which of those assignments happens first.
> This smells like those C bugs where the result of an expression depends on the order in which subexpressions are evaluated — something like “x = i + (i++)”. The C standard formally declares this as undefined behavior.
> The part I’m still confused by is how `acopy` got modified within the `foo` function, since it’s declared as `let`. After staring at this for a while longer, I’m forced to conclude that the compiler decided it could optimize the `b` parameter by actually passing a pointer to the Int and modifying it directly, and that this has the side effect of modifying the Array object that `acopy` is pointing to, even though it’s supposed to be immutable.
> In other words, this looks like a compiler bug. I can reproduce it with Swift 2.2 (which is what my `swift` CLI tool says it is, even though I have Xcode 7.3.1 and I thought that was Swift 2.3?)

Ah… I see what you mean about a `let` getting modified now… My mistake, I thought you were wondering why `arr` wasn’t `[4, 5, 99]` after foo returned. Yeah, I’m not sure about what’s happening within `foo`... Maybe someone who knows more will come along and provide an explanation, but at the moment I’m inclined to agree — both that you’ve found a bug, and with your guess and to how it’s happening. 

- Dave Sweeris
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