> On Jun 11, 2016, at 11:57 AM, David Sweeris via swift-users
> <swift-users@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.
> `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 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?)
—Jens
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