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. `b` is then dutifully modified in the function,
but there’s no mechanism for copying it back into `arr` when `foo` returns, so
the change isn’t reflected at the call site.
I *think* that the documentation you’re referring to in your last sentence is
talking about when the value is updated at the call site. That is, `arr` itself
isn’t updated until `foo` returns, but within `foo`, the argument `a` would
have whatever value you assign it, whenever you assign it.
I think.
HTH (also, I hope I’m right… the only thing worse that no info is bad info)
- Dave Sweeris
> On Jun 11, 2016, at 12:29 PM, Karl Pickett via swift-users
> <swift-users@swift.org> wrote:
>
> I don't believe the (spartan) docs address this case:
>
> func foo(inout a: [Int], inout b: Int) {
> let acopy = a
> a = [4, 5, 6]
> print(acopy) // prints "[1, 2, 3]"
> b = 99
> print(a) // prints "[4, 5, 6]"
> print(acopy) // prints "[1, 2, 99]" (e.g. a let variable changed!)
> }
>
> var arr = [1,2,3]
>
> foo(&arr, b: &arr[2])
>
> print(arr) // prints "[4, 5, 6]"
>
>
> Is this code "undefined", meaning the spec / doc doesn't specifically
> say what should be happening? For instance, I can't believe a let
> variable gets changed. The docs also say inout changes the original
> value when the function ends, not in the middle as is happening here.
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