---------
module a;

struct foo {}

deprecated alias bar = foo;

----------
module b;
struct bar {};


-----------
module c;

import a;
import b;

void baz(bar b) {}

Error: `a.bar` at source/a.d(5,1) conflicts with `b.bar` at .b.d(2,1)

I would have thought the undeprecated alias would have been used.

On Friday, May 11, 2018 14:02:22 Nicholas Wilson via Digitalmars-d-learn wrote:
> ---------
> module a;
>
> struct foo {}
>
> deprecated alias bar = foo;
>
> ----------
> module b;
> struct bar {};
>
>
> -----------
> module c;
>
> import a;
> import b;
>
> void baz(bar b) {}
>
> Error: `a.bar` at source/a.d(5,1) conflicts with `b.bar` at
> .b.d(2,1)
>
> I would have thought the undeprecated alias would have been used.

As I understand it, the only effect that deprecated really has is that you get a message printed telling you that the symbol is deprecated if it's used in undeprecated code. And that being the case, it definitely wouldn't affect something like overload resolution.

And if you think about it, making deprecated affecting things like overload resolution could get pretty risky. Remember that the whole idea here is that a symbol is deprecated so that folks will stop using it, but you don't want code to break because of it. If anything behaves differently, then it risks code breakage, and that's exactly what you were trying to avoid by deprecating the symbol rather than just removing it.

So, in general, when something is deprecated, we get to continue to live with the negatives of having it around - it's just that it discourages its use in new code and encourages people to stop using that symbol. So, when the symbol is removed, the only code that breaks should be unmaintained code that wasn't updated appropriately. But in the interim, it's still there causing symbol conflicts and whatnot, and I don't see how it could be otherwise without risking code breakage simply due to deprecation, and we really don't want that. It's disruptive enough as it is just to tell folks to stop using a symbol even if nothing breaks as long as the symbol is still around.

- Jonathan M Davis