Can't pass private symbols as alias parameter

[deadalnix via Digitalmars-d]

On Monday, 17 November 2014 at 14:13:21 UTC, Steven Schveighoffer
wrote:
> I note here, the above doesn't is incorrect regardless, I 
> assume you didn't actually try to compile this post example :)
>
> To fix, I did this:
>
> struct A(alias foo) {
>   auto bar() {
>      return foo();
>   }
> }
>

Sorry for that, I made a mistake while typing the sample code.

> The alias cannot override the protection attributes, and I'm 
> pretty sure that is intentional.
>
> I tried some workarounds:
>
> void main() {
>   public alias b = bar;
>   auto a = A!b();
> }
>
> It would be nice if that worked, because it does not expose bar 
> except in this one case, and it identifies that you know bar 
> has become public for this one case. But unfortunately, it has 
> the same issue.
>

:'(

> This works:
>
> void main() {
>   static auto b() { return bar();}
>   auto a = A!b();
> }
>
> But of course, it's not ideal, as you are relying on the 
> inliner to make this performant. Shortened version (I think 
> this implies static):
>
> auto a = A!(()=>bar())();
>
> Yuck. I really think what you wish should be allowed, even if 
> using some syntax to declare you know what you are doing.
>
> I tried some other goofy stuff, but I could not get around the 
> requirement that the alias not change the protection of the 
> symbol.
>
> -Steve

This is not an option to me. In my specific use case, the code
instantiating the template is itself a template, and they call
each other. That mean I get infinite template instantiation.