Eponymous/anonymous mixin templates

[Artur Skawina via Digitalmars-d]

On 06/08/15 17:14, Jeffrey Tsang via Digitalmars-d wrote:
> On Sunday, 7 June 2015 at 14:17:45 UTC, Artur Skawina wrote:
>> On 06/07/15 11:05, Jeffrey Tsang via Digitalmars-d wrote:
>>> I use a mixin template to define exactly one symbol, and at instantiation I wish to use that symbol immediately, once.
>>
>> AFAICT you're asking for the commented-out line in
>>
>>    auto Tmpl() = l;
>>
>>    void main(string[] argv) {
>>       auto l = argv.length;
>>       mixin Tmpl!() a;
>>       assert(a.Tmpl==l);
>>       //assert(a==l);
>>    }
>>
>> to work. That would probably be enough, make sense and have
>> no serious backward compat issues.
> 
> There are three separate things I would like to have:
> 
> 1. Eponymous trick
> 
> mixin T foo(T)() {
>     return bar;
> }
> 
> as pure syntactic sugar for
> 
> mixin template foo(T) {
>     T foo() {
>         return bar;
>     }
> }

This part already works, if you leave out the 'mixin' annotation
(which only prevents non-mixin use). "Normal" templates can be
mixed in too, see my `Tmpl` example above.

   T foo(T)() {
      return bar;
   }

   auto f(int bar) {
      mixin foo!double blah;
      return blah.foo();
   }

   void main() {
      assert (f(42)==42.0);
   }


> 2. Eponymous trick, calling end
> 
> mixin foo!T;
> 
> to also include as syntactic support
> 
> alias foo = foo!T.foo;
> 
> as well as the named version you listed.

No, that would hide the `foo` template symbol.
There's no such problem with the named version (other than
the eponymous look-up not working).


> 3. Inline/anonymous mixins
> 
> Some way of writing the equivalent of
> 
> bar = (mixin foo!T)() + 3;

This part I'm not sure about. Can't think of an interesting
use case, that wouldn't be better handled in some other way.
The obvious workaround would be:
      
   bar = { mixin foo!T f; return f()+3; }(); // today: `f.foo()+3`

[`{...}()` might result in a lambda/closure right now, but that
 should really be fixed (ie defined as a special case)]

artur