alias c=mixin(expr); disallowed, why?

[Artur Skawina]

On 06/23/13 13:23, Timon Gehr wrote:
> On 06/23/2013 12:19 PM, Artur Skawina wrote:
>> On 06/22/13 21:52, Timothee Cour wrote:
>>> Is there a reason the language spec disallows this?
>>>
>>> ----
>>> void main(){
>>>     auto a=mixin("1");//OK
>>>     alias b=a;//OK
>>>     mixin("alias c=a;");//OK
>>>     // alias c=mixin("a");//NG : Error: basic type expected, not mixin
>>> }
>>
>> How would that be different from "auto c=mixin("a");"?
>>
>> It's probably clear, but that error message is misleading, so i'll say
>> it anyway - the reason why your 'alias' line does not work is because
>> alias requires a symbol, but 'mixin()' is an expression.
>> Special-casing mixin-expressions (so that they propagate the symbol when
>> that is possible would be a bad idea); the other possibility is to allow
>> aliasing /expressions/. But that's a bad idea too, and would likely not
>> do what you expect it to do. A mixin-less version could be made to work,
>> but there are already other ways to get the same effect.
>> Hence the above question.
> 
> mixin template T(alias x){ }
> 
> void foo(){ import std.stdio; writeln("foo"); }
> 
> void main(){
>     auto a=mixin("1");
>     mixin T!a; // ok
>     mixin T!(mixin("a")); // ok
>     mixin T!1; // ok (!)
>     mixin T!(mixin("foo")); // ok (no implicit call)
> }

Yes, template parms and literals are special (literals are useful as template
parms, obviously).

What would be the arguments for allowing these two:

>     mixin T!(mixin("a")); // ok
>     mixin T!(mixin("foo")); // ok (no implicit call)

?
I didn't realize they were currently accepted. Is this actually
documented somewhere? The fact that `mixin("foo")` and `mixin("foo+1")`
mean subtly different things when used as a template parameter is a problem.

artur