Why can't templates with default arguments be instantiated without the bang syntax?

[Simen Kjaeraas]

On Thu, 15 Sep 2011 17:54:19 +0200, Christophe  
<travert at phare.normalesup.org> wrote:

> "Simen Kjaeraas" , dans le message (digitalmars.D.learn:29539), a
>  écrit :
>> On Thu, 15 Sep 2011 16:46:24 +0200, Andrej Mitrovic
>> <andrej.mitrovich at gmail.com> wrote:
>>
>>> struct Foo(T = int) {}
>>>
>>> void main()
>>> {
>>>     Foo foo;  // fail
>>>     Foo!() bar;  // ok
>>> }
>>>
>>> It would be very convenient to be able to default to one type like  
>>> this.
>>>
>>> For example, in CairoD there's a Point structure which takes doubles
>>> as its storage type, and then there's PointInt that takes ints. The
>>> reason they're not both a template Point() that takes a type argument
>>> is because in most cases the user will use the Point structure with
>>> doubles, and only in rare cases Point with ints. So to simplify code
>>> one doesn't have to write Point!double in all of their code, but
>>> simply Point.
>>>
>>> If the bang syntax wasn't required in presence of default arguments
>>> then these workarounds wouldn't be needed.
>>
>> How would you then pass a single-argument template as a template alias
>> parameter?
>>
>> Example:
>>
>> template Foo( ) {
>>      template Bar( ) {
>>      }
>> }
>>
>> template Baz(alias A) {
>>      mixin A!();
>> }
>>
>> void main( ) {
>>      mixin Baz!Foo;
>> }
>>
>> Does this mixin Foo or Bar to main's scope?
>
> I don't get the problem. Maybe I am not used to mixin enough. Can you
> mixin normal templates, and not only mixin templates ?
>
> Anyway, why would this mixin Bar ?
> As I understand the proposition, only "mixin Baz!(Foo.Bar);" and of
> course "mixin Baz!(Foo!().Bar)" should mixin Bar.

Sorry, you're right. I meant:

template Foo( ) {
     template Foo( ) {
     }
}

-- 
   Simen