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January 13, 2013
Aliasing lambdas
Is there any reason why this isn't allowed:

alias f = (x => x);

but this is:

template A(alias B) { alias A = B; }
alias f = A!(x => x);

???
January 13, 2013
Re: Aliasing lambdas
On 01/13/2013 08:55 PM, Peter Alexander wrote:
> Is there any reason why this isn't allowed:
>
> alias f = (x => x);
>
> but this is:
>
> template A(alias B) { alias A = B; }
> alias f = A!(x => x);
>
> ???

No. It is a gratuitous grammar limitation.
January 13, 2013
Re: Aliasing lambdas
On Sun, Jan 13, 2013 at 9:07 PM, Timon Gehr <timon.gehr@gmx.ch> wrote:
> On 01/13/2013 08:55 PM, Peter Alexander wrote:
>>
>> Is there any reason why this isn't allowed:
>>
>> alias f = (x => x);
>>
>> but this is:
>>
>> template A(alias B) { alias A = B; }
>> alias f = A!(x => x);
>>
>> ???
>
>
> No. It is a gratuitous grammar limitation.

The same limitation forbids:

alias __traits(whatever, ...) result;

The grammar says:

AliasDeclaration:
   alias BasicType Declarator

(and others, but this is the one limiting you). BasicType should be
replaced by PrimaryExrpession, or something like that.
January 13, 2013
Re: Aliasing lambdas
On 01/13/2013 09:38 PM, Philippe Sigaud wrote:
> On Sun, Jan 13, 2013 at 9:07 PM, Timon Gehr <timon.gehr@gmx.ch> wrote:
>> On 01/13/2013 08:55 PM, Peter Alexander wrote:
>>>
>>> Is there any reason why this isn't allowed:
>>>
>>> alias f = (x => x);
>>>
>>> but this is:
>>>
>>> template A(alias B) { alias A = B; }
>>> alias f = A!(x => x);
>>>
>>> ???
>>
>>
>> No. It is a gratuitous grammar limitation.
>
> The same limitation forbids:
>
> alias __traits(whatever, ...) result;
>
> The grammar says:
>
> AliasDeclaration:
>      alias BasicType Declarator
>
> (and others, but this is the one limiting you). BasicType should be
> replaced by PrimaryExrpession, or something like that.
>

I think it should be roughly:

alias TemplateArgument Identifier;
alias Identifier=TemplateArgument;


Note: The current grammar allows

alias int x = 2;

Which does not make any sense.
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