The is expression

[enuhtac]

Am 02.04.2011 04:00, schrieb Caligo:
> On Fri, Apr 1, 2011 at 5:14 PM, enuhtac <enuhtac_lists at gmx.de> wrote:
>> Hello,
>>
>> the "is" expression is a great feature of D - but its use is not very
>> intuitive, at least for me.
>> I'm trying to write a template that figures out if the template
>> parameter is of a given type.
>> This is the type I would like to check for:
>>
>> struct A( T, string s )
>> { ... };
>>
>> One possibility to accomplish this check is explicit template
>> specialization:
>>
>> template isA( T )
>> {
>>    enum bool isA = false;
>> };
>>
>> template isA( T : A!( U, s ), U, string s )
>> {
>>    enum bool isA = true;
>> };
>>
>> This more or less the C++ approach. But in D this could also be done
>> with static if and the "is" expression. As I understand "is" it should
>> be done like this:
>>
>> template isA( T )
>> {
>>    static if( is( T U == A!( U, s ), string s ) )
>>        enum bool isA = true;
>>    else
>>        enum bool isA = false;
>> };
>>
>> But this does not work. So what am I doing wrong?
>>
>> Regards,
>> enuhtac
>>
>>
> I'm new too, but I think it should be like this:
>
> template isA( T ){
>
>   enum bool isA = is(T : A)
> }
>
> if the name of enum is same as the template then you could use it as such:
>
> if( isA( T ) ){ }
>
> instead of
>
> if( isA( T ).isA ){ }
>
> Also note that : allows implicit conversion, while == requires the
> types to be exactly the same.
Your right. In your example it is possible to circumvent the "static if"
construct. I was not aware of this, so thanks for the hint.
But in my case this does not work as I'm using the following "is" form:

*is (* /Type/ /Identifier/ *==* /TypeSpecialization/ *,*
/TemplateParameterList/ *)

*Obviously "is" forms that include an "Identifier" can only be used in
"static if" constructs. Actually I could do without the "Identifier" but
I need the "TemplateParameterList". But there is no "is" form where you
get a "TemplateParameterList" without an "Identifier".
Actually that would be nice as this would look like the explicit
specialization:

template isA( T )
{
   static if( is( T == A!( U, s ), U, string s ) )
       enum bool isA = true;
   else
       enum bool isA = false;
};

Or with your simplification:

template isA( T )
{
   enum bool isA = is( T == A!( U, s ), U, string s );
};


But this is not possible as the "Identifier" is missing.

So still I do not know what I'm doing wrong in my original code.
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