Detecting array type without template specialization?

[Bruno Medeiros]

Sean Kelly wrote:
> Jarrett Billingsley wrote:
>> Say I want a template to do different things if it's passed an array 
>> versus a non-array type.  I can do this easily with template 
>> specialization:
>>
>> template Spoon(T)
>> {
>>     void knife()
>>     {
>>         writefln("single");
>>     }
>> }
>>
>> template Spoon(T : T[])
>> {
>>     void knife()
>>     {
>>         writefln("array");
>>     }
>> }
>>
>> ...
>>
>> mixin Spoon!(int) i;
>> mixin Spoon!(int[]) a;
>>
>> i.knife();
>> a.knife();
>>
>>
>> That prints
>>
>> single
>> array
>>
>> As I'd like it to.
>>
>> That works great, but what if I'm writing a much larger class, where 
>> most of the functionality would be the same between the array and 
>> non-array versions, and I only want to change, say, one function to 
>> work differently for arrays?  It would be redundant and cumbersome to 
>> have to copy most of the contents of one template to the other.
>>
>> Something like this can be done for classes versus non-classes:
>>
>> template Spoon(T)
>> {
>>     void knife()
>>     {
>>         static if(is(T : Object))
>>             writefln("class");
>>         else
>>             writefln("non-class");
>>     }
>> }
>>
>> But is there any way to do this for array types? 
> 
> template isArrayType( T )       { const bool isArrayType = false; }
> template isArrayType( T : T[] ) { const bool isArrayType = true;  }
> 
> template Spoon( T )
> {
>     void knife()
>     {
>         static if( isArrayType( T ) )
>             writefln( "array" );
>         else
>             writefln( "not array" );
>     }
> }
> 
> This sort of thing is why I created std.traits in Ares.
> 
> 
> Sean

It is kinda odd that the 'is expression' doesn't feature arrays as one 
of it's testable archetypes (it has: typedef struct union class 
interface enum function delegate)

-- 
Bruno Medeiros - CS/E student
http://www.prowiki.org/wiki4d/wiki.cgi?BrunoMedeiros#D