is expression on templated types

[Simen Kjaeraas]

On Sat, 25 Jun 2011 15:57:03 +0200, simendsjo <simen.endsjo at pandavre.com>  
wrote:

> I have a templated struct, and I'd like to check if a given type is this  
> struct. I have no idea how I should write the is expression..
>
> struct S(int C, int R, T) {
>      T[C][R] data;
> }
> template isS(T) {
>      enum isS = is(T : S); // How can I see if T is a kind of S no  
> matter what values S is parameterized on?
> }
> unittest {
>      static assert(isS!(S!(1,1,float)));
>      static assert(!isS!(float[1][1]));
> }

template isS(T) {
     static if ( is( T t == S!(C,R,U), int C, int R, U ) ) {
         enum isS = true;
     } else {
         enum isS = false;
     }
}

This works, but I thought the following had been added. Apparently not:

template isS(T) {
     static if ( is( T t == S!U, U... ) ) {
         enum isS = true;
     } else {
         enum isS = false;
     }
}

As for the weird way to write this, it is caused by some variants of the
isExpression not being available outside static if.

-- 
   Simen