is expression on templated types

[simendsjo]

On 25.06.2011 16:32, Simen Kjaeraas wrote:
> 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.
>

Thanks - works like a charm.
The is expression is quite a complex beast though.
Has anyone written any articles on it, or some more examples than the 
documentation?
Think I need many examples before I grok it.