is(T : A!T) tells if T can automatically be converted to A!T.  The
last line below is doing just that, yet the template constraint does not work.

class A(T) if (is(T : A!T))
{
}

// if (is(T : A!T)) gives this error:
// Error: template instance x.A!(B) does not match
//        template declaration A(T) if (is(T : A!T))
//        while looking for match for A!(B)

class B : A!(B)
{
}

void main(string[] args)
{
    B b = new B();
    A!B a = b; // compiles fine
}

-Eric

On Thursday, 12 May 2016 at 15:33:24 UTC, Eric wrote:
>
> is(T : A!T) tells if T can automatically be converted to A!T.  The
> last line below is doing just that, yet the template constraint does not work.
>
> class A(T) if (is(T : A!T))
> {
> }
>

Yes, it's a bug. Please file an issue.

Meanwhile try this workaround:
class A(T)
{
  static assert(is(T : A!T), "...");
}

>
> Yes, it's a bug. Please file an issue.
>
> Meanwhile try this workaround:
> class A(T)
> {
>   static assert(is(T : A!T), "...");
> }

Bug report filed, and thanks for the workaround.

-Eric