Re: Why type specialization is defined differently than is expression type specialization ?

[Timon Gehr]

On 06/27/2012 08:21 PM, Alex Rønne Petersen wrote:

On 27-06-2012 19:13, Timon Gehr wrote:

On 06/27/2012 03:56 PM, deadalnix wrote:

All is in the title.

I can do is(T : class) but not template(T : class) . Is this intended or
is it because it is ?

You cannot do is(T : class).

I think what's really worrying is that this is legal:

void foo(T : class)()
{
}

This is not legal.

But this is not:

template Foo(T : class)
{
}