On Friday, 11 October 2013 at 04:35:38 UTC, Ali Çehreli wrote:
On 10/10/2013 09:13 PM, Agustin wrote:
> I have a function that needs to check if the template
provided inherit a
> class.
>
> For example:
>
> public void function(T, A...)(auto ref A values)
function happens to be a keyword. :)
> {
> // static assert(IsBaseOf(L, T));
> }
>
> Check if T inherit class "L". Same result that
std::is_base_of<L,
> T>::value using C++. Any clean way to do it, without a dirty
hack.
One of the uses of the is expression determines "whether
implicitly convertible to". It may work for you:
public void foo(T, A...)(auto ref A values)
{
static assert(is (T : L));
}
class L
{}
class LL : L
{}
void main()
{
foo!LL(1, 2.3, "4");
}
Ali
On Friday, 11 October 2013 at 05:45:00 UTC, Jonathan M Davis
wrote:
On Thursday, October 10, 2013 21:35:37 Ali Çehreli wrote:
One of the uses of the is expression determines "whether
implicitly
convertible to". It may work for you:
public void foo(T, A...)(auto ref A values)
{
static assert(is (T : L));
}
Actually, checking for implicit conversion will work directly
in the template
signature without a template constraint or static assertion.
e.g.
public void foo(T : L, A...)auto ref A values)
{
}
- Jonathan M Davis
Those work great, thanks a lot!
