On 12/11/22 05:54, Salih Dincer wrote:
> On Sunday, 11 December 2022 at 09:43:34 UTC, Andrey Zherikov wrote:
>> Note that callback parameter must be compile-time parameter as it
>> represents a member of a type during introspection done by `foo`.
>
> I can't quite understand the question, this already works:
I think the OP is trying to create an anonymous template. I don't think
it's possible.
The following works because the template has the name 'print':
// Named template:
static void print(int i)() {
import std;
writeln(i);
}
void main() {
foo!print;
}
The following does not work when one attempts to use the template
anonymously. Just trying to pass a template just like a lambda:
foo!((int i)() {
import std;
writeln(i);
});
Note how '(int i)()' is hoping to define a template.
Ali
