call member function alias

[Jacob Carlborg]

On 2012-08-23 21:51, Ellery Newcomer wrote:
> if I have a member function alias and corresponding object and
> arguments, is there any way to turn them into a member function call?
>
> e.g.
>
> class X{
>   void a();
> }
>
> auto profit(alias fn, T, Args...)(T t, Args args) {
>    ???
> }
>
> profit!(X.fn, X)(x);
>
> Constraints are:
>
> 1) must conserve ability to omit default arguments
> 2) if x is a subclass of X which overrides a, must not call overriden a.
>
> I have mutually exclusive solutions for (1) and (2).
>
> .. wait, nevermind. I can probably just wrap the two. It's an
> interesting problem, though, so I guess I'll post it.
>
> For 1) just parse out the parameter list from typeof(&fn).stringof and
> mix it in as profit's arg list, and then just mixin x.a(paramids), but
> that won't counter D's virtual functions
>
> For 2) hack together a delegate
> dg.ptr = x;
> dg.func_ptr = &fn;
>
> but delegates don't support default arguments.

How about this:

import std.stdio;

class Foo
{
     auto forward (alias fn, Args...) (Args args)
     {
         return fn(args);
     }

     void bar (int a = 3)
     {
         writeln("bar ", a);
     }
}

auto call (alias fn, T, Args...) (T t, Args args)
{
     t.forward!(fn)(args);
}

void main ()
{
     auto foo = new Foo;
     call!(Foo.bar)(foo);
     call!(Foo.bar)(foo, 4);
}

Prints:

bar 3
bar 4

Could this work for you?

-- 
/Jacob Carlborg