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August 23, 2012
call member function alias
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.
August 24, 2012
Re: call member function alias
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
August 24, 2012
Re: call member function alias
On 08/23/2012 11:47 PM, Jacob Carlborg wrote:
>
> 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?
>

Nope :)

class Zoo: Foo
{
    override void bar(int a = 3) {
        writeln("Zoobar: ", a);
    }
}


auto zoo = new Zoo;
call!(Foo.bar)(zoo,4); // prints Zoobar: 4

And anyways, my two solutions composed together quite nicely.
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