Member function pointers

[Jacob Carlborg]

On 2013-06-10 14:36, Manu wrote:

> You supply 'this' at the time of calling. Read my OP.

Yes, exactly. It needs "this". It's the same thing as a delegate. It's 
just that with the delegate the context pointer and function pointer is 
combined. A function pointer (member or free) is useless if it's not called.

> It's Don's work of art. It's also how I came to find out about D in the
> first place ;)

I see.

> There's no way to specify to use the 'thiscall' calling convention.
> What I propose is a syntax that would describe a member function
> pointer, and imply the appropriate calling convention.

Right. I suggested a different syntax, but you want to reserve that 
syntax for something that match a library implementation, that's not 
even in the standard.

It's like saying I want int[] to match MyIntArray implemented in C++.

> funcptr pretends to be typed, but the type is just wrong. In your
> example, the type is 'void function()', it should be 'void function(Foo
> this)'.

"void function()" is part of the complete type. It becomes complete with 
the context pointer.

> So it's actually a lie. You can't call it. I'm not sure why it's typed
> at all... just a crash waiting to happen.

You can put a free function in a delegate and call it through the 
delegate. I guess you don't want that to be possible either.

> So what I'm suggesting is a syntax to express a member function pointer,
> and then it could be properly typed.

I don't think there's anything wrong with supporting C++ member function 
pointers but I don't think we need to add new syntax for it.

-- 
/Jacob Carlborg