[C++-sig] void* questions\thoughts

[Niall Douglas]

On 31 Oct 2005 at 15:27, Roman Yakovenko wrote:

> Hi. There is nice explanation in boost.python FAQs how to export
> function that takes\returns
> void*. I have 2 questions:
> 
> 1. Why void_ should be defined in each translation unit? If I understand right
>    "translation  unit" is cpp file?

void_ is a placeholder type for void as BPL doesn't support the void 
type itself, but does support opaque types. Hence the unfortunate 
proxy use of void_ * for void *.

> If there is no special reason to define void_ in every translation
> unit, then next question:
> 
> 2. Why boost.python does not define void_ class and does not registrate it?
>     BOOST_PYTHON_OPAQUE_SPECIALIZED_TYPE_ID( void_ )

I would personally wonder why BPL doesn't automatically treat void as 
an opaque type, and thus you can use void *. A huge amount of C uses 
void *, and a good proportion of C++ so lacking it makes producing 
bindings for such C and C++ trickier than it needs to be - most 
importantly in wrapping virtual functions, where a void * is passed 
in and must be passed out as-is to the C++ default implementation.

Dave, what was the rationale for not supporting void as an opaque 
type?

Cheers,
Niall