Re: Declaring overloads (and template functions).

Roman Yakovenko wrote:
> On 7/26/07, Rene Rivera <[email protected]> wrote:
>> ===C++
>> class_<A>()
>>      .def("f", overload<void(int)>(&A::f))
>> ===
>>
>> Though, I'm not sure if one can make such an "overload<>()" function
>> smart enough to automatically turn the bare function type into the
>> corresponding member function pointer type.
> 
> I could be wrong but you can check this with boost::function library.
> http://boost.org/doc/html/function/tutorial.html#id1186900 .
> If that library can do this, than you can do it too.

That is equivalent to using is_member_function_pointer 
<http://boost.org/doc/html/boost_typetraits/reference.html#boost_typetraits.is_member_function_pointer>. 
The problems with those is that we don't have a member function pointer 
yet. To be clear the function might be declared as:

template <typename FunctionType, typename FunctionPointerType>
FunctionPointerType overload( FunctionPointerType p )
{
     return p;
}

The hard part is in calculating the FunctionPointerType such that 
FunctionType doesn't have include the "A::*" member pointer 
qualification. I personally dont't see a way of doing it :-(


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