[C++-sig] wrapping typedef equivalent types

[Neal Becker]

David Abrahams wrote:

> Neal Becker <ndbecker2 at gmail.com> writes:
> 
>> David Abrahams wrote:
>>
>>> Neal Becker <ndbecker2 at gmail.com> writes:
>>> 
>>>> What happens if, using boost::python, I try to wrap 2 type equivalent
>>>> classes?
>>>>
>>>> For example (this is a trivial example):
>>>>
>>>> class_<int>...
>>>> class_<int32_t>...
>>>> where int == int32_t
>>>>
>>>> My real objective is to wrap classes
>>>>
>>>> template<int_t> class X {};
>>>>
>>>> where int_t is
>>>> int (natural int size)
>>>> int32_t (force 32-bit)
>>>> int64_t (force 64-bit)
>>>>
>>>> Can wrapped X<int>, X<int32_t>, X<int64_t> co-exist in the same python
>>>> module?
>>> 
>>> No, all wrapped classes have to have different types.  You could add a
>>> dummy parameter to X in order to distinguish them.
>>> 
>>
>> I don't understand the dummy parameter suggestion.  Could you elaborate?
> 
> template <class int_t, int bits> class X {};
> 
>   X<int, 16>
>   X<int32_t, 32>
>   X<int64_t, 64>
> 
> or
> 
>   template <int bits>
>   class X
>   {
>      typedef boost::int_t<bits>::fast int_t;
>      ...
>   };
>   
>> I found a different solution.  I just expose the int32_t and int64_t
>> versions, then use python code to set the equivalence.  In python, I
>> write something like:
>>
>> if (intsize() == 4): X = X32
>>
>> I guess there's nothing wrong with that approach :)
> 
> Not unless X32 and X64 are trying to be wrappers for the same
> underlying C++ class.
> 

Maybe I didn't explain this very well.  What I want in python is

Xint
Xint32
Xint64

Where:
Xint32 -> X<int32_t>
Xint64 -> X<int64_t>
Xint -> X<int>

So that Xint is type equivalent to Xint32 or Xint64, depending on the
platform.