On 16/04/12 19:13, Dave Abrahams wrote:
>
> on Mon Apr 16 2012, John Reid <j.reid-AT-mail.cryst.bbk.ac.uk> wrote:
>
>> On 15/04/12 03:23, Dave Abrahams wrote:
>>>
>>> You can't do this; don't even try. Each C++ class has to have a unique
>>> Python identity. If you just want to refer to the same class by a
>>> different name, you can of course:
>>>
>>
>>> BOOST_PYTHON_MODULE( _sandbox )
>>> {
>>> namespace bp = ::boost::python;
>>> object inner;
>>> {
>>> bp::scope scope = bp::class_< Outer1>( "Outer1" );
>>> inner = bp::class_< Inner>( "Inner" );
>>> }
>>>
>>> {
>>> object outer2 = bp::class_< Outer2>( "Outer2" );
>>> outer2.attr("Inner") = inner;
>>> }
>>> }
>>>
>>
>> I didn't know you could do that and it is useful but it is not quite
>> what I had in mind. I would rather not pass the inner object around
>> all the parts of my code that might need it. Is it possible to get it
>> out of the registry in the second block?
>
> Why don't you read to the end of my posting? The answer's there at the
> bottom :-)
>
Thanks! I was guilty of reading your post too quickly. And thx to Jim
too for what looks like a similar method.
John.
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