From: David Abrahams (abrahams_at_[hidden])
Date: 2001-03-31 19:32:07
This has come up several times. There is a limitation in Boost.Python which
makes trying to do what you want dangerous at best. I posted a proposed
solution at http://groups.yahoo.com/group/boost/message/10144, but given the
deafening silence I got as a response, I just keep directing people at the
section titled "If you can't (afford to) copy the referent, or the pointer
is non-const" on this page:
I would happily make the proposed change to Boost.Python, but would like to
hear /something/ from users before undertaking any redesign.
P.S. we don't say "BPL" anymore... we at boost are afraid of running out of
----- Original Message -----
Sent: Saturday, March 31, 2001 5:20 PM
Subject: [boost] creating a BPL instance from an existing C++ object.
> I'm using VC6 SP4 on Win2K SP1, with Boost 1.21.1 & Python 2.0.
> I want to create a 'BPL extension instance' object from an existing
> C++ object.
> Let me start with a conventional use of the BPL that leads to my
> My extension module creates a BPL extension class based on a C++
> class (call it CB).
> The user instantiates it ...
> >>> B = CB();
> Now BPL code instantiates the associated C++ class and ties it to the
> new BPL extension instance object. Correct?
> To get to my problem, let's say that the C++ class CB contains
> another class CBa. Then CBa will automatically get instantiated when
> the BPL code instantiates CB.
> The CBa class, by the way, is wrapped, so there is an associated
> Python (BPL extension instance) type, but since CBa got instantiated
> from C++, there is no associated Python object.
> I want to provide Python access to the existing C++ instance of CBa,
> and I don't know how. I'm going to start looking around in the BPL
> source code in the hopes of figuring out how to do this, but if
> anyone knows offhand that it is or isn't possible, please let me know.
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