Boost logo

Boost :

From: David Abrahams (david.abrahams_at_[hidden])
Date: 2001-06-15 19:53:07

----- Original Message -----
From: "John Max Skaller" <skaller_at_[hidden]>

> David Abrahams wrote:
> > C++ functions which are exposed to Python get an argument conversion
> > which is based on the argument types (using template type deduction).
So, if
> > there is a conversion from the passed Python type to the C++ argument
> > the function call succeeds with that conversion. If you want your C++
> > function to see a "raw" python type, there are several options:
> >
> > 1. Use a PyObject* argument. Its type conversion is the identity
> > 2. Use a boost::python::ref, which does the reference-counting part for
> > 3. Use a specific python type wrapper (e.g. boost::python::tuple) from
> > boost/python/objects.hpp.
> I see. Thanks! I'd never considered something
> other that (1). So I guess there is a predefined mapping
> from Python types to C++ types?

For some types, yes. The user can add additional mappings, and of course
classes that the user exposes to Python also get automatically mapped.


Boost list run by bdawes at, gregod at, cpdaniel at, john at