Oded Padon wrote:
There is another reason why I think this has to be possible.
> It is possible using ctypes. Using ctypes with the same C++
> code written above (except for extern "C"), the following python code:
import ctypes;
lib_cpp = ctypes.CDLL('./test_cpp.so')
result = ctypes.c_int(0);
On 08/02/2010 01:46 PM, Oded Padon wrote:
import ctypes;
lib_cpp = ctypes.CDLL('./test_cpp.so')
result = ctypes.c_int(0);
lib_cpp.add(1, 2, ctypes.byref(result));
print result;
prints: "c_long(3)"
I do prefer however to use boost.python, as I would at a later stage want to
expose C++ classes t
Thank you for your rapid response.
I had a typo in my code. It should have been:
void add(int const a, int const b, int & c)
{
c = a + b;
}
i.e. the last argument is passed by non-const reference.
One solution is indeed to rewrite the function to accept either a
python object (class, dict, etc
Oded Padon wrote:
The C++ code is:
void add(int const a, int const b, int const& c)
{
c = a + b;
}
This doesn't even compile under g++ 4.0.1 (Mac):
error: assignment of read-only reference 'c'
I hope you're not using a compiler that accepts such code as legal?
I must emphasize that I w
Hello everybody,
I'm very new to Boost.Python, and I just spent hours trying to do something
trivial, and couldn't get it. The C++ code is:
void add(int const a, int const b, int const& c)
{
c = a + b;
}
BOOST_PYTHON_MODULE(hello_ext)
{
using namespace boost::python;
def("add", add)
Hi,
I've found bug in boost python with division operator. Division operator with
custom types doesn't work with python 3.1. It is because there were some
changes
in this regard in python 3. I've created ticket with patch
https://svn.boost.org/trac/boost/ticket/4497 . Can somebody look at i
