Procedure for building and importing a Boost.python module using GCC?

Question

I've managed to compile a Boost.Python 'first try' but am unsure how to import it into python and call the methods it contains. My source file is as follows:

#include <stdlib.h>
#include <string>
#include <boost/python.hpp>

using namespace boost::python;

int test(int i)
{
       fprintf(stderr, "%s:\n", __FUNCTION__);
       return i * 5;
}

BOOST_PYTHON_MODULE(ipg)
{
       using namespace boost::python;
       def("test", test);
}

My makefile contains:

# Which compiler?
CC=c++

# Which flags for object files?
OFLAGS=-c -Wall -fPIC

# Which flags for the output binary?
BFLAGS=-Wall -shared -o ipg

# Which flags for boost python?
BPFLAGS=-I/usr/include/python2.7
BLIBS=-lpython2.7 -lboost_python -lboost_system

# Make.
all: source/python.cpp
    $(CC) $(BOUT) $(BFLAGS) $(BPFLAGS) $? $(BLIBS)

and my test script:

import sys

# My modules.
import ipg

ipg.test()

The output binary is placed alongside the test script, and the test script is run. This results in the following error:

Traceback (most recent call last): File "test.py", line 4, in import ipg ImportError: No module named ipg

What flags should I be using to compile my output binary, and how should I go about importing it into python? I've used boost.Python on Windows before, but that was quite a while ago.

Nathan Binkert · Accepted Answer · 2012-12-13 22:23:45Z

6

On Linux, if your module is called ipg, then you need to be creating a file called ipg.so. Here's a simple makefile;

ipg.o:
    g++ -o ipg.o -c ipg.cc -Wall -fPIC -I/usr/include/python2.7
ipg.so: ipg.o
    g++ -shared -o ipg.so ipg.o -lpython2.7 -lboost_python -lboost_system

edited Dec 13, 2012 at 22:23

answered Dec 13, 2012 at 22:14

Nathan Binkert

9,2041 gold badge32 silver badges37 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Liam M Over a year ago

Wow, that was pretty simple, thanks. Could you explain to me why I need the .so on the end? Does Python perform some sort of behind-the-scenes magic beforehand, requiring the extension as a filtering mechanism?

Nathan Binkert Over a year ago

It's just simply part of the lookup mechanism. Depending on the architecture, the importer will look for foo.py (source), foo.pyc (bytecode compiled), foo.pyo (bytecode optimized), foo.pyd (windows extension), foo.so (unix extension), foo.dylib (mac extension). I'm not certain on the ordering.

Collectives™ on Stack Overflow

Procedure for building and importing a Boost.python module using GCC?

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related