3

I have a module modA, which contains a synthesized submodule modB (created with PyModule_New); now importing the module:

  1. from modA import modB it is OK
  2. import modA.modB fails.

What am I missing?

  • modA.cpp (using boost::python, but it would be very likely the same with pure c-API of python):

    #include<boost/python.hpp> namespace py=boost::python; BOOST_PYTHON_MODULE(modA){ py::object modB=py::object(py::handle<>(PyModule_New("modB"))); modB.attr("__file__")="<synthetic>"; py::scope().attr("modB")=modB; };

  • compile with (g++ instead of clang++ works the same)

    clang++ -o modA.so modA.cpp -fPIC -shared -lboost_python `pkg-config python --cflags --libs`

  • test.py:

    import sys sys.path.append('.') from modA import modB import modA.modB

  • python test.py (note the first import is just fine, it is the second one which fails):

    Traceback (most recent call last): File "test.py", line 4, in <module> import modA.modB ImportError: No module named modB
Improve this question
2
  • You should add your solution as an answer and accept it. Commented Jun 16, 2012 at 12:34
  • I just did that, thanks for suggestion. Have to wait 2 days to accept it :-) Commented Jun 16, 2012 at 12:39

3 Answers 3

Reset to default
1

Thanks to this answer, I found the solution, which consists in sys.modules['modA.modB']=modB, but written in c++ in the module initialization function:

#include<boost/python.hpp> namespace py=boost::python; BOOST_PYTHON_MODULE(modA){ py::object modB=py::object(py::handle<>(PyModule_New("modA.modB"))); modB.attr("__file__")="<synthetic>"; py::scope().attr("modB")=modB; // this was the missing piece: sys.modules['modA.modB']=modB py::extract<py::dict>(py::getattr(py::import("sys"),"modules"))()["modA.modB"]=modB; };
Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

1

Here is the pattern I use, hopefully it improves previous answers...

module.h:

... #define STR(str) #str #define MAKE_SUBMODULE(mod, submod) object submod##_module(handle<>(borrowed(PyImport_AddModule(STR(mod.submod)))));\ scope().attr(STR(submod)) = submod##_module;\ scope submod##_scope = submod##_module;\ ... export_submodule(); ...

module.cpp:

BOOST_PYTHON_MODULE(modulename) { export_submodule(); }

submodule.cpp:

export_submodule() { // create submodule MAKE_SUBMODULE(modulename, submodulename) // all code below is now in submodule's scope ... }

The "generated" code looks like this:

object submodulename_module(handle<>(borrowed(PyImport_AddModule("modulename.submodulename")))); scope().attr("submodulename") = submodulename_module; scope submodulename_scope = submodulename_module;

It looks similar to eudoxos answer, but the differencs seems to be in detail: Instead of PyImportNew() I use PyImportAdd(). Therefor I don't need the last line to get the from module.submodule import * statement working.

Improve this answer

Comments

0

modB is not a file (i.e. a module), but some object in modA, hence it cannot be imported.

You may want to see Python's modules docs.

Improve this answer

4 Comments

modB is a module, it just happens to be created on the fly.
I don't understand. Please specify - is modB a file or not?
No, modB is a module object which is being created at runtime.
imp.find_module only searches files, according to documentation. Sorry for confusion, though, I found the solution meanwhile.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.