Calling Python functions from C++ with namespace

Question

I'm trying to execute Python code from c++ that will define a Python function and pass it back to c++ so it can be called back from there. That works fine but the problem is I can't provide the Python function with the namespace it had when it was originally defined.

struct MyClass {
    void log(const std::string & s)
    {
        cout << s << endl;
    }
    void callFnct(PyObject * fnct)
    {
        bp::call<void>(fnct);
        bp::call<void>(fnct);
    }
};

bp::class_<MyClass, boost::noncopyable> plugin("Plugin", bp::no_init);
plugin.def("callFnct", &MyClass::callFnct);

std::unique_ptr<MyClass> cls(new MyClass());

bp::object main_module = bp::import("__main__");
bp::object main_namespace = main_module.attr("__dict__");
bp::dict locals;
locals["plugin"] = bp::object(bp::ptr(cls.get()));

std::string scriptSource = 
                       "a=5\n"
                       "def my_func():\n"
                       "    a+=1\n"
                       "    plugin.log('won't work %d' % a)\n"
                       "plugin.log('this works')\n"
                       "plugin.callFnct(my_func)";
bp::object obj = bp::exec(bp::str(scriptSource), main_namespace, locals);

The initial call to plugin.log() works but once we call the python function in callFnct(), the namespace is gone so it can't see the variable a or the plugin module.

Does anyone know how to do bp::call<void>(fnct) by preserving the namespace and keep the variable a in scope?

StenSoft · Accepted Answer · 2015-02-06 18:37:00Z

4

That is because variables in non-local scopes cannot be rebound. It won't work even without calling to C++:

a = 5
def my_func():
    a += 5
    print(a)
my_func()

UnboundLocalError: local variable 'a' referenced before assignment

You need to import it first:

a = 5
def my_func():
    global a
    a += 5
    print(a)
my_func()

answered Feb 6, 2015 at 18:37

StenSoft

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