Boost Python - Global and Local dictionary

问题

I'm using BoostPython for embedding Python in my C++ project but I don't understand all stuffs about Python, especially the namespace system.

Actually, I used this code:

byte_code = Py_CompileString(filedata, filename, Py_file_input);

// [...]

PyObject* res = NULL;

PyObject* main_module = PyImport_AddModule("__main__");
PyObject* global_dict = PyModule_GetDict(main_module);
PyObject* local_dict = PyDict_New();
py::object local_namespace(py::handle<>(py::borrowed(local_dict)));

// Set a user object (only for this execution)
local_namespace["user_object"] = py::ptr(&MyObject);

res = PyEval_EvalCode( byte_code, global_dict, local_dict );

Py_XDECREF(res);
Py_XDECREF(local_dict);

But When I execute a python script like:

def testB():
    print("B")

def testA():
    testB() # NameError: global name 'testB' is not defined

testA() # Works
testB() # Works

Okay I could used

res = PyEval_EvalCode( byte_code, global_dict, global_dict );

instead of

res = PyEval_EvalCode( byte_code, global_dict, local_dict );

But I want preserve the global_dict from any new function definition (Because when I will launch a new script, I don't want that previous function definition from a very old execution can be called !)

This is a problem about namespace, isn't it ?

回答1:

Yes, this is a namespace problem. You should look into the way Python handles scope (specifically the "self" keyword).

In short, class member variables and functions are prefixed with "self." in order to specify that they are a member of the current object's scope. "Self" is loosely similar to the "this" keyword in C++ and C#.

来源：https://stackoverflow.com/questions/19772058/boost-python-global-and-local-dictionary