python - C++ 嵌入式解释器和对象

Question

I have a simple C++ program that starts an embedded python intepreter, imports a module and instantiates a class defined in that module.

I want to understand why the address of the python object (from the python point of view) and the address of the C++ object are different.

How are the python instance and the c++ view of that instance different memory addresses?

Here's some working code, using pybind11 :

#include <iostream>
#include <pybind11/pybind11.h>
#include <pybind11/embed.h>

using namespace std;
namespace py = pybind11;

int main()
{
    {
        py::scoped_interpreter guard{};
        py::module m = py::module::import("code");
        py::object o = m.attr("SomeClass")();
        cout << "[C++   ] object lives in " << &o << endl;
    }
    return 0;
}

def message_from_python(*args):
    print('[PYTHON]', *args)


class SomeClass:
    def __init__(self):
        message_from_python(self, 'being created')

    def __del__(self):
        message_from_python(self, 'being deleted')

Compilation, execution and stdout:

$ g++ -O3 -Wall -std=c++14 `python3 -m pybind11 --includes` code.cc -o code -lpython3.6m
$ ./code
[PYTHON] <code.SomeClass object at 0x7fb6926f1da0> being created
[C++   ] object lives in 0x7fffbed02588                   
[PYTHON] <code.SomeClass object at 0x7fb6926f1da0> being deleted

Answer 1

If we take a quick look at what a py::object is made of in its declaration here , you can see that the class holds a PyObject * as a protected member (inherited from py::handle ). This pointer contains the address that you see printed out from the python side. If you could access it, you would see that

cout << o.mptr << endl;

produces the same address that the python shell indicates.