如何在Pybind11/Nanobind中返回C++类对象并在Python调用其方法

需求概述 #

需要实现：C++中类B的bar()方法返回类A的对象，在Python中调用该返回对象的foo()方法。以下是三种主流绑定框架的实现方案，均无需使用lambda表达式。

Pybind11 实现 #

C++绑定代码 #

#include <pybind11/pybind11.h>
#include <pybind11/print.h>

namespace py = pybind11;

class A {
public:
    void foo() {
        py::print("Hello world");
    }
};

class B {
public:
    A bar() {
        A a;
        return a; // 直接返回栈上的A对象即可
    }
};

PYBIND11_MODULE(example, m) {
    // 绑定类A
    py::class_<A>(m, "A")
        .def(py::init<>())
        .def("foo", &A::foo);

    // 绑定类B，无需lambda，直接绑定bar方法
    py::class_<B>(m, "B")
        .def(py::init<>())
        .def("bar", &B::bar);
}

Python调用代码 #

import example

b_object = example.B()
a = b_object.bar()
a.foo()  # 输出: Hello world

Nanobind 实现 #

Nanobind语法与Pybind11高度相似，实现逻辑一致：

C++绑定代码 #

#include <nanobind/nanobind.h>
#include <nanobind/print.h>

namespace nb = nanobind;

class A {
public:
    void foo() {
        nb::print("Hello world");
    }
};

class B {
public:
    A bar() {
        A a;
        return a;
    }
};

NB_MODULE(example, m) {
    nb::class_<A>(m, "A")
        .def(nb::init<>())
        .def("foo", &A::foo);

    nb::class_<B>(m, "B")
        .def(nb::init<>())
        .def("bar", &B::bar);
}

Python调用代码 #

import example

b_object = example.B()
a = b_object.bar()
a.foo()  # 输出: Hello world

Boost.Python 实现 #

C++绑定代码 #

#include <boost/python.hpp>
#include <iostream>

namespace bp = boost::python;

class A {
public:
    void foo() {
        std::cout << "Hello world" << std::endl;
    }
};

class B {
public:
    A bar() {
        A a;
        return a;
    }
};

BOOST_PYTHON_MODULE(example) {
    bp::class_<A>("A")
        .def(bp::init<>())
        .def("foo", &A::foo);

    bp::class_<B>("B")
        .def(bp::init<>())
        .def("bar", &B::bar);
}

Python调用代码 #

import example

b_object = example.B()
a = b_object.bar()
a.foo()  # 输出: Hello world

内容的提问来源于stack exchange，提问作者Tommy Wolfheart