python - 类型错误 : No to_python (by-value) converter found for C++ type

Question

我正在尝试使用 Boost.Python 将我的 C++ 类公开给 Python。这是我正在尝试做的事情的简单版本:

struct Base {
    virtual ~Base() {};
    virtual char const *Hello() {
        printf("Base.Hello\n");
        return "Hello. I'm Base.";
    };
};

struct Derived : Base {
    char const *Hello() {
        printf("Derived.Hello\n");
        return "Hello. I'm Derived.";
    };

    Base &test() {
        printf("Derived.test\n");
        // ...
        // After some calculation, we get result reference `instance'
        // `instance' can be an instance of Base or Derived.
        // ...
        return instance;
    }
};

我想在 python 中使用上面的类如下:

instance = Derived()

// If method test returns an instance of Base
instance.test().Hello() // Result: "Hello. I'm Base."

// If method test returns an instance of Derived
instance.test().Hello() // Result: "Hello. I'm Derived."

我不知道这个问题有什么好的解决方案。我刚试过这个:

struct BaseWrapper : Base, wrapper<Base> {
    char const *Hello() {
        printf("BaseWrapper.Hello\n");
        if (override Hello = this->get_override("Hello")) {
            return Hello();
        }
        return Base::Hello();
    }

    char const *default_Hello() {
        printf("BaseWrapper.default_Hello\n");
        return this->Base::Hello();
    }
};

struct DerivedWrapper : Derived, wrapper<Derived> {
    char const *Hello() {
        printf("DerivedWrapper.Hello\n");
        if (override Hello = this->get_override("Hello")) {
            return Hello();
        }
        return Derived::Hello();
    }

    char const *default_Hello() {
        printf("DerivedWrapper.default_Hello\n");
        return this->Derived::Hello();
    }

    Base &test() {
        printf("DerivedWrapper.test\n");
        if (override Hello = this->get_override("test")) {
            return Hello();
        }
        return Derived::test();
    }

    Base &default_test() {
        printf("DerivedWrapper.default_test\n");
        return this->Derived::test();
    }
};

而他们，我使用以下代码:

BOOST_PYTHON_MODULE(Wrapper) {
    class_<BaseWrapper, boost::noncopyable>("Base")
                .def("Hello", &Base::Hello, &BaseWrapper::default_Hello);

    class_<DerivedWrapper, boost::noncopyable, bases<Base> >("Derived")
            .def("Hello", &Derived::Hello, &DerivedWrapper::default_Hello)
            .def("test", &Derived::test,  return_value_policy<copy_non_const_reference>());
}

但是当我将上面的代码编译成一个.so文件，并在python中使用时

derived = Wrapper.Derived() 
derived.test()

它抛出一个异常:

TypeError: No to_python (by-value) converter found for C++ type: Base

1. 这篇文章和我的有同样的错误，但是以不同的方式，它对我没有太大帮助。 Boost.Python call by reference : TypeError: No to_python (by-value) converter found for C++ type:

2. 这篇文章解决了类似的问题，但也没有帮助我。 https://github.com/BVLC/caffe/issues/3494

我有两个问题:

1. 如果我尝试的方式是正确的方式，如何解决TypeError问题？
2. 如果我尝试了错误的方法，那么使用 boost.python 解决问题的最佳方法是什么？

Answer 1

这段代码对我有用:

struct Base {
    virtual ~Base() {};
    virtual char const *hello() {
        return "Hello. I'm Base.";
    };
};

struct Derived : Base {
    char const *hello() {
        return "Hello. I'm Derived.";
    };

    Base &test(bool derived) {
        static Base b;
        static Derived d;
        if (derived) {
            return d;
        } else {
            return b;
        }
    }
};

BOOST_PYTHON_MODULE(wrapper)
{
    using namespace boost::python;
    class_<Base>("Base")
        .def("hello", &Base::hello)
        ;

    class_<Derived, bases<Base>>("Derived")
        .def("test", &Derived::test, return_internal_reference<>())
        ;
}

测试模块:

>>> import wrapper
>>> d = wrapper.Derived()
>>> d.test(True).hello()
"Hello. I'm Derived."
>>> d.test(False).hello()
"Hello. I'm Base."
>>>