TypeError：没有为C++ Typ找到的TyPython（ByValm）转换器

Question

我试图将我的C++类暴露给PythonBoost.Python. 下面是我要做的事情的简化版本：

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：

^{pr2}$

我不知道解决这个问题的好办法。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();
    }
};

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>());
}

但当我把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()

它抛出了一个例外：

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

1. 这篇文章和我的错误是一样的，但不同的是，它对我帮助不大。 Boost.Python call by reference : TypeError: No to_python (by-value) converter found for C++ type:

2. 这篇文章解决了一个类似的问题，但也没有帮助我。 https://github.com/BVLC/caffe/issues/3494

我有两个问题：

1. 如果我尝试的方法是正确的，如何解决打字错误的问题？在
2. 如果我尝试了一个错误的方法，那么什么是解决问题的最佳方法boost.python？在

Answer 1

这个代码适用于我：

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：

^{pr2}$