Boost.Python call by reference:TypeError: No to_python (by-v

问题描述

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化版本:

I'm trying to expose my C++ Classes to Python using Boost.Python. Here is a simplyfied version of what i'm trying to do:

我有一个从 boost::noncopyable 派生的类 A 和第二个类 B，其方法将 A 的引用作为参数.

I have a class A deriving from boost::noncopyable and a second class B with a method that takes a reference to A as an argument.

class A : boost::noncopyable { /*...*/ };

class B {

public:

    virtual void do_something(A& a) {
        /*...*/
    }
};

我公开的类如下:

/* Wrapper for B, so B can be extended in python */
struct BWrap : public B, wrapper<B> {

    void do_something(A &a) {

        if (override do_something = this->get_override("do_something")) {
            do_something(a);
            return;
        }
        else {
            B::do_something(a);
        }
    }

    void default_do_something(A& a) { this->B::do_something(a); }
};

BOOST_PYTHON_MODULE(SomeModule) {

    class_<A, boost::noncopyable>("A");

    class_<BWrap, boost::noncopyable>("B")
        .def("do_something", &B::do_something, &BWrap::default_do_something)
    ;
}

我在 python 中扩展 B 是这样的:

I extend B in python like this:

test.py:

import SomeModule


class BDerived(SomeModule.B):

    def do_something(self, a):
        pass

并像这样调用扩展的 B:

and call the extended B like this:

try {
    py::object main = py::import("__main__"); 
    py::object global(main.attr("__dict__")); 
    py::object result = py::exec_file("test.py", global, global); 
    py::object pluginClass = global["BDerived"]; 
    py::object plugin_base = pluginClass(); 

    B& plugin = py::extract<B&>(plugin_base) BOOST_EXTRACT_WORKAROUND;

    A a;
    B.do_something(a);
}
catch (py::error_already_set) { 
    PyErr_Print();
}

但是这会导致错误消息:

However this results in an error message:

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

如果 A 不是从 boost::noncopyable 派生的，则代码运行没有任何错误，但是 do_something(A& a) 中的参数 acode> 在函数调用期间被复制，即使它是通过引用传入的.但是仅仅删除 A 上的不可复制要求不是一个选项，因为它存在是有原因的.

If A isn't derived from boost::noncopyable the code runs without any errors but the argument a in do_something(A& a) gets copied during the function call even though it's passed in by reference. But just removing the noncopyable requirement on A isn't an option since it's there for a reason.

有什么解决问题的建议吗?

Any suggestions how to solve the problem?

谢谢.

推荐答案

将 B.do_something(a); 更改为 B.do_something(boost::ref(a));.