c++ - boost.Python 如何公开 boost::shared_ptr 的 typedef？

Question

我有一个 C++ 类定义为:

class MyFuture {
public:
    virtual bool isDone() = 0;
    virtual const std::string& get() = 0;
    virtual void onDone(MyCallBack& callBack) = 0;
    virtual ~MyFuture() { /* empty */ } 
};

typedef boost::shared_ptr<MyFuture> MyFuturePtr;

我使用 boost.python 将类公开给 Python，因为(此类从未在 Python 中创建，而是从 API 调用返回，因此返回 noncopyable):

BOOST_PYTHON_MODULE(MySDK)
{
    class_<MyFuture, noncopyable>("MyFuture", no_init)
        .def("isDone", &MyFuture::isDone)
        .def("get", &MyFuture::get, return_value_policy<copy_const_reference>())
        .def("onDone", &MyFuture::onDone)
    ;
}

在 python 中，我像这样使用它:

import MySDK

def main():
    # more code here ...
    future = session.submit()
    response = future.get
    print response

if __name__ == "__main__":
    main()      

但这会导致 Python 错误:

File "main.py", line 14, in main
    future = session.submit()
TypeError: No to_python (by-value) converter found for C++ type: class boost::shared_ptr<class MySDK::MyFuture>     

如何公开 typedef typedef boost::shared_ptr<MyFuture> MyFuturePtr; ？

更新

将使用 boost.Python 公开的类更改为:

class_<MyFuture, boost::shared_ptr<MyFuture> >("MyFuture", no_init)

导致编译错误:

boost\python\converter\as_to_python_function.hpp(21): 
    error C2259: 'MySDK::MyFuture' : cannot instantiate abstract class

Answer 1

class_<MyFuture, boost::shared_ptr<MyFuture>, boost::noncopyable>("MyFuture")

根据文档

http://www.boost.org/doc/libs/1_62_0/libs/python/doc/html/reference/high_level_components.html#high_level_components.boost_python_class_hpp.class_template_class_t_bases_hel