如何为 boost python 中作为元组的一部分返回的对象指定返回策略

发布于 2024-10-15 04:25:26 字数 834 浏览 1 评论 0原文

我有一个创建 C++ 对象的函数。在重构之前，我从 C++ 函数返回一个指针，并且在 boost python 包装器声明中，我将使用 boost::python::return_value_policy;。像这样的事情：

MyObject* create_object_from_description(std::string& description) 
{
    ...
    return myObject;
}
BOOST_PYTHON_MODULE(pymol) {
    boost::python::def("create_obj_from_desc", create_object_from_description,
        (boost::python::arg("description")),
        "",
        boost::python::return_value_policy<boost::python::manage_new_object>()
);

现在我需要返回的不仅仅是对象，还需要一些错误消息，重构基本上改变了 C++ 函数以让它返回一个元组：

boost::python::tuple create_object_from_description(std::string& description) 
{
    ...
    return boost::python::make_tuple(myObject, errmsg);
}

我应该如何通过此更改指定返回策略？

原文

I have a function that creates a C++ object. Before the refactor, I was returning a pointer from the C++ function, and in the boost python wrapper declaration, I would use boost::python::return_value_policy<boost::python::manage_new_object>. Something like this:

MyObject* create_object_from_description(std::string& description) 
{
    ...
    return myObject;
}
BOOST_PYTHON_MODULE(pymol) {
    boost::python::def("create_obj_from_desc", create_object_from_description,
        (boost::python::arg("description")),
        "",
        boost::python::return_value_policy<boost::python::manage_new_object>()
);

Now I need to return more than just the object, but also some error messages, and the refactor basically changes the C++ function to let it return a tuple:

boost::python::tuple create_object_from_description(std::string& description) 
{
    ...
    return boost::python::make_tuple(myObject, errmsg);
}

How should I specify the return policy with this change?

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零度℉ 2024-10-22 04:25:26

我认为这是不可能的。您可能需要做的是更改函数签名并在 python 中重新包装该函数。像这样：

boost::python::tuple create_object_from_description(std::string& description, 
                                                    MyObject& myObject) 
{
    myObject = mo; 
    return errmsg 
}

然后在 python 中有一个像这样的函数：

def create_object_from_description(description):
    mo = MyObject()
    errmsg = pymol.create_object_from_description(description, mo)
    return mo, errmsg

当然，你真正应该做的是抛出异常。 Boost::python 在将 c++ 异常转换为 python 异常方面做得非常好，反之亦然

I don't think that this is possible. What you probably need to do is change the function signature and re-wrap the function in python. Like this:

boost::python::tuple create_object_from_description(std::string& description, 
                                                    MyObject& myObject) 
{
    myObject = mo; 
    return errmsg 
}

Then in python have a function like this:

def create_object_from_description(description):
    mo = MyObject()
    errmsg = pymol.create_object_from_description(description, mo)
    return mo, errmsg

Of course, what you should REALLY do is throw an exception. Boost::python does a very nice job of turning c++ exceptions in to python exception and vice-versa

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