Pybind11:用指针成员包装结构? [英] Pybind11: Wrap a struct with a pointer member?

问题描述

我需要包装以下结构

struct dataStruct {
    const std::vector<int>& data;
    bool valid_data = true;
}

在我的包装文件中，在 PYBIND11_MODULE 下等等我有

In my wrapper file, under PYBIND11_MODULE and so forth I have

py::class_<dataStruct>(m, "dataStruct")
   .def(py::init<>())
   .def_readwrite("data", &dataStruct::data)
   .def_readwrite("valid_data", &dataStruct::valid_data);

我收到两个错误

no instance of function template "pybind11::class_<type_, options...>::def_readwrite [with type_=dataStruct, options=<>]" matches the argument list -- argument types are: (const char [7], <error-type>) -- object type is: pybind11::class_<dataStruct>

pointer to member of type "const std::vector<int, std::allocator<int>> &" is not allowed

我知道这是由于结构中的指针造成的，但是在查看 pybind11 文档时，我无法找到如何处理结构中的指针.

I know this is due to the pointer in the struct, but when looking at the pybind11 documentation I could not find out how to deal with pointers when they are within a struct.

推荐答案

我之前所做的是围绕 dataStruct 编写一个包装结构，您可以将其暴露给 Python.这将允许您保留向量的额外副本作为实际的 Python 列表.

What I have done before is to write a wrapper struct around dataStruct that you expose to Python. This will allow you to keep an extra copy of the vector as an actual Python list.

struct dataStructPy : dataStruct {
    const py::list & data_py;
}

然后您可以将 data_py 设置为属性而不是读写，并且您可以编写在 std::vector 和 之间进行转换的 get/set 函数py::list.这允许您在 C++ 端使用 std::vector 而在 Python 端使用标准 Python list.此类函数可能如下所示:

You can then set data_py as a property instead of a readwrite and you can write get/set functions that do a conversion between std::vector and py::list. This allows you to use a std::vector on the C++ side while using the standard Python list on the Python side. Such functions might look like:

py::list dataStructPy::data_py_get()
{
    py::list list;
    for(const auto & x : this->data) {
        list.append(x);
    }

    return list;
}

void dataStructPy::data_py_set(const py::list & data_py)
{
    this->data.clear();
    for(int i = 0; i < py::len(data_py); ++i) {
        this->data.emplace_back(data_py[i]);
    }
}

然后您可以将 dataStructPy 暴露给 Python 并将其称为 dataStruct(您可以看到我们仍然暴露了 &dataStruct::valid_data):

You can then expose dataStructPy to Python and call it dataStruct (you can see we still expose &dataStruct::valid_data):

py::class_<dataStructPy>(m, "dataStruct")
   .def(py::init<>())
   .def_readwrite("valid_data", &dataStruct::valid_data);
   .add_property("data", &dataStructPy::data_py_get, &dataStructPy::data_py_set);

在 Python 中，执行以下操作

In Python, doing the following

x = dataStruct()
x.data = [1, 2, 3, 4]

将导致 C++ x.data_py = py::list{1, 2, 3, 4}; 和 x.data = std::vector{1, 2,3, 4};

will result in the C++ x.data_py = py::list{1, 2, 3, 4}; and x.data = std::vector{1, 2, 3, 4};

注意:这仅适用于赋值运算符 (=).如果在 Python 方面，如果你执行 x.data[3] = 5 它不会更新 C++ 向量，那么你需要创建一个对象，该对象使用 [] 运算符从 Python 访问向量的内存.

NOTE: This only works for the assignment operator (=). If on the Python side, if you do x.data[3] = 5 it will not update the C++ vector, to do that you would need to create an object that accesses the vector's memory from Python using the [] operator.

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