如何在Python C API中定义class方法？

Question

我试图在 C++ 代码中创建 Python class 定义并在 Python 中访问它。然而，函数被调用但参数没有被正确接收。请帮助我正确地做到这一点。

#include <iostream>
#include <Python.h>

using namespace std;

#include <Python.h>

static PyObject* MyClass__Init(PyObject *self, PyObject *args)
{
    cout << "MyClass__Init Called" << endl;
    Py_INCREF(Py_None);
    return Py_None;
};

static PyObject* MyModule__Start(PyObject *self, PyObject *args)
{
    const char* zBuff;

    if (PyArg_ParseTuple(args, "s", &zBuff))
        cout << "MyModule Start Called with parameter " << zBuff << endl;
    else
        cout << "MyModule Start ERROR" << endl;

    Py_INCREF(Py_None);
    return Py_None;
};

static PyObject* MyClass__Start(PyObject *self, PyObject *args)
{
    const char* zBuff;

    if (PyArg_ParseTuple(args, "s", &zBuff))
        cout << "MyClass Start Called with parameter" << zBuff << endl;
    else
        cout << "MyClass Start ERROR" << endl;

    Py_INCREF(Py_None);
    return Py_None;
};

static PyMethodDef pModuleMethods[] =
{
    {"Start", MyModule__Start, METH_VARARGS, ""},
    {NULL, NULL, 0, NULL}
};

static PyMethodDef pClassMethods[] = 
{
    {"__init__", MyClass__Init, METH_VARARGS, ""},
    {"Start", MyClass__Start, METH_VARARGS, ""},
    {NULL, NULL, 0, NULL}
};

void Start()
{
    Py_Initialize();

    /* create a new module and class */
    PyObject *pClassDic = PyDict_New();
    PyObject *pClassName = PyString_FromString("MyClass");
    PyObject *pClass = PyClass_New(NULL, pClassDic, pClassName);

    PyObject *pModule = Py_InitModule("MyModule", pModuleMethods);
    PyObject *pModuleDic = PyModule_GetDict(pModule);

    /* add methods to class */
    for (PyMethodDef* pDef = pClassMethods; pDef->ml_name != NULL; pDef++)
    {
        PyObject *pFunc = PyCFunction_New(pDef, NULL);
        PyObject *pMethod = PyMethod_New(pFunc, NULL, pClass);
        PyDict_SetItemString(pClassDic, pDef->ml_name, pMethod);
    }

    PyDict_SetItemString(pModuleDic, "MyClass", pClass);

    PyRun_SimpleString("import MyModule\n"
        "MyModule.Start('Hello Module')\n"
        "myObj = MyModule.MyClass()\n"
        "myObj.Start('Hello Class')\n");

    Py_Finalize();
};

int main()
{
    Start();
};

输出是，

MyModule Start Called with parameter Hello Module
MyClass__Init Called
MyClass Start ERROR

调用 Module 函数没有任何问题，但调用 class 方法时没有正确的输入变量。

Answer 1

看来 self 的参数总是 NULL 而不是 - 对于 class 方法 - 在参数列表中传递对 self 的引用。因此，要解析 class 方法的参数，您还需要解析对 self 的引用。

从 Python 2.6 开始，您可以向 PyArg_ParseTuple 提供格式说明符列表。格式说明符的数量必须适合传递给函数的参数数量（请参阅 (items) (tuple) [matching-items] 下的 https://docs.python.org/2/c-api/arg.html）。

通过修改 MyClass__Start 函数来解析附加参数，您可以解析并打印两个参数以检查它们。对我来说，下面的代码

static PyObject* MyClass__Start(PyObject *self, PyObject *args)
{
    PyObject* argListSelf;
    const char* zBuff;

    if (PyArg_ParseTuple(args, "Os", &argListSelf, &zBuff)) {
        cout << "MyClass Start Called with parameters " <<  
        cout << PyString_AsString(PyObject_Str(argListSelf)) <<
        cout << " and " << zBuff << endl;

        cout << "self " << PyString_AsString(PyObject_Str(self)) << endl;
    }   
    else {
        if(PyErr_Occurred())
            PyErr_Print();

        cout << "MyClass Start ERROR" << endl;
    }   

    Py_INCREF(Py_None);
    return Py_None;
};

结果

MyModule Start Called with parameter Hello Module
MyClass__Init Called
MyClass Start Called with parameters \
    0x602428<?.MyClass instance at 0x7f484a333200>0x602428 and HelloClass
self <NULL>

注意我打印的是self的指针值，是NULL。我还添加了

if(PyErr_Occurred())
    PyErr_Print();

为了调试目的，我总是会添加它。

如何在Python C API中定义class方法？

How to define class methods in Python C API?

c

python

python-c-api