从 C++ 中查找 python 函数参数

Question

我正在从 C++ 调用 python 函数。我想知道是否可以确定参数的数量和这些参数的名称。我已阅读链接How to find the number of parameters to a Python function from C? 但是我真的不明白。

我有这个 C++ 函数，它从 pyFunction.py 调用函数“add”。'add' 接受两个参数并返回总和。

static float CallPythonFunc( float *parameters )
{
    PyObject *pName, *pModule, *pDict, *pFunc, *pValue, *pArgs;
    float ret;

    // Initialize the python interpreter
    Py_Initialize();

    // Make sure we are getting the module from the correct place
    // ### This is where we will put the path input
    PyRun_SimpleString("import sys");
    PyRun_SimpleString("sys.path.append(\"/Developer/IsadoraSDK/IsadoraDemoMathFunction/\")");

    // Build the name object
    // ### This is where we will put the function input
    pName = PyString_FromString("pyFunction");

    // Load the module object
    pModule = PyImport_Import(pName);

    // pDict is a borrowed reference
    pDict = PyModule_GetDict(pModule);

    // pFunc is a borrowed reference
    pFunc = PyDict_GetItemString(pDict, "add");

    //
    // Somehow get the number of arguments and possible the arguments names from 'add'
    //

    if (PyCallable_Check(pFunc)) 
    {       
        // Set the number of arguments
                // This is where I would like to pass in number of arguments
        pArgs = PyTuple_New( 2 /*number of arguments*/ );

        //
        // Instead of the following if I had the arguments I could loop through them
        // and pass the correct number in
        //

        // Argument 1
        pValue = PyFloat_FromDouble((double)parameters[0]);
        PyTuple_SetItem(pArgs, 0, pValue);

        // Argument 2
        pValue = PyFloat_FromDouble((double)parameters[1]);
        PyTuple_SetItem(pArgs, 1, pValue);

            // Make the call to the function
        pValue = PyObject_CallObject(pFunc, pArgs);

        // Set return value
        ret = (float)PyFloat_AsDouble(pValue);

        // Clean up
        Py_DECREF(pArgs);
        Py_DECREF(pValue);
    }

// Clean up
Py_DECREF(pModule);
Py_DECREF(pName);

// Finish the Python Interpreter
Py_Finalize();

return ret;
}

我真的不太熟悉 C/C++，所以任何帮助都会非常有帮助。感谢大家的时间！

编辑：所以像下面这样？

PyObject *tuple, *arglist;
tuple = PyObject_CallMethod(pFunc,"inspect.getargspec","add");
arglist = PyTuple_GetItem(tuple,0);
int size = PyObject_Size(arglist);

Answer 1

您链接到的问题的这个答案似乎是您想要的。inspect.getargspec在 Python 方面完全按照您的要求执行操作，正如答案所述，您可以使用PyObject_CallMethod或该链接目标中描述的相关函数之一inspect.getargspec从您的 C++ 代码中调用，将返回的元组作为 a获取PyObject，用于PyTuple_GetItem(returned_tuple, 0)获取参数列表，然后使用列表中的PyObject_Size()或PyObject_Length()来获取参数的数量。您还需要检查返回的元组的第二个和第三个元素，并为不是 的两个元素中的每一个将参数数量增加 1 Py_None。请参阅下面的代码片段了解原因。

>>> import inspect
>>> def testfunc(a, b, c, *d, **e):
    pass

>>> inspect.getargspec(testfunc)
ArgSpec(args=['a', 'b', 'c'], varargs='d', keywords='e', defaults=None)

这是您应该做什么的示例（并非所有可能的错误都可能被检查，但应该是所有可能需要的 NULL 检查）：

PyObject *pName, *pInspect, *argspec_tuple, *arglist;
int size;

pName = PyString_FromString("inspect");

if (pName)
{
    pInspect = PyImport_Import(pName);
    Py_DECREF(pName);


    if (pInspect)
    {
        pName = PyString_FromString("getargspec");

        if (pName)
        {
            argspec_tuple = PyObject_CallMethodObjArgs(pInspect, pName, pFunc, NULL);
            Py_DECREF(pName);

            if (argspec_tuple)
            {
                arglist = PyTuple_GetItem(argspec_tuple, 0);

                if (arglist)
                {
                    size = PyObject_Size(arglist)
                         + (PyTuple_GetItem(argspec_tuple, 1) == Py_None ? 0 : 1)
                         + (PyTuple_GetItem(argspec_tuple, 2) == Py_None ? 0 : 1);  // Haven't actually tested this, but it should work
                }
            }
        }
    }
}