我正在开发Python 3 C扩展。
我可以获得等效或任意位置或关键字参数吗?
例如,在Python中,我可以写:
def fun(name, parent, *args, **kwargs):
# do something with name and parent
# do something with args and kwargs
pass
但我在C中找不到简单的等价物。虽然我们可以用PyObject* args
和PyObject* kwargs
完美地编写函数,但我不能轻易地从任何一个(args / kwargs)“解析”名称和父级它来了。
采取:
static PyObject* myFunction(PyObject* self, PyObject* args, PyObject* kwargs) {
char* kwds[] = {"parent", "name", NULL};
PyObject* name = NULL;
PyObject* parent = NULL;
if (!PyArg_ParseTupleAndKeywords(args, kwargs, "OO", kwds, &parent, &name)) {
goto errorParseTupleAndKeywords;
}
/* Do something with name and parent */
/* parent and name maybe have appeared either in args or kwargs */
/* But I don't have any extra positional (*args) or keyword (**kwargs) here */
}
我能想到的“手动”方法看起来大致如下:
static PyObject* myFunction(PyObject* self, PyObject* args, PyObject* kwargs) {
PyObject* name = NULL;
PyObject* parent = NULL;
int inKwargs = 0;
// Pretend to do something with parent
if (PyDict_GetItemString(kwargs, "parent")) {
inKwargs++;
PyDict_DelItemString(kwargs, "parent");
}
// Pretend to do something with name
if (PyDict_GetItemString(kwargs, "name")) {
inKwargs++;
PyDict_DelItemString(kwargs, "name");
}
// Not sure if -1 works here
PyObject* newArgs = PyTuple_GetSlice(args, inKwargs, -1); // this is *args
// the remaining kwargs can be used as **kwargs
}
答案 0 :(得分:2)
在C API中,PyObject* args
实际上是一个Python元组,而PyObject* kwargs
实际上是一个Python字典。至少这是PyArg_ParseTupleAndKeywords
internally所要求的:
int PyArg_ParseTupleAndKeywords(PyObject *args, PyObject *keywords, const char *format, char **kwlist, ...)
{
// …
if ((args == NULL || !PyTuple_Check(args)) ||
(keywords != NULL && !PyDict_Check(keywords)) ||
format == NULL ||
kwlist == NULL)
{
PyErr_BadInternalCall();
return 0;
}
// …
}
vgetargskeywords
中该函数的实际实现也会再次声明这一点,因此您应该可以通过从对象中手动提取来替换PyArg_ParseTupleAndKeywords
调用。
这意味着您可以同时使用tuple和dict API,也可以使用iterator protocol来迭代这些对象中的项目。