Add keyword argument support to itertools.count().
diff --git a/Modules/itertoolsmodule.c b/Modules/itertoolsmodule.c
index 75c6e7e..a44d2ae 100644
--- a/Modules/itertoolsmodule.c
+++ b/Modules/itertoolsmodule.c
@@ -2916,11 +2916,10 @@
Py_ssize_t cnt = 0;
PyObject *long_cnt = NULL;
PyObject *long_step = NULL;
+ static char *kwlist[] = {"start", "step", 0};
- if (type == &count_type && !_PyArg_NoKeywords("count()", kwds))
- return NULL;
-
- if (!PyArg_UnpackTuple(args, "count", 0, 2, &long_cnt, &long_step))
+ if (!PyArg_ParseTupleAndKeywords(args, kwds, "|OO:count",
+ kwlist, &long_cnt, &long_step))
return NULL;
if (long_cnt != NULL && !PyNumber_Check(long_cnt) ||
@@ -3027,10 +3026,10 @@
}
PyDoc_STRVAR(count_doc,
- "count([firstval[, step]]) --> count object\n\
+ "count([start[, step]]) --> count object\n\
\n\
Return a count object whose .__next__() method returns consecutive\n\
-integers starting from zero or, if specified, from firstval.\n\
+integers starting from zero or, if specified, from start.\n\
If step is specified, counts by that interval. Equivalent to:\n\n\
def count(firstval=0, step=1):\n\
x = firstval\n\