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Add keyword argument support to itertools.count().

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Raymond Hettinger 2009-02-14 00:25:51 +00:00
parent 544c3e19e6
commit a4038038c6
3 changed files with 10 additions and 9 deletions
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11
Modules/itertoolsmodule.c
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@ -3233,11 +3233,10 @@ count_new(PyTypeObject *type, PyObject *args, PyObject *kwds)
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) ||
@ -3353,10 +3352,10 @@ count_repr(countobject *lz)
}
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\