Give itertools.repeat() a length method.

This commit is contained in:
Raymond Hettinger 2004-02-10 09:25:40 +00:00
parent 27da291b2c
commit 5cab2e3a88
2 changed files with 21 additions and 2 deletions

View File

@ -607,6 +607,12 @@ class TestVariousIteratorArgs(unittest.TestCase):
self.assertRaises(TypeError, list, tee(N(s))[0])
self.assertRaises(ZeroDivisionError, list, tee(E(s))[0])
class LengthTransparency(unittest.TestCase):
def test_repeat(self):
self.assertEqual(len(repeat(None, 50)), 50)
self.assertRaises(TypeError, len, repeat(None))
class RegressionTests(unittest.TestCase):
def test_sf_793826(self):
@ -826,7 +832,7 @@ __test__ = {'libreftest' : libreftest}
def test_main(verbose=None):
test_classes = (TestBasicOps, TestVariousIteratorArgs, TestGC,
RegressionTests)
RegressionTests, LengthTransparency)
test_support.run_unittest(*test_classes)
# verify reference counting

View File

@ -2347,6 +2347,19 @@ repeat_next(repeatobject *ro)
return ro->element;
}
static int
repeat_len(repeatobject *ro)
{
if (ro->cnt == -1)
PyErr_SetString(PyExc_TypeError, "len() of unsized object");
return (int)(ro->cnt);
}
static PySequenceMethods repeat_as_sequence = {
(inquiry)repeat_len, /* sq_length */
0, /* sq_concat */
};
PyDoc_STRVAR(repeat_doc,
"repeat(element [,times]) -> create an iterator which returns the element\n\
for the specified number of times. If not specified, returns the element\n\
@ -2366,7 +2379,7 @@ static PyTypeObject repeat_type = {
0, /* tp_compare */
0, /* tp_repr */
0, /* tp_as_number */
0, /* tp_as_sequence */
&repeat_as_sequence, /* tp_as_sequence */
0, /* tp_as_mapping */
0, /* tp_hash */
0, /* tp_call */