def test_list_iteration(ut):
s = '''
l1 = [1, 2, 3]
def iter_over_list():
for e in l1:
return e
r = iter_over_list()
def iter_over_list2():
for e in list(l1):
return e
r2 = iter_over_list2()
'''
ps.checkString(s)
r = find_in_history('r', ps)
ut.assertNums([1, 2, 3], r[0:3])
ut.assertCont(r[3])
r = find_in_history('r2', ps)
# unknown type due to unknown list function
ut._assertType(TypeError, r[0])
def test_dict_iteration_by_keys_using_explicit_call_to_keys_method(ut):
'''
Support the simplest case - one iteration over dictionary using explicit keys() call
For(target=Name(id='k', ctx=Store()),
iter=Call(func=Attribute(value=Name(id='dict_', ctx=Load()),
attr='keys', ctx=Load()), args=[], keywords=[], starargs=None, kwargs=None),
body=[...], orelse=[])
'''
s = '''
dict_ = {1: 2, 3: 4}
def iter_over_dict_keys():
for k in dict_.keys():
# which element returned is undefined
# (actually, it is defined by the order in which keys are iterated,
# which in turn is defined by the hash function, but it's
# too complicated)
# we'll assume that any value can be returned
return dict_.get(k)
new_values = iter_over_dict_keys()
'''
ps.checkString(s)
r = find_in_history('new_values', ps)
ut.assertEqual(3, len(r))
ut.assertNums([2, 4], r[0:2])
ut.assertCont(r[2]) # represents implicitly returned None