Mocking assert_called_with in Python

Question

I'm having some trouble understanding why the following code does not pass:

test.py

import mock
import unittest

from foo import Foo


class TestFoo(unittest.TestCase):
    @mock.patch('foo.Bar')
    def test_foo_add(self, Bar):
        foo = Foo()
        foo.add(2, 2)
        Bar.add.assert_called_with(2, 2)


if __name__ == '__main__':
    unittest.main()

foo.py

from bar import Bar

class Foo(object):
    def add(self, x, y):
        bar = Bar()
        return bar.add(x, y)

bar.py

class Bar(object):
    def add(self, x, y):
        print('b.Bar --> Adding {} + {}'.format(x, y))
        return x + y

In the code, Foo.add creates an instance of Bar and returns the result of Bar.add when invoked. Why does testing assert_called_with for Bar.add fail? I believe I am mocking the Bar in the correct place (I am mocking foo.Bar because that is the namespace which it is being looked up, rather than bar.Bar).

Traceback (most recent call last): File "/Users/iain/PycharmProjects/testing/venv/lib/python2.7/site-packages/mock.py", line 1201, in patched return func(*args, **keywargs) File "test.py", line 12, in test_a_b fake_Bar.add.assert_called_with(2, 2) File "/Users/iain/PycharmProjects/testing/venv/lib/python2.7/site-packages/mock.py", line 831, in assert_called_with raise AssertionError('Expected call: %s\nNot called' % (expected,)) AssertionError: Expected call: add(2, 2) Not called

Felipe · Accepted Answer · 2015-06-09 00:23:10Z

13

You are mocking the method call in the right place. However, since you are calling the method from an instance, it is a bound method, and thus receives the instance as the first argument (the self parameter), in addition to all the other arguments.

Edit: Since Bar is replaced with a Mock instance, Bar().add does not know it's a method (and hence is not bound to anything). In other words, Bar is a Mock, Bar() is a Mock, and Bar().add is also a Mock. bar.add is thus a newly created mock, called with arguments (2, 2). One way of asserting this call would be:

@mock.patch('foo.Bar')
def test_foo_add(self, Bar):
    foo = Foo()
    foo.add(2, 2)
    Bar.return_value.add.assert_called_with(2, 2)

Depending on how your actual code looks, you may want to instead mock the method rather than the class:

@mock.patch('foo.Bar.add')
def test_foo_add(self, bar_add):
    foo = Foo()
    foo.add(2, 2)
    bar_add.assert_called_with(2, 2)

edited Jun 9, 2015 at 0:23

answered Jun 8, 2015 at 20:35

Felipe

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4 Comments

Iain Over a year ago

I've modified the test so that foo is passed as the first arg to Bar.add.assert_called_with() but still the test fails? The output reads the same (AssertionError: Expected call: add(<foo.Foo object at 0x102108a50>, 2, 2) Not called)

Felipe Over a year ago

Whoops, my sample code was totally wrong. Will update in a few with the proper answer. Sorry!

Iain Over a year ago

Thanks! I think I fell made the mistake of thinking that if I mocked foo.Bar then all of its attributes and methods remained intact and unchanged unless I explicitly changed their behaviour . Would this be a case where one might use the mock.patch.object method instead?

Felipe Over a year ago

patch.object and patch have identical functionality; they simply use a different way of referring to the object to patch. To patch Bar in the foo module, you would use mock.patch(foo, 'Bar'), and the results would be the same. I don't know of a pre-built way of getting the behavior you want, unfortunately.

Graham · Accepted Answer · 2017-08-08 05:20:17Z

-2

I believe your problem would be solved like this:

from bar import Bar

class Foo(object):
    def add(self, x, y):
        self.bar = Bar()
        return self.bar.add(x, y)

edited Aug 8, 2017 at 5:20

Graham

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answered Aug 8, 2017 at 3:24

wakooo

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