Python returns MagicMock object instead of return_value

python python-unittest magicmock
Mehdi Jafarnia Jahromi picture Mehdi Jafarnia Jahromi · Jul 5, 2016 · Viewed 47.5k times · Source

I have a python file a.py which contains two classes A and B.

class A(object):
    def method_a(self):
        return "Class A method a"

class B(object):
    def method_b(self):
        a = A()
        print a.method_a()

I would like to unittest method_b in class B by mocking A. Here is the content of the file testa.py for this purpose:

import unittest
import mock
import a


class TestB(unittest.TestCase):

    @mock.patch('a.A')
    def test_method_b(self, mock_a):
        mock_a.method_a.return_value = 'Mocked A'
        b = a.B()
        b.method_b()


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

I expect to get Mocked A in the output. But what I get is:

<MagicMock name='A().method_a()' id='4326621392'>

Where am I doing wrong?

Answer

jonrsharpe picture jonrsharpe · Jul 5, 2016

When you @mock.patch('a.A'), you are replacing the class A in the code under test with mock_a.

In B.method_b you then set a = A(), which is now a = mock_a() - i.e. a is the return_value of mock_a. As you haven't specified this value, it's a regular MagicMock; this isn't configured either, so you get the default response (yet another MagicMock) when calling methods on it.

Instead, you want to configure the return_value of mock_a to have the appropriate method, which you can do as either:

mock_a().method_a.return_value = 'Mocked A' 
    # ^ note parentheses

or, perhaps more explicitly:

mock_a.return_value.method_a.return_value = 'Mocked A'

Your code would have worked in the case a = A (assigning the class, not creating an instance), as then a.method_a() would have triggered your mock method.

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