将一个MagicMock实例分配给另一个MagicMock实例的return_value

Question

我试图模拟一个MagicMock实例的函数的返回值，但结果并不像我预期的那样：

>>> f = mock.MagicMock() # => <MagicMock id='139903823124048'>
>>> g = mock.MagicMock() # => <MagicMock id='139903823522512'>
>>> f.goo.return_value = g 
>>> g                    # => <MagicMock name='mock.goo()' id='139903823522512'>

实例g没有改变，但是它的名称改变了？当我尝试的时候：

>>> f.goo(1,2)
>>> g.zoo('a')
>>> f.goo(3,4)
>>> f.goo.assert_has_calls([call(1,2), call(3,4)])
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/home/lando/.local/lib/python2.7/site-packages/mock/mock.py", line 969, in assert_has_calls
    ), cause)
  File "/home/lando/.local/lib/python2.7/site-packages/six.py", line 737, in raise_from
    raise value
AssertionError: Calls not found.
Expected: [call(1, 2), call(3, 4)]
Actual: [call(1, 2), call().zoo('a'), call(3, 4)]

为什么g的呼叫成为f.goo呼叫的一部分？甚至：

>>> f.goo.call_args_list # => [call(1, 2), call(3, 4)]

Answer 1

这种行为不是很直观，但却是按预期进行的。请看-

https://docs.python.org/3/library/unittest.mock.html#unittest.mock.Mock.assert_has_calls https://docs.python.org/3/library/unittest.mock.html#unittest.mock.Mock.mock_calls

对返回值的调用也会在mock_calls中进行跟踪。

使用any_order标志。