修补函数的 __call__
问题描述
我需要在测试中修补当前日期时间.我正在使用这个解决方案:
I need to patch current datetime in tests. I am using this solution:
def _utcnow():
return datetime.datetime.utcnow()
def utcnow():
"""A proxy which can be patched in tests.
"""
# another level of indirection, because some modules import utcnow
return _utcnow()
然后在我的测试中,我会执行以下操作:
Then in my tests I do something like:
with mock.patch('***.utils._utcnow', return_value=***):
...
但今天我想到了一个想法,我可以通过修补函数 utcnow
的 __call__
来简化实现,而不是额外添加一个 _utcnow
.
But today an idea came to me, that I could make the implementation simpler by patching __call__
of function utcnow
instead of having an additional _utcnow
.
这对我不起作用:
from ***.utils import utcnow
with mock.patch.object(utcnow, '__call__', return_value=***):
...
如何优雅地做到这一点?
How to do this elegantly?
解决方案
当你修补一个函数的 __call__
时,你是在设置那个 的 __call__
属性实例.Python 实际上调用了类上定义的 __call__
方法.
When you patch __call__
of a function, you are setting the __call__
attribute of that instance. Python actually calls the __call__
method defined on the class.
例如:
>>> class A(object):
... def __call__(self):
... print 'a'
...
>>> a = A()
>>> a()
a
>>> def b(): print 'b'
...
>>> b()
b
>>> a.__call__ = b
>>> a()
a
>>> a.__call__ = b.__call__
>>> a()
a
将任何东西分配给 a.__call__
是没有意义的.
Assigning anything to a.__call__
is pointless.
但是:
>>> A.__call__ = b.__call__
>>> a()
b
TLDR;
a()
不调用 a.__call__
.它调用 type(a).__call__(a)
.
TLDR;
a()
does not call a.__call__
. It calls type(a).__call__(a)
.
回答为什么type(x).__enter__(x)
而不是 Python 标准 contextlib 中的 x.__enter__()
?".
There is a good explanation of why that happens in answer to "Why type(x).__enter__(x)
instead of x.__enter__()
in Python standard contextlib?".
特殊方法查找的 Python 文档中记录了此行为.
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