Question

我正在考虑在我的测试套件中重命名一些内置函数，但是我发现这样做具有全局效果（当我预期它们只在本地产生效果时）。例如：

import time
def test():
    time.sleep = "hello" #woah there! time is mutable so this won't just apply locally!

print time.sleep #prints <built-in function sleep>
test()
print time.sleep #prints hello (!)

我必须将time.sleep恢复到test()结束之前的状态吗？

这是气馁 ...... 应该我做这种测试的方式？

Answer 1

如果您想要以这种方式测试对象，则应使用dependency injection和mocking。从程序的“顶部”传入一个对象（在本例中为时间）。然后你可以通过传入一个模拟版本对单个函数或对象进行单元测试。

示例：

# Function to be tested
def callSleep(timer):
    timer.sleep(5)

# Example usage
def main():
    import time
    timer = time

    callSleep(timer)

# Example test
def testFunction():


    class MockTimer:
        numCalled = 0
        withValue = 0
        def sleep(self, val):
            self.numCalled += 1
            self.withValue = val

    mockTimer = MockTimer()

    callSleep(mockTimer)

    print "Num called:", mockTimer.numCalled, "with value", mockTimer.withValue

Answer 2

我会按照上面@ Joe的建议，但下面是你的问题的快速解决方法。至于为什么会发生这种情况，对time.sleep的引用在全局范围内，因此替换它的效果不仅限于本地范围。

import time
def test():
    old_sleep = time.sleep # Save a reference to the builtin
    time.sleep = "hello" #shouldn't this just set time.sleep locally?
    print 'Inside test:', time.sleep
    time.sleep = old_sleep # replace the reference

print time.sleep #prints <built-in function sleep>
test()
print time.sleep  #prints <built-in function sleep>

重命名内置函数以进行测试

2 个答案: