为什么测试用例中的'yield'让测试通过

Question

我试图理解以下python代码片段,取自SublimeText3插件开发单元测试示例.

def test_delayed_insert(self):
    sublime.set_timeout(
        lambda: self.view.run_command("delayed_insert_hello_world"),
        100)
    # `delayed_insert_hello_world` will be execulated after the timeout
    # `yield 1000` will yield the runtime to main thread and continue
    # the execution 1 second later
    yield 1000
    row = self.getRow(0)
    self.assertEqual(row, "hello world")

这怎么办？如果unittest不支持这个(将测试称为生成器),代码将不会被执行,对吧？我目前的理解是unittest框架采用屈服值'x',并将其自身暂停为'x'ms,在这种情况下为1000ms,类似于c ++线程中yield/sleep的语义.但是,这取决于'unittest'以这种方式真正使用产生的价值.

如果确实如此,为什么这个测试不会失败(在st3 = python 3.3.6和python 3.5.2中尝试过)？

from unittest import TestCase

class TestMe(TestCase):
    def test_failtest(self):
        self.assertTrue(False)
        yield 0

运行它给出:

~$ python3 -m unittest test_me.py --verbose
test_failtest (test_me.TestMe) ... ok

----------------------------------------------------------------------
Ran 1 test in 0.000s

OK

Answer 1

TL; DR您链接的代码是子类化自定义TestCase类(DeferrableTestCase),这可能会改变"正常"行为unittest.

让我们退后一步,不要使用unittest.

def foo():
    assert True is False

foo()

这提高了AssertionError预期.然而,

def foo():
    assert True is False
    yield

foo()

才不是.

原因是yield关键字转向foo生成器.为了获得,AssertionError我们将不得不使用foo返回的生成器:

def foo():
    assert True is False
    yield

next(foo())
  File "main.py", line 48, in <module>
next(foo())
  File "main.py", line 45, in foo
assert True is False
AssertionError

此行为是继承到生成器,与如何unittest或assert工作无关:

def foo():
    1/0
    yield

gen = foo()
print('No exception yet')
next(gen)

输出

'No exception yet'
Traceback (most recent call last):
File "main.py", line 50, in <module>
  next(gen)
File "main.py", line 45, in foo
  1/0
ZeroDivisionError: division by zero

但是,您链接的代码是子类化自定义TestCase类(DeferrableTestCase),这可能会改变行为.