创建无法腌制的对象

Question

如何在我的rpc代码中轻松创建一个无法用于测试边缘情况的对象？

它需要是:

1. 简单
2. 可靠(预计未来版本的python或pickle不会中断)
3. 跨平台

编辑:预期用途如下所示:

class TestRPCServer:
    def foo(self):
        return MagicalUnpicklableObject()

def test():
    with run_rpc_server_and_connect_to_it() as proxy:
        with nose.assert_raises(pickle.PickleError):
            proxy.foo()

Answer 1

如果你需要的只是一个在你腌制它时会引发异常的对象,那么对于测试的目的,你可以放弃这个__getstate__方法.

>>> class C:
...     def __getstate__(self):
...         raise Exception
... 
>>> pickle.dumps(C())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1374, in dumps
    Pickler(file, protocol).dump(obj)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 224, in dump
    self.save(obj)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 286, in save
    f(self, obj) # Call unbound method with explicit self
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 723, in save_inst
    stuff = getstate()
  File "<stdin>", line 3, in __getstate__
Exception

如果您想要一个不太人为的场景,请考虑使用OS资源的对象,如文件句柄,套接字或线程等.

>>> with open('spam.txt', 'w') as f:
...     pickle.dumps(f)
... 
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 1374, in dumps
    Pickler(file, protocol).dump(obj)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 224, in dump
    self.save(obj)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/pickle.py", line 306, in save
    rv = reduce(self.proto)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/copy_reg.py", line 70, in _reduce_ex
    raise TypeError, "can't pickle %s objects" % base.__name__
TypeError: can't pickle file objects