有办法在 subclass 中修改 parent class 的方法，使其返回某个变量吗？（ parent class 的该方法无返回值）

# parent class class A(object): # ... bala bala ... def test(self, url): # ... bala bala ... rsp = self.http.send(url=url) rsp = json.loads(rsp.content) self._session = self._session_type(rsp['a'], rsp['b']) # subclass class B(A): def test(self, url): # ... bala bala ... rsp = self.http.send(url=url) rsp = json.loads(rsp.content) self._session = self._session_type(rsp['a'], rsp['b']) self._xx = rsp['xx']

如果 A.test 有将 rsp 返回的话，那 subclass 就只需要

class B(A): def test(self, url): rsp = super(self.__class__, self).test(url) self._xx = rsp['xx']

所以...是否有办法可以让 parent class 的 test 方法返回 rsp 呢？（排除修改 parent class 源码的方法）

第 1 条附言 2017-06-28 23:54:56 +08:00

2楼的方法测试可行，实际代码片段测试，tracert 的覆盖方法会比复制父类代码的覆盖方法慢上9~10倍（timerit repeat number = 100000）

import sys # parent class class A(object): # ... bala bala ... def test(self, url): # ... bala bala ... rsp = self.http.send(url=url) rsp = json.loads(rsp.content) self._session = self._session_type(rsp['a'], rsp['b']) # subclass class B(A): def test(self, url): _locals = {} def tracert(frame, event, arg): if event == 'return': _locals.update(frame.f_locals.copy()) sys.setprofile(tracert) try: super(self.__class__, self).test(url) finally: sys.setprofile(None) rsp = _locals['rsp'] self._xx = rsp['xx']

6 条回复 2017-06-29 07:41:59 +08:00