super，decorators和gettattribute

如果类child和parent定义如下，它应该像 一样工作： >> x =儿童（） >> x.foo（） I我是父母的foo 我是孩子的foo。 """ name = fun .__ name __ def装饰（自我，* args，** kwargs）： 尝试： super_object = super（self .__ class__，self） 这里有一个明显常见的错误：你不想传递超级 self .__ class__，但是他把这个方法绑定到了这个类。两个 是不一样的，因为子类的实例将子类 作为self .__ class__，而不是当前的类。所以super会返回 当前类或它的子类，这意味着（因为你从self调用了这个 方法），你最终会递归地调用这个方法。 所有这些都意味着你的装饰师可能不得不把这个类作为一个参数。 ＃现在我想要达到相当于调用的东西 ＃parent.foo（* args，** kwargs） ＃如果我想将它限制在这个例子中 ＃这不起作用：在这个例子中，它调用了孩子的foo， ＃进入一个永恒的循环（而不是调用父母的 ＃foo，正如我所料）。 ＃super_object .__ getattribute __（name）（* args，** kwargs） ＃这确实有效，但我觉得它很丑陋 eval（''super_object。％s（* args，** kwargs） ''％name） 除了AttributeError： 传递＃如果父母没有实现乐趣，我们不在乎 ＃关于它 返回乐趣（自我，* args，** kwargs）＃hopefully none 装饰.__ name__ = name 返回装饰 班级父母（对象）： def foo（self）： print''我是父母''foo'' 班级孩子（父母）： @endmethod def foo（self）： print"我很好foo。 if __name __ ==''__ main__''： x = child（） x.foo（） 有人能告诉我如何调用超类方法知道它的名字吗？ 如果您知道它的名字，就像调用任何对象的方法一样： getattr （super_object，name）（* args，** kwargs） 代码似乎按照你想要的方式工作： >> x.foo（） 我是父母的'foo 我是foo'的foo。 < mike - Mike Meyer< mw*@mired.org> http://www.mired.org/consulting.html 独立网络/ Unix / Perforce顾问，电子邮件以获取更多信息。 1月12日9:47 pm，Mike Meyer< mwm-keyword- python.b4b ... @ mired.org> 写道： 如果你调用任何对象的方法的方式相同知道它的名字"： getattr（super_object，name）（* args，** kwargs） 非常感谢你的回答！ 然而，我很惊讶地得知 super_object .__ getattr __（name）（* args， ** kwargs） getattr（super_object，name）（* args，** kwargs） 不等价。这很奇怪，至少在使用len（） 和.__ len __，str（）和.__ str__时。你可能知道没有遵循getattr的惯例背后的理由是什么？ 祝你好运， - Richard Hello all, I am playing around w/ Python''s object system and decorators and Idecided to write (as an exercise) a decorator that (if applied to amethod) would call the superclass'' method of the same name beforedoing anything (initially I wanted to do something like CLOS[1] :before and :end methods, but that turned out to be toodifficult). However, I cannot get it right (specially, get rid of the eval). Isuspect that I may be misunderstanding something happening betweensuper objects and __getattribute__ methods. Here''s my code: def endmethod(fun):"""Decorator to call a superclass'' fun first. If the classes child and parent are defined as below, it shouldwork like: >>x = child()x.foo() I am parent''s fooI am child''s foo."""name = fun.__name__def decorated(self, *args, **kwargs):try:super_object = super(self.__class__, self) # now I want to achieve something equivalent to calling# parent.foo(*args, **kwargs)# if I wanted to limit it only to this example # this doesn''t work: in the example, it calls child''s foo,# entering in an eternal loop (instead of calling parent''s# foo, as I would expect). # super_object.__getattribute__(name)(*args, **kwargs) # this does work, but I feel it''s uglyeval(''super_object.%s(*args, **kwargs)'' % name)except AttributeError:pass # if parent doesn''t implement fun, we don''t care# about itreturn fun(self, *args, **kwargs) # hopefully none decorated.__name__ = namereturn decoratedclass parent(object):def foo(self):print ''I am parent\''s foo'' class child(parent):@endmethoddef foo(self):print "I am foo\''s foo." if __name__==''__main__'':x = child()x.foo() Can anybody tell me how to call a superclass method knowing its name? Thanks in advance,-- Richard [1] http://en.wikipedia.org/wiki/Common_Lisp_Object_System 解决方案 On Jan 12, 7:45 pm, Richard Szopa <[email protected]: doing anything (initially I wanted to do something like CLOS[1] :before and :end methods, but that turned out to be toodifficult).Erm, I meant :before and :after methods. -- Richard On Sat, 12 Jan 2008 10:45:25 -0800 (PST) Richard Szopa <ry***********@gmail.comwrote: Hello all,I am playing around w/ Python''s object system and decorators and Idecided to write (as an exercise) a decorator that (if applied to amethod) would call the superclass'' method of the same name beforedoing anything (initially I wanted to do something like CLOS[1] :before and :end methods, but that turned out to be toodifficult).However, I cannot get it right (specially, get rid of the eval). Isuspect that I may be misunderstanding something happening betweensuper objects and __getattribute__ methods.Here''s my code:def endmethod(fun): """Decorator to call a superclass'' fun first. If the classes child and parent are defined as below, it should work like: >>x = child() >>x.foo() I am parent''s foo I am child''s foo. """ name = fun.__name__ def decorated(self, *args, **kwargs): try: super_object = super(self.__class__, self)There''s an apparently common bug here: you don''t want to pass superself.__class__, but the class that the method is bound to. The twoaren''t the same, as an instance of a subclass will have the subclassas self.__class__, and not the current class. So super will return thecurrent class or a subclass of it, meaning (since you invoked thismethod from self) you''ll wind up invoking this method recursively.All of which means your decorator is probably going to have to takethe class as an argument. # now I want to achieve something equivalent to calling # parent.foo(*args, **kwargs) # if I wanted to limit it only to this example # this doesn''t work: in the example, it calls child''s foo, # entering in an eternal loop (instead of calling parent''s # foo, as I would expect). # super_object.__getattribute__(name)(*args, **kwargs) # this does work, but I feel it''s ugly eval(''super_object.%s(*args, **kwargs)'' % name) except AttributeError: pass # if parent doesn''t implement fun, we don''t care # about it return fun(self, *args, **kwargs) # hopefully none decorated.__name__ = name return decoratedclass parent(object): def foo(self): print ''I am parent\''s foo''class child(parent): @endmethod def foo(self): print "I am foo\''s foo."if __name__==''__main__'': x = child() x.foo()Can anybody tell me how to call a superclass method knowing its name?The same way you call any object''s methods if you know it''s name": getattr(super_object, name)(*args, **kwargs) The code seems to work the way you want: >>x.foo() I am parent''s fooI am foo''s foo. <mike--Mike Meyer <mw*@mired.org>http://www.mired.org/consulting.htmlIndependent Network/Unix/Perforce consultant, email for more information.On Jan 12, 9:47 pm, Mike Meyer <[email protected]>wrote: The same way you call any object''s methods if you know it''s name": getattr(super_object, name)(*args, **kwargs)Thanks a lot for your answer! However, I am very surprised to learn that super_object.__getattr__(name)(*args, **kwargs) getattr(super_object, name)(*args, **kwargs) are not equivalent. This is quite odd, at least when with len()and .__len__, str() and .__str__. Do you maybe know what''s therationale behind not following that convention by getattr? Best regards, -- Richard 这篇关于super，decorators和gettattribute的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！