开发者

Issue with making object callable in python

I wrote code like this

>>> class a(object):
        def __init__(self):
            self.__call__ = 开发者_开发技巧lambda x:x

>>> b = a()

I expected that object of class a should be callable object but eventually it is not.

>>> b()

Traceback (most recent call last):
    File "<pyshell#5>", line 1, in <module>
        b()
TypeError: 'a' object is not callable
>>> callable(b)
False

>>> hasattr(b,'__call__')
True
>>> 

I can't understand why. Please help me.


Special methods are looked up on the type (e.g., class) of the object being operated on, not on the specific instance. Think about it: otherwise, if a class defines __call__ for example, when the class is called that __call__ should get called... what a disaster! But fortunately the special method is instead looked up on the class's type, AKA metaclass, and all is well ("legacy classes" had very irregular behavior in this, which is why we're all better off with the new-style ones -- which are the only ones left in Python 3).

So if you need "per-instance overriding" of special methods, you have to ensure the instance has its own unique class. That's very easy:

class a(object):
    def __init__(self):
        self.__class__ = type(self.__class__.__name__, (self.__class__,), {})
        self.__class__.__call__ = lambda x:x

and you're there. Of course that would be silly in this case, as every instance ends up with just the same "so-called per-instance" (!) __call__, but it would be useful if you really needed overriding on a per-individual-instance basis.


__call__ needs to be defined on the class, not the instance

class a(object):
    def __init__(self):
        pass
    __call__ = lambda x:x

but most people probably find it more readable to define the method the usual way

class a(object):
    def __init__(self):
        pass
    def __call__(self):
        return self

If you need to have different behaviour for each instance you could do it like this

class a(object):
    def __init__(self):
        self.myfunc = lambda x:x

    def __call__(self):
        return self.myfunc(self)


What about this? Define a base class AllowDynamicCall:

class AllowDynamicCall(object):
     def __call__(self, *args, **kwargs):
         return self._callfunc(self, *args, **kwargs)

And then subclass AllowDynamicCall:

class Example(AllowDynamicCall):
     def __init__(self):
         self._callfunc = lambda s: s
0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜