覆盖实例上的特殊方法 [英] Overriding special methods on an instance

问题描述

我希望有人能回答这个对Python有深刻理解的问题:)

I hope someone can answer this that has a good deep understanding of Python :)

考虑以下代码:

>>> class A(object):
...     pass
...
>>> def __repr__(self):
...     return "A"
...
>>> from types import MethodType
>>> a = A()
>>> a
<__main__.A object at 0x00AC6990>
>>> repr(a)
'<__main__.A object at 0x00AC6990>'
>>> setattr(a, "__repr__", MethodType(__repr__, a, a.__class__))
>>> a
<__main__.A object at 0x00AC6990>
>>> repr(a)
'<__main__.A object at 0x00AC6990>'
>>>

请注意，repr(a)如何无法产生"A"的预期结果? 我想知道为什么会这样，是否有办法实现这一目标...

Notice how repr(a) does not yield the expected result of "A" ? I want to know why this is the case and if there is a way to achieve this...

相反，下面的示例可以工作(也许是因为我们不打算重写特殊方法?):

I contrast, the following example works however (Maybe because we're not trying to override a special method?):

>>> class A(object):
...     def foo(self):
...             return "foo"
...
>>> def bar(self):
...     return "bar"
...
>>> from types import MethodType
>>> a = A()
>>> a.foo()
'foo'
>>> setattr(a, "foo", MethodType(bar, a, a.__class__))
>>> a.foo()
'bar'
>>>

推荐答案

Python不会调用特殊方法，这些方法的名称在实例上由__包围，而仅在类上，显然可以提高性能.因此，无法直接在实例上覆盖__repr__()并使之正常工作.相反，您需要执行以下操作:

Python doesn't call the special methods, those with name surrounded by __ on the instance, but only on the class, apparently to improve performance. So there's no way to override __repr__() directly on an instance and make it work. Instead, you need to do something like so:

class A(object):
    def __repr__(self):
        return self._repr()
    def _repr(self):
        return object.__repr__(self)

现在，您可以通过替换_repr()来覆盖实例上的__repr__().

Now you can override __repr__() on an instance by substituting _repr().

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