如何避免使用super()进行无限递归? [英] How to avoid infinite recursion with super()?

问题描述

我有这样的代码:

class A(object):
    def __init__(self):
          self.a = 1

class B(A):
    def __init__(self):
        self.b = 2
        super(self.__class__, self).__init__()

class C(B):
    def __init__(self):
        self.c = 3
        super(self.__class__, self).__init__()

实例化B可以按预期工作，但是实例化C无限递归并导致堆栈溢出.我该如何解决?

Instantiating B works as expected but instantiating C recursed infinitely and causes a stack overflow. How can I solve this?

推荐答案

实例化C调用B.__init__时，self.__class__仍为C，因此super()调用将其带回B.

When instantiating C calls B.__init__, self.__class__ will still be C, so the super() call brings it back to B.

在调用super()时，请直接使用类名.因此，在B中，调用super(B, self)，而不是super(self.__class__, self)(并且最好在C中使用super(C, self)).从Python 3开始，您可以只使用不带参数的super()来实现相同的目的

When calling super(), use the class names directly. So in B, call super(B, self), rather than super(self.__class__, self) (and for good measure, use super(C, self) in C). From Python 3, you can just use super() with no arguments to achieve the same thing

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