调用super的init时，Python中的最大递归深度错误. [英] Maximum recursion depth error in Python when calling super's init.

问题描述

我有一个类层次结构A<-B<-C，在B中，我需要在构造函数中进行一些处理，因此我从这篇文章中得出了以下代码:

I have a class hierarchy A <- B <- C, in B, I need some processing in the constructor, so I came up with this code from this post: Understanding Python super() with __init__() methods

#!/usr/bin/python

class A(object):
    def __init__(self, v, v2):
        self.v = v
        self.v2 = v2

class B(A):
    def __init__(self, v, v2):
        # Do some processing
        super(self.__class__, self).__init__(v, v2)

class C(B):
    def hello():
        print v, v2


b = B(3, 5)
print b.v
print b.v2

c = C(1,2)
print c

但是，由于最大递归超出了我，我遇到了运行时错误

However, I have an runtime error from maximum recursion exceeded

  File "evenmore.py", line 12, in __init__
    super(self.__class__, self).__init__(v, v2)
RuntimeError: maximum recursion depth exceeded while calling a Python object

怎么了?

推荐答案

首先要考虑的是:C从B继承构造函数(因为它不在C中定义).

First thing to consider: C inherits constructor from B (because it's not defined in C).

要考虑的第二件事: C类中__init__调用中的self.__class__是C，而不是B.

Second thing to consider: self.__class__ in __init__ invocation in C class is C, not B.

让我们分析一下:

• C().__init__调用super(self.__class__, self).__init__(v, v2)，解析为 super(C, self).__init__(v, v2)表示B.__init__(self, v, v2).
• 传递给B.__init__的第一个参数的类型为C. super(self.__class__, self).__init__(v, v2)再次解析为B.__init__(self, v, v2).
• 一次又一次，一次又一次.还有无限的递归.

• C().__init__ calls super(self.__class__, self).__init__(v, v2) which is resolved to super(C, self).__init__(v, v2) which means B.__init__(self, v, v2).
• First argument passed to B.__init__ has a type C. super(self.__class__, self).__init__(v, v2) is again resolved to B.__init__(self, v, v2).
• And again, and again, and again. And there is your infinite recursion.

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