如何初始化基类(超级)? [英] How do I initialize the base (super) class?

问题描述

在Python中，请考虑以下代码:

In Python, consider I have the following code:

>>> class SuperClass(object):
    def __init__(self, x):
        self.x = x

>>> class SubClass(SuperClass):
    def __init__(self, y):
        self.y = y
        # how do I initialize the SuperClass __init__ here?

如何在子类中初始化SuperClass __init__?我正在关注Python教程，但并未涵盖该内容.当我在Google上搜索时，发现了不止一种方法.处理此问题的标准方法是什么?

How do I initialize the SuperClass __init__ in the subclass? I am following the Python tutorial and it doesn't cover that. When I searched on Google, I found more than one way of doing. What is the standard way of handling this?

推荐答案

Python(直到第3版)支持旧样式"和新样式类.新样式类是从object派生的并且是您正在使用的类，并通过super()调用它们的基类，例如

Python (until version 3) supports "old-style" and new-style classes. New-style classes are derived from object and are what you are using, and invoke their base class through super(), e.g.

class X(object):
  def __init__(self, x):
    pass

  def doit(self, bar):
    pass

class Y(X):
  def __init__(self):
    super(Y, self).__init__(123)

  def doit(self, foo):
    return super(Y, self).doit(foo)

由于python了解旧样式和新样式类，因此有多种方法可以调用基本方法，这就是为什么您找到了多种方法的原因.

Because python knows about old- and new-style classes, there are different ways to invoke a base method, which is why you've found multiple ways of doing so.

出于完整性考虑，老式类使用基类显式调用基方法，即

For completeness sake, old-style classes call base methods explicitly using the base class, i.e.

def doit(self, foo):
  return X.doit(self, foo)

但是，既然您不应该再使用旧样式，那么我就不会对此太在意.

But since you shouldn't be using old-style anymore, I wouldn't care about this too much.

Python 3仅了解新型类(无论是否从object派生).

Python 3 only knows about new-style classes (no matter if you derive from object or not).

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