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扯下Python的super()

时间:2020-02-28 12:21:38      阅读:70      评论:0      收藏:0      [点我收藏+]

标签:style   重定义   overflow   run   class   tun   运行   线程   没有   

注: Python 2.7.x 环境下

今晚搜东西无意中看到这篇Understanding Python super() with __init__() methods.

其实这篇老早就看过了, 不过有一篇很好的回答之前没有注意到.

首先说下super(), 我只在类的单继承时的__init__()中使用过.

注意super只能用在新式类(new-style class)中, 也就是继承自object类对象的子类:

class A(object):
    ....

以前遇到过一个问题, 排查了半天, 才发现是老式类定义.

传统的super使用方法如:

class Base(object):
    def __init__(self, id):
        self.id = id

class Child(Base):
    def __init__(self, id, name):
        super(Child, self).__init__(id)
        self.name = name

这个是Python2.2之后才支持的特性, 在之前只能:

class Child(Base):
    def __init__(self, id, name):
        Base.__init__(self, id)
        self.name = name

这样做的好处就是不需要显示的在初始化时指明Child的父类名是什么, 在复杂的继承环境下, 以致会牵一发动一身.

不过就像那篇帖子top1的回答里所说:

But the main advantage comes with multiple inheritance

super在多继承这种更复杂的环境下, 才能发挥真正的威力, 这也是python文档中提到的第二个使用场景. 当然至今没遇到过这种复杂环境, 所以没有发言权.


上面扯了一些super的基本情况, 接着该扯下帖子里top2的回答了.

里面提到了这个用法:

super(self.__class__, self).__init__()

关于__class__:

instance.__class__ : The class to which a class instance belongs.

因为前阵子在使用多线程(threading.Thread)时, 写了一个基类, 然后有两个类分别继承自这个基类, 设置线程名就是类名, 这时就用到了__class__:

class base_thread(threading.Thread):
    def __init__(self, **kwargs):
        threading.Thread.__init__(self)
        self.name = self.__class__.__name__

所以对这个比较敏感, 才留意了下这个回答, 没想到却发现了一些坑...

按照帖子里的那个回复:

This unfortunately does not necessarily work if you want to inherit the constructor from the superclass.

例子:

#!/usr/bin/env python
# -*- coding: utf-8 -*-

class Polygon(object):
    def __init__(self, id):
        self.id = id

class Rectangle(Polygon):
    def __init__(self, id, width, height):
        super(self.__class__, self).__init__(id)
        self.shape = (width, height)

class Square(Rectangle):
    pass

p = Polygon(10)
print p.id

r = Rectangle(5, 1, 2)
print r.id

s = Square(20, 2, 4)
print s.id

运行结果:

% python test.py
10
5
Traceback (most recent call last):
  File "test.py", line 65, in <module>
    s = Square(20, 2, 4)
  File "test.py", line 53, in __init__
    super(self.__class__, self).__init__(id)
TypeError: __init__() takes exactly 4 arguments (2 given)

执行到Square类时, 报错说应该有4个参数, 但是实际上只有两个.

简化下代码, 并加一些调试输出:

#!/usr/bin/env python
# -*- coding: utf-8 -*-

class Polygon(object):
    def __init__(self, id):
        print('in Polygon, self.__class__ is %s' % self. 大专栏  扯下Python的super()__class__)
        self.id = id

class Rectangle(Polygon):
    def __init__(self, id, width, height):
        super(self.__class__, self).__init__(id)
        #super(Rectangle, self).__init__(id)
        print('in Rectangle, self.__class__ is %s' % self.__class__)
        self.shape = (width, height)

p = Polygon(10)
print p.id

r = Rectangle(5, 1, 2)
print r.id

结果是:

% python test.py
in Polygon, self.__class__ is <class '__main__.Polygon'>
10
in Polygon, self.__class__ is <class '__main__.Rectangle'>
in Rectangle, self.__class__ is <class '__main__.Rectangle'>
5

可以看出来, 在Rectangle初始化时, 通过super调用父类Polygon进行初始化, 而 __class__还是Rectangle.

所以在上一个例子中, Square因为和Rectangle的初始化方法一样, 所以初始化时会调用:

super(Square, self).__init__(id)

即:

Rectangle.__init__(id)

但是实际上Rectangle接收4个参数的初始化, 所以这里报错.

接着考虑, 解决参数个数不一致的问题? 那么就让参数多一致:

#!/usr/bin/env python
# -*- coding: utf-8 -*-

class Polygon(object):
    def __init__(self, id, width, weight):
        print('in Polygon, self.__class__ is %s' % self.__class__)
        self.id = id

class Rectangle(Polygon):
    def __init__(self, id, width, height):
        super(self.__class__, self).__init__(id, width, height)
        #super(Rectangle, self).__init__(id)
        print('in Rectangle, self.__class__ is %s' % self.__class__)
        self.shape = (width, height)

class Square(Rectangle):
    def __init__(self, id, width, height):
        super(self.__class__, self).__init__(id, width, height)
        #super(Rectangle, self).__init__(id)
        print('in Square, self.__class__ is %s' % self.__class__)
        self.shape = (width, height)


p = Polygon(10, 3, 6)
print p.id

r = Rectangle(5, 1, 2)
print r.id

s = Square(20, 2, 4)
print s.id

运行报错:

% python test.py
in Polygon, self.__class__ is <class '__main__.Polygon'>
10
in Polygon, self.__class__ is <class '__main__.Rectangle'>
in Rectangle, self.__class__ is <class '__main__.Rectangle'>
5
Traceback (most recent call last):
  File "test.py", line 30, in <module>
    s = Square(20, 2, 4)
  File "test.py", line 18, in __init__
    super(self.__class__, self).__init__(id, width, height)
  File "test.py", line 11, in __init__
    super(self.__class__, self).__init__(id, width, height)
  File "test.py", line 11, in __init__

  ...

  File "test.py", line 11, in __init__
    super(self.__class__, self).__init__(id, width, height)
  File "test.py", line 11, in __init__
    super(self.__class__, self).__init__(id, width, height)
  File "test.py", line 11, in __init__
    super(self.__class__, self).__init__(id, width, height)
RuntimeError: maximum recursion depth exceeded while calling a Python object

在Rectangle的super这一样发生了无限循环.

在Square的super函数里:

super(self.__class__, self).__init__(id, width, height)

相当于:

Rectangle.__init__(id, width, height)

而此时在Retangle的super函数里, __class__还是等于Square, 所以super(self.__class__, self)就是Rectangle自身, 所以在这里发生了死循环.

唯一的做法就是在Square中重定义__init__, 并且和__class__无关.

这块有点绕, 需要理解下.

扯下Python的super()

标签:style   重定义   overflow   run   class   tun   运行   线程   没有   

原文地址:https://www.cnblogs.com/lijianming180/p/12376348.html

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