0 
class A(object):
    aalist = []
    ai = 0
    def __init__(self):
    self.ablist = list()
    def add(self):
        self.aalist.append(["a","a"])
        self.ablist.append(["b","b"])
        self.ai = self.ai + 1
class B(A):
    def __init__(self):
        A.__init__(self)
        self.bblist = list()
    def badd(self):
        self.bblist.append(['c','c'])
for i in range(1,4):
    c = B()
    c.add()
    c.badd()
    print c.aalist,c.ablist,c.bblist,c.ai

运行这些代码，结果提示：

[['a', 'a']] [['b', 'b']] [['c', 'c']] 1
[['a', 'a'], ['a', 'a']] [['b', 'b']] [['c', 'c']] 1
[['a', 'a'], ['a', 'a'], ['a', 'a']] [['b', 'b']] [['c', 'c']] 1

我不知道为什么 list 在循环中更新，而 int 似乎保持静态

4

1 回答 1

2

You're doing completely different operations with them. Note that you assign a new integer to self.ai when you do self.ai = self.ai + 1. When you work with the list, you work on it in place using append. Also note that since you're working with an int (which is immutable), you can't change it's value in place.

There's also evidence that you're exploring class variables vs. instance variables here. Note that in your example, ablist and bblist are instance variable whereas aalist is a class variable. You can access them all the same way within a method (self.XXlist), but when you mutate self.aalist, you'll mutate the version on the class. In other words, self.aalist is A.aalist will return True. This allows all instances of your class to share the same list. When one updates the list, all the other instances will know immediately (unless they bind an instance level attribute with the same name -- That variable will then take precedence in the attribute lookup).

于 2012-10-25T14:37:37.120 回答