Python类中的属性是否共享? [英] Are the attributes in a Python class shared or not?

问题描述

以下代码困扰我:-

class mytest:
    name="test1"
    tricks=list()
    def __init__(self,name):
        self.name=name
        #self.tricks=[name]
        self.tricks.append(name)

t1=mytest("hello world")
t2=mytest("bye world")
print t1.name,t2.name
print t1.tricks,t2.tricks

输出为:-

hello world bye world
['hello world', 'bye world'] ['hello world', 'bye world']

表示列表tricks由两个实例t1和t2共享，这在

meaning that the list tricks is shared by the two instances t1 and t2, which has been explained in the section 9.3.5 of https://docs.python.org/3/tutorial/classes.html

但是，如果我执行以下代码:-

However, if I execute the following code:-

class mytest:
    name="test1"
    tricks=list()
    def __init__(self,name):
        self.name=name
        self.tricks=[name]
        self.tricks.append(name)

t1=mytest("hello world")
t2=mytest("bye world")
x=t1.tricks
if type(x) is list:
    print 'a list'
elif type(x) is tuple:
    print 'a tuple'
else:
    print 'neither a tuple or a list'
print t1.name,t2.name
print t1.tricks,t2.tricks

输出如下:-

a list
hello world bye world
['hello world', 'hello world'] ['bye world', 'bye world']

现在，列表tricks似乎不再由这两个实例t1和t2共享. 我的问题是，这里的机制是什么?

Now it seems that the list tricks is no longer shared by those two instances t1 and t2. My question is, what are the mechanics here?

推荐答案

区别在于，在第二个示例中，您将创建一个新列表self.tricks作为对象的属性:

The difference is that in your second example you are creating a new list, self.tricks, as an attribute of the object:

def __init__(self,name):
    self.name=name
    self.tricks=[name]    # <-- this is creating the new attribute for the object
    self.tricks.append(name)

第一个示例之所以有效是因为Python解析名称的方式:如果在对象中找不到self.tricks(因为尚未创建)，那么它将尝试将其作为类的成员来查找.由于存在技巧，因此您可以访问它.

The first example works because of Python's way of resolving the names: If self.tricks cannot be found in the object (because it hasn't been created), then it tries to find it as a member of the class. Since tricks is there, then you can access it.

如果您在第二个示例中尝试使用mytest.tricks，这可能对您很清楚:

It may become clear to you if you try to use mytest.tricks in your second example:

def __init__(self,name):
    self.name=name
    mytest.tricks=[name]    # <-- this is accesing the class attribute instead
    self.tricks.append(name)

这将输出您的实际期望.

That will output what you are actually expecting.

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