了解Python中的可变性 [英] Understanding mutability in Python

问题描述

我有以下代码段:

a = [1,2,3]
b = a
b = [4,5,6] # a doesnt change

和这个:

a = [1,2,3]
b = a
b[0] = 5 # a also changes

b的初始化如何决定a的可变性?

How is b's initialization playing a part in deciding the mutability of a?

推荐答案

创建列表并将其分配给变量时，例如

When you create a list and assign it to a variable, like

a = [1, 2, 3]

您创建一个对象，变量a包含对该对象的引用.当你做

You create an object, and the variable a holds a reference to that object. When you do

b = a

您在b中分配了相同的引用，因此a和b现在都指向您的列表.因此，当您执行

You assign the same reference in b, so now both a and b point to your list. So when you execute

b[0] = 5

您更改了同一列表.

您可以通过使用id()函数看到此操作，该函数返回对象的内存地址:

You can see this in action by using the id() function, which returns the memory address of the object:

>>> a = [1, 2, 3]
>>> b = a
>>> id(a), id(b)
(140454191340720, 140454191340720)

它们是相同的.关键是，a和b本身不是列表，它们指向列表.

They are identical. The point is, a and b are not lists themselves, they point to a list.

当您执行以下操作时:

a = [1, 2, 3]
b = a
b = [2, 3, 4]

您首先为b分配了指向它所指向的列表的引用，然后又为其分配了新引用.

You first assigned to b a reference to the list that a points to, but then you assign a new reference to it.

顺便说一句，当您按照以下方式操作时，这可能会在后面咬住您

By the way, this can bite you in the behind when you do something along the lines of

def foo (a=[]):
    a.append(42)
    return a

由于参数a在每次调用时都指向相同的列表，因此，如果您在没有参数的情况下调用此函数5次，则该列表将包含5x 42:

since the argument a points to the same list on every invocation, if you call this function 5 times without arguments, the list will contain 5x 42:

foo(); foo(); foo(); foo();
print(foo())
>>> [42, 42, 42, 42]

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