Python多重分配和引用 [英] Python multiple assignment and references

问题描述

为什么多重赋值为int而不是列表或其他对象提供不同的引用?

Why does multiple assignment make distinct references for ints, but not lists or other objects?

>>> a = b = 1
>>> a += 1
>>> a is b
>>>     False
>>> a = b = [1]
>>> a.append(1)
>>> a is b
>>>     True

推荐答案

在int示例中，您首先将相同的对象分配给a和b，然后将a重新分配给另一个对象(结果a+1). a现在引用了另一个对象.

In the int example, you first assign the same object to both a and b, but then reassign a with another object (the result of a+1). a now refers to a different object.

在列表示例中，您为a和b分配了相同的对象，但是您并没有做任何更改. append仅更改列表对象的内部状态，而不更改其标识.因此它们保持不变.

In the list example, you assign the same object to both a and b, but then you don't do anything to change that. append only changes the interal state of the list object, not its identity. Thus they remain the same.

如果将a.append(1)替换为a = a + [1]，则会得到不同的对象，因为再次将新对象(a+[1]的结果)分配给a.

If you replace a.append(1) with a = a + [1], you end up with different object, because, again, you assign a new object (the result of a+[1]) to a.

请注意，a+=[1]的行为将有所不同，但这是一个完整的其他问题.

Note that a+=[1] will behave differently, but that's a whole other question.

这篇关于Python多重分配和引用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！