Python：通过引用和切片分配传递 [英] Python: Pass by reference and slice assignment

问题描述

在Python中，列表通过引用传递给函数，对吧？

In Python, lists are passed by reference to functions, right?

如果是这样，那么发生了什么？

If that is so, what's happening here?

>>> def f(a):
...     print(a)
...     a = a[:2]
...     print(a)
...
>>> b = [1,2,3]
>>> f(b)
[1, 2, 3]
[1, 2]
>>> print(b)
[1, 2, 3]
>>>

推荐答案

事实上，对象是通过引用传递的，但 a = a [：2] 基本上创建一个新的局部变量，指向列表的切片。

Indeed the objects are passed by reference but a = a[:2] basically creates a new local variable that points to slice of the list.

修改列表对象到位后，您可以将其分配给其切片（切片分配）。

To modify the list object in place you can assign it to its slice(slice assignment).

考虑 a 和 b 在这里相当于你的全局 b 和本地 a ，这里赋值 a 不会影响 b ：

Consider a and b here equivalent to your global b and local a, here assigning a to new object doesn't affect b:

>>> a = b = [1, 2, 3]    
>>> a = a[:2]  # The identifier `a` now points to a new object, nothing changes for `b`.
>>> a, b
([1, 2], [1, 2, 3])
>>> id(a), id(b)
(4370921480, 4369473992)  # `a` now points to a different object

切片分配按预期工作：

Slice assignment work as expected:

>>> a = b = [1, 2, 3]    
>>> a[:] = a[:2]  # Updates the object in-place, hence affects all references.
>>> a, b
([1, 2], [1, 2])
>>> id(a), id(b)
(4370940488, 4370940488)  # Both still point to the same object

---

相关：切片整个列表和直接分配的切片分配有什么区别？

这篇关于Python：通过引用和切片分配传递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！