list.__iadd__ 和 list.__add__ 的不同行为 [英] Different behaviour for list.__iadd__ and list.__add__

问题描述

考虑以下代码:

<预><代码>>>>x = y = [1, 2, 3, 4]>>>x += [4]>>>X[1, 2, 3, 4, 4]>>>是[1, 2, 3, 4, 4]

然后考虑这个:

<预><代码>>>>x = y = [1, 2, 3, 4]>>>x = x + [4]>>>X[1, 2, 3, 4, 4]>>>是[1, 2, 3, 4]

为什么这两者有区别?

(是的，我试过搜索这个).

解决方案

__iadd__ 改变列表，而 __add__ 返回一个 new 列表，如演示.

x += y 的表达式首先尝试调用 __iadd__，如果失败，则调用 __add__ 跟随一个赋值(见 Sven 的评论进行小幅修正).由于 list 有 __iadd__ 那么它就做了一点o 变异魔法".

consider the following code:

>>> x = y = [1, 2, 3, 4]
>>> x += [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4, 4]

and then consider this:

>>> x = y = [1, 2, 3, 4]
>>> x = x + [4]
>>> x
[1, 2, 3, 4, 4]
>>> y
[1, 2, 3, 4]

Why is there a difference these two?

(And yes, I tried searching for this).

解决方案

__iadd__ mutates the list, whereas __add__ returns a new list, as demonstrated.

An expression of x += y first tries to call __iadd__ and, failing that, calls __add__ followed an assignment (see Sven's comment for a minor correction). Since list has __iadd__ then it does this little bit 'o mutation magic.

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