首先，您需要了解 __add __ 和 __ iadd __ 。

对象的 __ add __ 方法是常规添加：它接受两个参数，返回它们的总和，并且不修改任何参数。

An object's __add__ method is regular addition: it takes two parameters, returns their sum, and doesn't modify either parameter.

一个对象的 __iadd __ 方法也需要两个参数，但是在原地进行更改，修改第一个参数的内容。因为这需要对象变异，所以不可变类型（如标准数字类型）不应该有 __ iadd __ 方法。

An object's __iadd__ method also takes two parameters, but makes the change in-place, modifying the contents of the first parameter. Because this requires object mutation, immutable types (like the standard number types) shouldn't have an __iadd__ method.

a + b 使用 __ add __ 。如果它存在， a + = b 使用 __ iadd __ 如果没有，它通过 __ add __ 来模拟它，如 tmp = a + b; a = tmp 。 operator.add 和 operator.iadd 以相同的方式不同。

a + b uses __add__. a += b uses __iadd__ if it exists; if it doesn't, it emulates it via __add__, as in tmp = a + b; a = tmp. operator.add and operator.iadd differ in the same way.

另一个问题是： operator.iadd（x，y）不等于 z = x; z + = y ，因为如果没有 __ iadd __ 存在 __ add __ 将被使用。您需要指定值以确保结果存储在两种情况下： x = operator.iadd（x，y）。

To the other question: operator.iadd(x, y) isn't equivalent to z = x; z += y, because if no __iadd__ exists __add__ will be used instead. You need to assign the value to ensure that the result is stored in both cases: x = operator.iadd(x, y).

你可以很容易地看到它：

You can see this yourself easily enough:

import operator
a = 1
operator.iadd(a, 2)
# a is still 1, because ints don't have __iadd__; iadd returned 3

b = ['a']
operator.iadd(b, ['b'])
# lists do have __iadd__, so b is now ['a', 'b']