Python就地操作符函数如何与标准操作符函数不同？ [英] How are Python in-place operator functions different than the standard operator functions?
问题描述
为什么不是 operator.iadd（x，y）等于 z = x; z + = y ？
Why isn't operator.iadd(x, y) equivalent to z = x; z += y? And how does operator.iadd(x, y) differ from operator.add(x, y)?
从 docs ：
From the docs:
许多操作都有一个in放置
版本。以下函数
提供了比原来的
语法更原始的对
就地操作符的访问;例如，
语句x + = y等价于x =
operator.iadd（x，y）。
的另一种解释是，z =
operator.iadd（x，y）相当于
复合语句z = x; z + = y。
Many operations have an "in-place" version. The following functions provide a more primitive access to in-place operators than the usual syntax does; for example, the statement x += y is equivalent to x = operator.iadd(x, y). Another way to put it is to say that z = operator.iadd(x, y) is equivalent to the compound statement z = x; z += y.
相关问题，但我对Python类方法不感兴趣;只是对内置Python类型的常规操作符而已。
Related question, but I am not interested in Python class methods; just regular operators on built-in Python types.
推荐答案
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']
这篇关于Python就地操作符函数如何与标准操作符函数不同？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！
