Python 就地运算符函数与标准运算符函数有何不同? [英] How are Python in-place operator functions different than the standard operator functions?

问题描述

来自文档:

许多操作都有一个就地"版本.以下功能提供更原始的访问就地操作员比平时语法确实；例如，语句 x += y 等价于 x =operator.iadd(x, y).另一种方式就是说z =operator.iadd(x, y) 等价于复合语句 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.

问题:

• 为什么 operator.iadd(x, y) 不等于 z = x;z += y?

operator.iadd(x, y) 与 operator.add(x, y) 有何不同?

相关问题，但我对 Python 类方法不感兴趣；只是内置 Python 类型的常规运算符.

Related question, but I'm not interested in Python class methods; just regular operators on built-in Python types.

推荐答案

首先，你需要了解__add__和__iadd__的区别.

First, you need to understand the difference between __add__ and __iadd__.

对象的 __add__ 方法是常规加法:它接受两个参数，返回它们的和，并且不修改任何一个参数.

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__ 方法也接受两个参数，但进行了就地更改，修改了第一个参数的内容.因为这需要对象突变，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__.a += b 使用 __iadd__ 如果它存在；如果没有，它通过 __add__ 模拟它，如 tmp = a + b;a = tmp.operator.add 和 operator.iadd 的区别相同.

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:operator.iadd(x, y) 不等于 z = x;z += y，因为如果没有 __iadd__ 存在，则将使用 __add__ 代替.您需要分配值以确保在两种情况下都存储结果: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这一点:

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']

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