英文:
What is the "&^" operator in golang?
问题
我无法真正使用"AND NOT"进行谷歌搜索并获得任何有用的结果，这个运算符到底是什么，我该如何在像C语言这样的语言中实现它？我查看了规范，但里面没有什么有用的信息，只有一个列表，上面写着&^（AND NOT）。
英文:
I can't really google the name AND NOT and get any useful results, what exactly is this operator, and how could I do this in a language like C? I checked the specification, and there is nothing helpful in there but a list that says it's &^ (AND NOT).
答案1
得分: 56
Go表达式x &^ y的C等效表达式只是x & ~y。这实际上是“x与（y的按位取反）”的意思。
在规范的算术运算符部分中，描述了&^作为“位清除”操作，这给出了你想要使用它的一个想法。作为两个单独的操作，~y将每个位转换为零，然后清除x中对应的位。每个零位将被转换为一，这将保留x中对应的位。
因此，如果你将x | y视为根据掩码常量y打开x的特定位的方法，那么x &^ y则相反，关闭那些相同的位。
英文:
The C equivalent of the Go expression x &^ y is just x & ~y. That is literally "x AND (bitwise NOT of y)".
In the arithmetic operators section of the spec describes &^ as a "bit clear" operation, which gives an idea of what you'd want to use it for. As two separate operations, ~y will convert each one bit to a zero, which will then clear the corresponding bit in x. Each zero bit will be converted to a one, which will preserve the corresponding bit in x.
So if you think of x | y as a way to turn on certain bits of x based on a mask constant y, then x &^ y is doing the opposite and turns those same bits off.
答案2
得分: 6
> &^ 运算符是位清除（AND NOT）运算符：在表达式 z = x &^ y 中，如果 y 的对应位为 1，则 z 的对应位为 0；否则，z 的对应位等于 x 的对应位。
来自《The Go Programming Language》
示例：
package main
import "fmt"
func main(){
var x uint8 = 1
var y uint8 = 1 << 2
fmt.Printf("%08b\n", x &^ y);
}
结果：
> 00000001
英文:
> The &^ operator is bit clear (AND NOT): in the expression z = x &^ y,
> each bit of z is 0 if the corresponding bit of y is 1; otherwise it
> equals the corresponding bit of x.
From The Go Programming Language
Example:
package main
import "fmt"
func main(){
var x uint8 = 1
var y uint8 = 1 << 2
fmt.Printf("%08b\n", x &^ y);
}
Result:
> 00000001
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