&^= 运算符有什么作用？

Question

I came across this line in the source code for the Go language itself:

small &^= 4096 - 1

I have never seen the &^= operator before (either in Go or in C) - what does it do?

Answer 1

As the comments have mentioned, &^ is the "bit clear" ( AND NOT ) operator, and a &^= b means a = (a &^= b)

Playground :

package main

import (
    "fmt"
)

func main() {
    var x int64 = 0xdeadbeefface
    fmt.Printf("   x = %064b %012x\n", x, x)

    var y int64 = 0xffff0000
    fmt.Printf("   y = %064b %012x\n", y, y)

    x &^= y
    fmt.Printf("x&^y = %064b %012x\n", x, x)
}

   x = 0000000000000000110111101010110110111110111011111111101011001110 deadbeefface
   y = 0000000000000000000000000000000011111111111111110000000000000000 0000ffff0000
x&^y = 0000000000000000110111101010110100000000000000001111101011001110 dead0000face

And small &^= 4096 - 1 ?

Since 4095 - 1 == 0b0000111111111111 , it would mean "set the 12 least significant bits of a number to zero".

func main() {
    small := 0xdeadbeefface
    fmt.Printf("             small : %064b %012x\n", small, small)
    small &^= 4096 - 1
    fmt.Printf("small &^= 4096 - 1 : %064b %012x\n", small, small)
}

             small : 0000000000000000110111101010110110111110111011111111101011001110 deadbeefface
small &^= 4096 - 1 : 0000000000000000110111101010110110111110111011111111000000000000 deadbeeff000