&^ 这个运算符最早是没有的
测试代码:
```
package main
import "fmt"
const (
hasMonotonic = 1 << 63
nsecMask = 1<<30 - 1
nsecShift = 30
)
func main() {
sec := 1632376722
nsec := 2324512
ynsec := uint64(nsec)
ysec := uint64(sec) << nsecShift
wall := hasMonotonic | uint64(sec)<<nsecShift | uint64(nsec)
fmt.Printf("%b\n", ysec)
fmt.Printf("%b\n", ynsec)
D := int64(3600 * 10e9)
oldnsec := wall & nsecMask
fmt.Printf("%v\n", oldnsec)
newNsec := int32(oldnsec) + int32(D%10e9)
fmt.Printf("%v\n", newNsec)
fmt.Printf("wall :%b\n", wall)
fmt.Printf("nsecMask. :%b\n", nsecMask)
fmt.Printf("wall_move :%b\n", wall>>30<<30)
fmt.Printf("wall&^nsecMask:%b\n", wall&^nsecMask)
wall1 := wall&^nsecMask | uint64(nsec)
fmt.Printf("wall1:%b\n", wall1)
wall2 := wall>>30<<30 | uint64(nsec)
fmt.Printf("wall2:%b\n", wall2)
fmt.Printf("equal:%v\n", wall1 == wall2)
}
```
以上代码我是变通处理,因为wall1 := wall&^nsecMask | uint64(nsec) 中 nsecMask 二进制是全1的,我直接用位移运算>>30<<30代替了。计算结果也是一样的。
但是如果不是 比如 1000000011000000000,就没法处理了。有人知道怎么拆分用& | ^其他语言也有的位运算符来达到 &^一样的结果?
czyt
云在青天水在瓶
这个操作叫做 bit clear ,搜了下,希望对你有所帮助。
1、Setting a bit. Use the bitwise OR operator ( | ) to set a bit. number |= 1 << x; That will set a bit x .
2、Clearing a bit. Use the bitwise AND operator ( & ) to clear a bit. number &= ~(1 << x); That will clear bit x . ...
3、Toggling a bit. The XOR operator ( ^ ) can be used to toggle a bit. number ^= 1 << x;
#1
更多评论
wandercn
https://hotbuild.rustpub.com
<a href="/user/dagongrenzzZ" title="@dagongrenzzZ">@dagongrenzzZ</a> 谢谢,可以了
#3