英文:
How to reverse a binary number?
问题
我是你的中文翻译助手，以下是你要翻译的内容：
我是Golang的新手，对于有经验的Golang开发人员来说，这应该是一个简单的问题。我尝试做来自Spotify的相同测试，看看在Golang中我们能有多快 
英文:
I'm newbie in Golan, this should be an easy question for experienced golang devs. I try to do the same test from Spotify to see how fast we can go in Golang
答案1
得分: 6
通常的位操作的 C 解决方案可以直接转换为 Go 语言。
package main
import "fmt"
func BitReverse32(x uint32) uint32 {
x = (x&0x55555555)<<1 | (x&0xAAAAAAAA)>>1
x = (x&0x33333333)<<2 | (x&0xCCCCCCCC)>>2
x = (x&0x0F0F0F0F)<<4 | (x&0xF0F0F0F0)>>4
x = (x&0x00FF00FF)<<8 | (x&0xFF00FF00)>>8
return (x&0x0000FFFF)<<16 | (x&0xFFFF0000)>>16
}
func main() {
cases := []uint32{0x1, 0x100, 0x1000, 0x1000000, 0x10000000, 0x80000000, 0x89abcdef}
for _, c := range cases {
fmt.Printf("%08x -> %08x\n", c, BitReverse32(c))
}
}
英文:
The usual bit-twiddling C solutions translate immediately to Go.
package main
import "fmt"
func BitReverse32(x uint32) uint32 {
x = (x&0x55555555)<<1 | (x&0xAAAAAAAA)>>1
x = (x&0x33333333)<<2 | (x&0xCCCCCCCC)>>2
x = (x&0x0F0F0F0F)<<4 | (x&0xF0F0F0F0)>>4
x = (x&0x00FF00FF)<<8 | (x&0xFF00FF00)>>8
return (x&0x0000FFFF)<<16 | (x&0xFFFF0000)>>16
}
func main() {
cases := []uint32{0x1, 0x100, 0x1000, 0x1000000, 0x10000000, 0x80000000, 0x89abcdef}
for _, c := range cases {
fmt.Printf("%08x -> %08x\n", c, BitReverse32(c))
}
}
答案2
得分: 2
注意：自2013年以来，您现在有一个专用的math/bits包，它是在Go 1.9（2017年8月）中引入的。
它提供了一系列的Reverse()和ReverseBytes()函数：不再需要自己实现了。
此外，在大多数架构上，编译器还会将此包中的函数识别为内部函数，以提供额外的性能。
英文:
Note: since 2013, you now have a dedicate math/bits package with Go 1.9 (August 2017).
And it does come with a collection of Reverse() and ReverseBytes() functions: no need to implement one anymore.
Plus, on most architectures, functions in this package are additionally recognized by the compiler and treated as intrinsics for additional performance.
答案3
得分: 1
最直接的解决方案是使用strconv将位转换为数字，然后通过位移操作将数字反转。我不确定它的速度有多快，但应该可以工作。
package main
import "fmt"
import "strconv"
func main() {
bits := "10100001"
bits_number := 8
number, _ := strconv.ParseUint(bits, 2, bits_number)
r_number := number - number // reserve type
for i := 0; i < bits_number; i++ {
r_number <<= 1
r_number |= number & 1
number >>= 1
}
fmt.Printf("%s [%d]\n", strconv.FormatUint(r_number, 2), r_number)
}
英文:
The most straight-forward solution would be converting the bits into a number with strconv and then reversing the number by shifting the bits. I'm not sure how fast it would be, but it should work.
package main
import "fmt"
import "strconv"
func main() {
bits := "10100001"
bits_number := 8
number, _ := strconv.ParseUint(bits, 2, bits_number)
r_number := number - number // reserve type
for i := 0; i < bits_number; i++ {
r_number <<= 1
r_number |= number & 1
number >>= 1
}
fmt.Printf("%s [%d]\n", strconv.FormatUint(r_number, 2), r_number)
}
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