在golang中将[] uint32转换为[] byte，反之亦然

Question

什么是最有效的方式（性能）转换[]uint32，并从[]byte在Golang？

例如：

func main() {
   source := []uint32{1,2,3}
   dest := make([]byte, 4 * len(source))
   // source to dest
   // ?
   check := len(dest)/4
   // dest to check
   // ?
}

我有一个解决方案，但它包含div，mod和乘法

package main
import (
    "fmt"
)
func main() {
    source := []uint32{1,2,3}
    dest := make([]byte, 4*len(source))
    fmt.Println(source)
    for start, v := range source {
       dest[start*4+0] = byte(v % 256)
       dest[start*4+1] = byte(v / 256 % 256)
       dest[start*4+2] = byte(v / 256 / 256 % 256)
       dest[start*4+3] = byte(v / 256/ 256/ 256% 256)
    }
    fmt.Println(dest)
    check := make([]uint32,cap(dest)/4)
    for start := 0; start<len(check); start++ {
       check[start] = uint32(dest[start*4+0]) + uint32(dest[start*4+1]) * 256 + uint32(dest[start*4+2]) * 256 * 256 + uint32(dest[start*4+3]) * 256 * 256 * 256
    }  
    fmt.Println(check)
}

Answer 1

我怀疑你是在追逐这样的游乐场

调整LittleEndian为BigEndian适当的

package main

import (
    "bytes"
    "encoding/binary"
    "fmt"
)

func main() {
    buf := new(bytes.Buffer)
    source := []uint32{1, 2, 3}
    err := binary.Write(buf, binary.LittleEndian, source)
    if err != nil {
        fmt.Println("binary.Write failed:", err)
    }
    fmt.Printf("Encoded: % x\n", buf.Bytes())

    check := make([]uint32, 3)
    rbuf := bytes.NewReader(buf.Bytes())
    err = binary.Read(rbuf, binary.LittleEndian, &check)
    if err != nil {
        fmt.Println("binary.Read failed:", err)
    }
    fmt.Printf("Decoded: %v\n", check)

}