如何在Go中将[4] uint8转换为uint32？

问题描述：

how to convert go's type from uint8 to unit32?
Just code:

package main

import (
    "fmt"
)

func main() {
    uInt8 := []uint8{0,1,2,3}
    var uInt32 uint32
    uInt32 = uint32(uInt8)
    fmt.Printf("%v to %v
", uInt8, uInt32)
}

~>6g test.go && 6l -o test test.6 && ./test
test.go:10: cannot convert uInt8 (type []uint8) to type uint32

答

package main

import (
    "encoding/binary"
    "fmt"
)

func main() {
    u8 := []uint8{0, 1, 2, 3}
    u32LE := binary.LittleEndian.Uint32(u8)
    fmt.Println("little-endian:", u8, "to", u32LE)
    u32BE := binary.BigEndian.Uint32(u8)
    fmt.Println("big-endian:   ", u8, "to", u32BE)
}

Output:

little-endian: [0 1 2 3] to 50462976
big-endian:    [0 1 2 3] to 66051

The Go binary package functions are implemented as a series of shifts.

func (littleEndian) Uint32(b []byte) uint32 {
    return uint32(b[0]) | uint32(b[1])<<8 | uint32(b[2])<<16 | uint32(b[3])<<24
}

func (bigEndian) Uint32(b []byte) uint32 {
    return uint32(b[3]) | uint32(b[2])<<8 | uint32(b[1])<<16 | uint32(b[0])<<24
}

答

Are you trying to the following?

t := []int{1, 2, 3, 4}
s := make([]interface{}, len(t))
for i, v := range t {
    s[i] = v
}

答

Opposite question: How to convert uint32 to uint8 in Go?

I just want the lowest address byte, So, I meet the question. But There is no answer about it.

This question helps me a lot. I write my answer here.

var x uint32
x = 0x12345678
y := uint8(x & 0xff)

If you want get each byte of uint32, you may need this.

binary.BigEndian.PutUint32(buffer, Uint32Number)

如何在Go中将[4] uint8转换为uint32？

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