如何在 Go 中将 [4]uint8 转换为 uint32？

Question

如何将go的类型从uint8转换为unit32？
只需代码：

package main

import (
    "fmt"
)

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

~>6g test.go && 6l -o 测试 test.6 && ./测试
test.go:10: 无法将 uInt8（类型 []uint8）转换为 uint32 类型

Answer 1

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)
}

输出：

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

Go 二进制包 函数是通过一系列转换实现的。

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
}

Answer 2

您是否正在尝试以下？

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