How to convert [4]uint8 into uint32 in Go?

Question

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          


        

Answer 1

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
}

Answer 2

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
}