BMP檔案頭定義：

WORD 兩個位元組 16bit

DWORD 四個位元組 32bit

package main 
import (
 "encoding/binary"
 "fmt"
 "os"
)
 
func main() {
 file,err := os.Open("tim.bmp")
 if err != nil {
  fmt.Println(err)
  return
 }
 
 defer file.Close() 
 //type拆成兩個byte來讀
 var headA,headB byte
 //Read第二個引數位元組序一般windows/linux大部分都是LittleEndian,蘋果系統用BigEndian
 binary.Read(file,binary.LittleEndian,&headA)
 binary.Read(file,&headB)
 
 //檔案大小
 var size uint32
 binary.Read(file,&size)
 
 //預留位元組
 var reservedA,reservedB uint16
 binary.Read(file,&reservedA)
 binary.Read(file,&reservedB)
 
 //偏移位元組
 var offbits uint32
 binary.Read(file,&offbits) 
 fmt.Println(headA,headB,size,reservedA,reservedB,offbits) 
}

執行結果

66 77 196662 0 0 54

使用結構體方式

package main 
import (
 "encoding/binary"
 "fmt"
 "os"
)
 
type BitmapInfoHeader struct {
 Size   uint32
 Width   int32
 Height   int32
 Places   uint16
 BitCount  uint16
 Compression uint32
 SizeImage  uint32
 XperlsPerMeter int32
 YperlsPerMeter int32
 ClsrUsed  uint32
 ClrImportant uint32
}
 
func main() {
 file,&offbits)
 
 fmt.Println(headA,offbits)
 
 infoHeader := new(BitmapInfoHeader)
 binary.Read(file,infoHeader)
 fmt.Println(infoHeader) 
}

執行結果：

66 77 196662 0 0 54

&{40 256 256 1 24 0 196608 3100 3100 0 0}

補充：golang(Go語言) byte/[]byte 與 二進位制形式字串 互轉

效果

把某個位元組或位元組陣列轉換成字串01的形式，一個位元組用8個”0”或”1”字元表示。

比如：

byte(3) –> “00000011”
[]byte{1,2,3} –> “[00000001 00000010 00000011]”
“[00000011 10000000]” –> []byte{0x3,0x80}

開源庫 biu

實際上我已經將其封裝到一個開源庫了(biu)，其中的一個功能就能達到上述效果：

//byte/[]byte -> string
bs := []byte{1,3}
s := biu.BytesToBinaryString(bs)
fmt.Println(s) //[00000001 00000010 00000011]
fmt.Println(biu.ByteToBinaryString(byte(3))) //00000011
//string -> []byte
s := "[00000011 10000000]"
bs := biu.BinaryStringToBytes(s)
fmt.Printf("%#v\n",bs) //[]byte{0x3,0x80}

程式碼實現

const (
 zero = byte('0')
 one = byte('1')
 lsb = byte('[') // left square brackets
 rsb = byte(']') // right square brackets
 space = byte(' ')
)
var uint8arr [8]uint8
// ErrBadStringFormat represents a error of input string's format is illegal .
var ErrBadStringFormat = errors.New("bad string format")
// ErrEmptyString represents a error of empty input string.
var ErrEmptyString = errors.New("empty string")
func init() {
 uint8arr[0] = 128
 uint8arr[1] = 64
 uint8arr[2] = 32
 uint8arr[3] = 16
 uint8arr[4] = 8
 uint8arr[5] = 4
 uint8arr[6] = 2
 uint8arr[7] = 1
}
// append bytes of string in binary format.
func appendBinaryString(bs []byte,b byte) []byte {
 var a byte
 for i := 0; i < 8; i++ {
  a = b
  b <<= 1
  b >>= 1
  switch a {
  case b:
   bs = append(bs,zero)
  default:
   bs = append(bs,one)
  }
  b <<= 1
 }
 return bs
}
// ByteToBinaryString get the string in binary format of a byte or uint8.
func ByteToBinaryString(b byte) string {
 buf := make([]byte,8)
 buf = appendBinaryString(buf,b)
 return string(buf)
}
// BytesToBinaryString get the string in binary format of a []byte or []int8.
func BytesToBinaryString(bs []byte) string {
 l := len(bs)
 bl := l*8 + l + 1
 buf := make([]byte,bl)
 buf = append(buf,lsb)
 for _,b := range bs {
  buf = appendBinaryString(buf,b)
  buf = append(buf,space)
 }
 buf[bl-1] = rsb
 return string(buf)
}
// regex for delete useless string which is going to be in binary format.
var rbDel = regexp.MustCompile(`[^01]`)
// BinaryStringToBytes get the binary bytes according to the
// input string which is in binary format.
func BinaryStringToBytes(s string) (bs []byte) {
 if len(s) == 0 {
  panic(ErrEmptyString)
 }
 s = rbDel.ReplaceAllString(s,"")
 l := len(s)
 if l == 0 {
  panic(ErrBadStringFormat)
 }
 mo := l % 8
 l /= 8
 if mo != 0 {
  l++
 }
 bs = make([]byte,l)
 mo = 8 - mo
 var n uint8
 for i,b := range []byte(s) {
  m := (i + mo) % 8
  switch b {
  case one:
   n += uint8arr[m]
  }
  if m == 7 {
   bs = append(bs,n)
   n = 0
  }
 }
 return
}

以上為個人經驗，希望能給大家一個參考，也希望大家多多支援我們。如有錯誤或未考慮完全的地方，望不吝賜教。