如何解压各种形式的整数在Golang一个字节的缓冲区？(How do I unpack variou

我需要一个字节的缓冲区来提取各个领域。 我想出了这个解决方案：

func (fs *FileSystem) readSB() {
    // fs.f is a *os.File
    buf := make([]byte, 1024)
    fs.f.ReadAt(buf, 1024)

    // Offset: type
    var p *bytes.Buffer

    // 0: uint32
    p = bytes.NewBuffer(buf[0:])
    binary.Read(p, binary.LittleEndian, &fs.sb.inodeCount)
    // 4: uint32
    p = bytes.NewBuffer(buf[4:])
    binary.Read(p, binary.LittleEndian, &fs.sb.blockCount)
    // 20: uint32
    p = bytes.NewBuffer(buf[20:])
    binary.Read(p, binary.LittleEndian, &fs.sb.firstDataBlock)
    // 24: uint32
    p = bytes.NewBuffer(buf[24:])
    binary.Read(p, binary.LittleEndian, &fs.sb.blockSize)
    fs.sb.blockSize = 1024 << fs.sb.blockSize
    // 32: uint32
    p = bytes.NewBuffer(buf[32:])
    binary.Read(p, binary.LittleEndian, &fs.sb.blockPerGroup)
    // 40: uint32
    p = bytes.NewBuffer(buf[40:])
    binary.Read(p, binary.LittleEndian, &fs.sb.inodePerBlock)
}

有没有更美好/成语/直白这样的方式？

• 我想保留偏移明确
• 我想从字节的缓冲区读取，不求从文件中读取时可能。

Answer 1:

你可以通过避免使用每次创建一个新的缓冲.Next()跳过你不想读字节：

{
    // Offset: type
    p := bytes.NewBuffer(buf)

    // 0: uint32
    binary.Read(p, binary.LittleEndian, &fs.sb.inodeCount)

    // 4: uint32
    binary.Read(p, binary.LittleEndian, &fs.sb.blockCount)

    // Skip [8:20)
    p.Next(12)

    // 20: uint32
    binary.Read(p, binary.LittleEndian, &fs.sb.firstDataBlock)

    // 24: uint32
    binary.Read(p, binary.LittleEndian, &fs.sb.blockSize)
    fs.sb.blockSize = 1024 << fs.sb.blockSize

    // Skip [28:32)
    p.Next(4)

    // 32: uint32
    binary.Read(p, binary.LittleEndian, &fs.sb.blockPerGroup)

    // Skip [36:40)
    p.Next(4)

    // 40: uint32
    binary.Read(p, binary.LittleEndian, &fs.sb.inodePerBlock)
}

或者你可以通过避免大块大块的阅读和创建要直接读取用头结构binary.Read ：

type Head struct {
    InodeCount      uint32  //  0:4
    BlockCount      uint32  //  4:8
    Unknown1        uint32  //  8:12
    Unknown2        uint32  // 12:16
    Unknown3        uint32  // 16:20
    FirstBlock      uint32  // 20:24
    BlockSize       uint32  // 24:28
    Unknown4        uint32  // 28:32
    BlocksPerGroup  uint32  // 32:36
    Unknown5        uint32  // 36:40
    InodesPerBlock  uint32  // 40:44
}

func main() {
    var header Head

    err = binary.Read(file, binary.LittleEndian, &header)

    if err != nil {
        log.Fatal(err)
    }

    log.Printf("%#v\n", header)
}

Answer 2:

我有一个包binpacker来处理这些情况

例

示例数据：

buffer := new(bytes.Buffer)
packer := binpacker.NewPacker(buffer)
unpacker := binpacker.NewUnpacker(buffer)
packer.PushByte(0x01)
packer.PushUint16(math.MaxUint16)

解压缩：

var val1 byte
var val2 uint16
var err error
val1, err = unpacker.ShiftByte()
val2, err = unpacker.ShiftUint16()

要么

var val1 byte
var val2 uint16
var err error
unpacker.FetchByte(&val1).FetchUint16(&val2)
unpacker.Error() // Make sure error is nil

文章来源: How do I unpack various form of integers in a byte buffer in Golang?