Question

我必须反序列化来自另一个应用程序的一些二进制消息。我很乐意使用restruct.io，但消息结构中的某些字段使用“非标准”位数（5位，3位，...... 10位......）。

有没有办法处理这种类型的结构？我一直在寻找一段时间没有任何成功，所以任何帮助都会受到欢迎。

提前致谢

我会试着举个例子来澄清我的问题。鉴于代码：

package main

import (
    "encoding/binary"
    "fmt"

    restruct "gopkg.in/restruct.v1"
)

type MessageType uint8

const (
    MessageTypeOne MessageType = iota + 1
    MessageTypeTwo
    MessageTypeThree
)

// Message is the data to deserialize from the binary stream
type Message struct {
    Length     uint32      `struct:"uint32"` // message size in bytes (including length)
    Type       MessageType `struct:"uint8"`
    Version    uint8       `struct:"uint8:4"` // Just need 4 bits
    Subversion uint8       `struct:"uint8:2"` // just need 2 bits
    Optional   uint8       `struct:"uint8:1"` // just one bit --> '1' means next field is NOT present
    NodeName   string      ``
    ANumber    uint16      `struct:"uint16:10"` // just need 10 bits
}

// (length(4)+type(1)+(version(4bits)+Subversion(2bits)+Optional(1bit))) = 6 bytes
// need 32bit alignment
func main() {
    var inStream = []byte{0x08, // just 8 bytes needed
        0x01,       // messge type = MessageTypeOne
        0x4a,       // Version=0100 Subversion=10 Optional=1 ANumber = 0 (MSB bit)
        0x00, 0x60, // ANumber(000 0000 011) Padding = 0 0000 for 32 bits alignment
    }
    var msg Message

    err := restruct.Unpack(inStream, binary.BigEndian, &msg)
    if err != nil {
        panic(err)
    }
    fmt.Println(msg)
    // Expected:
    // msg.Length = 8
    // msg.Type = 1
    // msg.Version = 4
    // msg.Subversion = 2
    // msg.Optional = 1
    // msg.NodeName = ""
    // msg.ANumber = 3
}

我将从TCP连接接收inStream，并希望反序列化二进制数据并获得具有预期值的Message结构...

希望这能澄清我的问题。

再次感谢;）

Answer 1

虽然可能没有用于实现此自定义struct包装的通用包，但您可以轻松创建自己的方法，只提取每个字段所需的位。

func (m *Message) UnmarshalBinary(data []byte) error {
    m.Length = binary.BigEndian.Uint32(data[:4])

    if int(m.Length) > len(data) {
        return fmt.Errorf("not enough bytes")
    }

    m.Type = MessageType(data[4])

    m.Version = data[5] >> 4
    m.Subversion = data[5] >> 2 & 0x03
    m.Optional = data[5] >> 1 & 0x01

    // move the index for ANumber back if there's an optional string
    idx := 6
    if m.Optional == 0 {
        // remove the last two bytes for ANumber
        end := int(m.Length) - 2
        m.NodeName = string(data[6:end])
        idx = end
    }

    m.ANumber = uint16(data[idx]&0xc0)<<2 | uint16(data[idx]&0x3f<<2|data[idx+1]>>6)
    return nil

}

当然，您可以添加更多绑定检查以返回错误，而不是在索引越界时让这种恐慌。

我稍微修改了您的inStream切片以符合您的定义，您可以在此处查看示例输出：https://play.golang.org/p/FoNoazluOF

Answer 2

我一直在研究restruct.io的一些补丁，以便能够使用位域....仍未完全测试但似乎有效......

测试后会尝试发送拉取请求...

func (e *encoder) writeBits(f field, inBuf []byte) {

    var inputLength uint8 = uint8(len(inBuf))

    if f.BitSize == 0 {
        // Having problems with complex64 type ... so we asume we want to read all
        //f.BitSize = uint8(f.Type.Bits())
        f.BitSize = 8 * inputLength
    }

    // destPos: Destination position ( in the result ) of the first bit in the first byte
    var destPos uint8 = 8 - e.bitCounter

    // originPos: Original position of the first bit in the first byte
    var originPos uint8 = f.BitSize % 8
    if originPos == 0 {
        originPos = 8
    }

    // numBytes: number of complete bytes to hold the result
    var numBytes uint8 = f.BitSize / 8

    // numBits: number of remaining bits in the first non-complete byte of the result
    var numBits uint8 = f.BitSize % 8

    // number of positions we have to shift the bytes to get the result
    var shift uint8
    if originPos > destPos {
        shift = originPos - destPos
    } else {
        shift = destPos - originPos
    }
    shift = shift % 8

    var inputInitialIdx uint8 = inputLength - numBytes
    if numBits > 0 {
        inputInitialIdx = inputInitialIdx - 1
    }

    if originPos < destPos {
        // shift left
        carry := func(idx uint8) uint8 {
            if (idx + 1) < inputLength {
                return (inBuf[idx+1] >> (8 - shift))
            }
            return 0x00

        }
        mask := func(idx uint8) uint8 {
            if idx == 0 {
                return (0x01 << destPos) - 1
            }
            return 0xFF
        }
        var idx uint8 = 0
        for inIdx := inputInitialIdx; inIdx < inputLength; inIdx++ {
            e.buf[idx] |= ((inBuf[inIdx] << shift) | carry(inIdx)) & mask(idx)
            idx++
        }

    } else {
        // originPos >= destPos => shift right
        var idx uint8 = 0
        // carry : is a little bit tricky in this case because of the first case
        // when idx == 0 and there is no carry at all
        carry := func(idx uint8) uint8 {
            if idx == 0 {
                return 0x00
            }
            return (inBuf[idx-1] << (8 - shift))
        }
        mask := func(idx uint8) uint8 {
            if idx == 0 {
                return (0x01 << destPos) - 1
            }
            return 0xFF
        }
        inIdx := inputInitialIdx
        for ; inIdx < inputLength; inIdx++ {
            //note: Should the mask be done BEFORE the OR with carry?
            e.buf[idx] |= ((inBuf[inIdx] >> shift) | carry(inIdx)) & mask(idx)

            idx++
        }
        if ((e.bitCounter + f.BitSize) % 8) > 0 {
            e.buf[idx] |= carry(inIdx)
        }
    }

    //now we should update buffer and bitCounter
    e.bitCounter = (e.bitCounter + f.BitSize) % 8

    // move the head to the next non-complete byte used
    headerUpdate := func() uint8 {
        if (e.bitCounter == 0) && ((f.BitSize % 8) != 0) {
            return (numBytes + 1)
        }
        return numBytes
    }

    e.buf = e.buf[headerUpdate():]

    return
}

如何反序列化非标准大小的字段？

2 个答案: