Store UDP packet in Structure using C++

Question

I am new in C++ programming. I am trying to create a Software which communicates to Camera. I am able to communicate Camera to my Software. Through WireShark I can see that camera is sending me packet which is hex representation.

I want to store that packets in structure.

For Example:-

The packet I get is

 char packet_bytes[] = {
  0x00, 0x00, 0x00, 0x01, 0x01, 0x00, 0x00, 0x00,
  0x00, 0x00, 0x00, 0x01, 0x00, 0x00, 0x00, 0x00,
  0x00, 0x00, 0x00, 0x00, 0x01, 0x10, 
};

each value is of 1 byte I want to store the exact value in this struct

My Code to store in Struct

m_receivedBytes = recvfrom(sock, (char*)m_packetBuffer, sizeof(m_packetBuffer), 0, (sockaddr*)&cameraInfo, &m_socketLength);
        if (m_receivedBytes > 0)
        {

            switch (m_protocolType)
            {
            case StreamProtocol: ProtocolStruct.m_status = m_packetBuffer[0] + m_packetBuffer[1];
                            ProtocolStruct.m_blockID = m_packetBuffer[2] + m_packetBuffer[3];
                            ProtocolStruct.m_format = m_packetBuffer[4];
                            ProtocolStruct.m_packetID = m_packetBuffer[5] + m_packetBuffer[6] + m_packetBuffer[7];

                            switch (ProtocolStruct.m_format)
                            {
                            case 1: ProtocolStruct.m_leader->m_fieldInfo = m_packetBuffer[9];
                                    ProtocolStruct.m_leader->m_payloadType = m_packetBuffer[10] + m_packetBuffer[11];
                                    break;
                            default:
                                break;
                            }
                            break;

            default:        
                        break;
            }

The packet size is 22 so I store the value like this, I know this is wrong.

Example

If the 2 bytes are 10 01 the when I use + operator then result is 11 which is incorrect The correct answer should be 1001.

So can anyone please tell me how to put all that data in Structure

The addition is correct – 10 + 1 = 11 regardless of your numerical base – but addition is not what you're looking for. Read about bitwise operations and shifting ( `|`, `&`, `>>`, `<<`). — molbdnilo, Aug 04 '18 at 12:06
You can generally reassemble things by bit-shifting left (`<<`) and bitwise-ORing (`|`) - e.g. `x = (buf[n] << 24) | (buf[n+1] << 16) | (buf[n+2] << 8) | buf[n+3]` - but you need to make sure `m_packetBuffer` is an array of `unsigned char` / `uint8_t`. Alternatively, you could aim e.g. `uint32_t* p = reinterpret_cast(&m_packetBuffer[n])` then read `*p` (in which case leave `m_packetBuffer` as an array of `char`). Either way, you may need to convert for [endianness](https://en.wikipedia.org/wiki/Endianness) - see [`htonl`](https://linux.die.net/man/3/htonl) et al. — Tony Delroy, Aug 04 '18 at 12:58

score 1 · Answer 1 · answered Aug 04 '18 at 15:27

When dealing with telecoms packets, you have to insure that the same exact byte layout and order is shared between the two peers. You problem is that a struct definition is compiler specific ; one quick and dirty way is to use a "packed" layout :

struct __attribute__((__packed__)) ProtocolStruct
{
    __int16         m_status;
    __int16         m_blockID;
    __int8          m_format;
    __int32         m_packetID;
    struct Trailer *m_trailer;

}ProtocolStruct;

This solve the layout problem but not the byte order problem also name endianess.

However not always enough see Is gcc's __attribute__((packed)) / #pragma pack unsafe?

Store UDP packet in Structure using C++

1 Answers1