How C++ `recv` function acts at data receving? Could it receive a partial "packet"?

Question

static void HandlePackets(void* pParams)
{
   int iResult = 0;
   char recvbuf[MAX_PACKET_LENGTH];

   printf("Packet handling started\n");

   while((iResult = recv(lhSocket, recvbuf, MAX_PACKET_LENGTH, 0)) > 0)
      printf("Bytes received: %d\n", iResult);

   printf("Packet handling stopped with reason %i", WSAGetLastError());
}

For now, it only prints the amount of bytes received.

Could such things happen, that recv will receive only a half of packet? Or one full packet and half of next packet, if server sent them one by one fast?

For example, server sent a single packet with 512 bytes length, is it possible that recv first got 500 bytes, and the remain 12 will receive from second attempt?

If server sending a lot of packets with 512 bytes length for each, is it possible that recv will get 700 bytes from first executing and the remain bytes from second?

MAX_PACKET_LENGTH is 1024

(I talk here about application layer packets, not transport layer.)

The whole problem is - do I need to make some possibility for client to combine received bytes into one packet or split over-received bytes to different packets?

Answer 1

Is it possible that recv first got 500 bytes, and the remain 12 will receive from second attempt?

Yes, definitely.

You have no guarantee that, when the sending end sends a burst of X bytes, that the receiving end will pick them all up in a single call to recv. recv doesn't even know how many bytes are in your application-layer "packet".

The whole problem is - do I need to make some possibility for client to combine received bytes into one packet or split over-received bytes to different packets?

Your application will definitely have to accumulate data from possibly sequential reads, fill up a buffer, and implement parsing to look for a full packet.

TCP/IP doesn't know your application's protocol; as David said, if you want to split the incoming data stream into "packets" then you must do that yourself.