7 replies to this topic

[Ripiz]

Hello,

I've used UDP before and it was convenient as it receives packets. However I've experienced some issues (mostly out of order packets), so I thought TCP might be better solution than creating own system to fix UDP disadvantages. Sadly TCP came with own disadvantage: it receives bytes, not packets, so I'm looking into a way to put bytes back into packets, otherwise data cannot be processed. I have one idea in my mind, but not sure if this is correct way, so I thought could seek some help here.

Here's my idea:

// general packet structure
class Packet {
    union {
        short size;
        char data[1024]; // first two bytes contain whole packet's size
    };

    void send() {
        send(socket, data, size);
    }
};

// sender
Packet p;
p << 0x0057;
p << player.name;
p.send();

// receiver
char data[1024];
short totalSize = 0;
while(true) {
    totalSize += recv(data + totalSize);
    if(totalSize >= 2)
        if(totalSize >= (short)data) { // comparing with first 2 bytes (pseudo-cast)
            // if we received more data than packet's size we can process it
            threadPool.queueForProcessing(data, (short)data);
            // copy remaining bytes to the beginning of data
        }
}

Thank you in advance.

Edited by Ripiz, 29 May 2012 - 09:19 AM.

[frob]

Yes, TCP is stream based, and that is one of the many tradeoffs that are made.

Yes, this format is very common:

{ int messagesize, int messagetype, message }

I would not use the 1024 byte buffer. Prefer to use the actual size of the buffer as sent over the wire.

Be sure you handle situations where you don't have all the message, and situations where more than one message arrives at once.

[ProgrammerDX]

struct Packet {
	 union {
		  short size;
		  char data[1024];
	 };
};

.... in receiver (usually an independent thread):

Packet packet;
int bytesReceived;

//Try receive packet size (sizeof short)
bytesReceived = recv(socket, packet.data, sizeof(short), 0);

if (bytesReceived == 0) {
	 error;
}

//Make sure receive the sizeof short
while (bytesReceived < sizeof(short)) {
	 int len = recv(socket, packet.data + bytesReceived, sizeof(short) - bytesReceived, 0);

	 if (len == 0) {
		  error;
	 }

	 bytesReceived += len;
};

//Now we know size so we can receive the whole packet
while (bytesReceived < packet.size) {
	 int len = recv(socket, packet.data + bytesReceived, packet.size - bytesReceived, 0);

	 if (len == 0) {
		  error;
	 }

	 bytesReceived += len;
};

//Full packet received

Edited by ProgrammerDX, 29 May 2012 - 10:37 AM.

[hplus0603]

First: This is actually answered in the FAQ.

Second: You should not block on TCP receive until a full packet is received. Doing so means that your loop will be frozen if someone sends a half packet. This could be because of a crash, or because of sudden network disconnection, or because of a malicious user, or whatever.

In general, you want the network step of your main loop to be:

- If the socket has data, drain the TCP socket. Typically, this means calling recv() into some cyclic buffer with a big "ask" -- recv() will only return what's actually there.
- Look at the head of your drain buffer. If there is enough data there to tell what the packet header is, look at the packet header. If there is enough data there to cover the entire packet length as shown by the header, dequeue that packet and move the drain buffer read pointer forward by as much. Handle the packet, and repeat this step.

Specifically, you want to "drain" the network quickly, and then use pointer arithmetic and byte inspection to sort out whatever data you have. This is more efficient, more robust, and easier to implement than code that tries to only call recv() for the header / data parts. Consider if you have a length field of more than one byte, and you only recv() half of that field -- now what?
If the packet structure is <id-as-uint16, length-as-uint16, payload> then the decode when you read to a given buffer is simple:

  if (buffer.available() >= 4) { // have a header
    uint16 len = (buffer[2] << 8) | buffer[3]; // assume buffer is unsigned char
    if (buffer.available() >= 4 + len) { // have all the payload
      uint16 type = (buffer[0] << 8) | buffer[1];
      handle_packet(type, len, &buffer[4]); // or whatever
      buffer.remove_front(4 + len);
    }
  }

Edited by hplus0603, 29 May 2012 - 12:21 PM.

[ProgrammerDX]

Yeah try getting more data than just the header is probably faster although very minimal (especially minimum with large receive windows). It's a good suggestion nonetheless!

Though if you're trying to learn the subject and api it's probably a steeper learning curve like that.

[rip-off]

Yeah try getting more data than just the header is probably faster although very minimal (especially minimum with large receive windows).

The point of the scheme hplus is advocating is not (just) to speed things up, but to avoid blocking. If you avoid blocking, you can also avoid threading.

Though if you're trying to learn the subject and api it's probably a steeper learning curve like that.

If you're learning an API, make a sandbox project to do so in. If you are implementing production code, you need to have a reasonably solid understanding of the API you are using.

And as you were advocating threading - using threading to deal with the network also has a steep learning curve, with the significant downside that mistakes are less obvious than in a non-blocking solution.

[Rasterman]

If you are creating something that will ship I recommend using a higher level library, I use enet and have no problems with it, it solved the very problem you have run into, it basically simulates a TCP connection using UDP packets, providing the best of both.

[hplus0603]

it's probably a steeper learning curve like that

I don't think so. The pattern of "drain the socket in a single recv()" is actually a best practice, and actually leads to simpler, more robust code. I highly recommend doing that, rather than what the original code was trying to do. It will lead to a complete, working, robust program sooner, and thus will lead to a faster learning curve.