recv() function too slow

Question

Hi i'm quite a newbie to Python. I' writting a simple LAN game (not simple for me) using a pygame module.

Here's the problem - I have two computers (one old intel Atom netbook, the other intel i5 NTB). I want to achieve at least 5 FPS (the netbook is slowering the NTB, but not so much, now i have around 1,5 FPS), but calling recv() function twice a main loop takes total around 0,5 seconds on each machine. The wifi signal is strong and the router is 300Mbit/s and it sends a short roughly 500-character string. As you can see for measuring time i use time.clock().

Here's the part of the "server" code, which i usually run on the i5 NTB:

while 1:
    start = time.clock()
    messagelen = c.recv(4)      #length of the following message (fixed 4 character)
    if " " in messagelen:
        messagelen = messagelen.replace(" ","")
    message = cPickle.loads(c.recv(int(messagelen))) #list of the arrows, other player position and changes in the game map
    arrowsmod = message[0]
    modtankposan = message[1]
    removelistmod = message[2]
    for i in removelistmod:
        try:
             randopos.remove(i)
        except ValueError:
            randopossv.remove(i)

    print time.clock()-start


    tosendlist=[]
    if len(arrows) == 0:  #if there are no arrows it appends only an empty list
        tosendlist.append([])
    else:
        tosendlist.append(arrows)
    tosendlist.append([zeltankpos, 360-angle])
    if len(removelist) == 0:   #if there are no changes of the map it appends only an empty list
        tosendlist.append([])
    else:
        tosendlist.append(removelist)
        removelist=[]
    tosend=cPickle.dumps(tosendlist)
    tosendlen = str(len(tosend))
    while len(tosendlen)<4:
        tosendlen+=" "
    c.sendall(tosendlen)   #sends the length to client
    c.sendall(tosend)      #sends the actual message(dumped list of lists) to client

    ...something else which takes <0,05 sec on the NTB

Here's the part of the "client" game code (just inverted the beginning - sending/receiving parts):

while 1:
    tosendlist=[]
    if len(arrows) == 0:  #if there are no arrows it appends only an empty list
        tosendlist.append([])
    else:
        tosendlist.append(arrows)
    tosendlist.append([zeltankpos, 360-angle])
    if len(removelist) == 0:   #if there are no changes of the map it appends only an empty list
        tosendlist.append([])
    else:
        tosendlist.append(removelist)
        removelist=[]
    tosend=cPickle.dumps(tosendlist)
    tosendlen = str(len(tosend))
    while len(tosendlen)<4:
        tosendlen+=" "
    s.sendall(tosendlen)   #sends the length to server
    s.sendall(tosend)      #sends the actual message(dumped list of lists) to server

    start = time.clock()
    messagelen = s.recv(4)      #length of the following message (fixed 4 character)
    if " " in messagelen:
        messagelen = messagelen.replace(" ","")
    message = cPickle.loads(s.recv(int(messagelen))) #list of the arrows, other player position and changes in the game map
    arrowsmod = message[0]
    modtankposan = message[1]
    removelistmod = message[2]
    for i in removelistmod:
        try:
             randopos.remove(i)
        except ValueError:
            randopossv.remove(i)

    print time.clock()-start
    ... rest which takes on the old netbook <0,17 sec

When I run let's say a single player version of the game on one machine (without the socket module) on the i5 NTB it has 50 FPS in the up left corner of the map and 25 FPS in the down right corner (the 1000x1000 pixel map contains 5x5 pixel squares, i think it's slower because of the bigger coordinates, but i can't believe that so much. BTW recv while ran as a LAN game in the down right corner of the map takes approx. the same time) on the Atom netbook it has 4-8 FPS.

So could you please tell me, why it's so slow? The computers are not synchronized, one is faster, the other slower, but it can't be that they are waiting for each other, it would be max 0,17 secs delay, right? And plus the long recv calling would be only on the faster computer? Also I don't exactly know how the send/recv function work. It's weird the sendall takes literally no time while receiving takes 0,5 secs. Maybe sendall is trying to send in the background while the rest of the program continues forward.

Answer 1

As mentioned by Armin Rigo, recv will return after packets are received by the socket, but packets don't necessarily need to be transmitted immediately after calling send. While send returns immediately, OS caches the data internally and might wait some time for more data being written to the the socket before actually transmitting it; this is called Nagle's algorithm and avoids sending lots of small packets over the network. You can disable it and push packets quicker to the wire; try enabling TCP_NODELAY options on the sending socket (or both if your communication is bidirectional), by calling this:

sock.setsockopt(socket.IPPROTO_TCP, socket.TCP_NODELAY, 1)

This could potentially reduce amount of time recv is sleeping due to no data.

As the Wikipedia states:

This algorithm interacts badly with TCP delayed acknowledgments, a feature introduced into TCP at roughly the same time in the early 1980s, but by a different group. With both algorithms enabled, applications that do two successive writes to a TCP connection, followed by a read that will not be fulfilled until after the data from the second write has reached the destination, experience a constant delay of up to 500 milliseconds, the "ACK delay". For this reason, TCP implementations usually provide applications with an interface to disable the Nagle algorithm. This is typically called the TCP_NODELAY option.

There is a mention of 0.5s which you're seeing in your benchmark, so this might be a reason.

Answer 2

Yes, send() or sendall() will occur in the background (unless the connexion is saturated right now, i.e. there is already too much data waiting to be sent). By contrast, recv() will immediately get the data only if it arrived already, but if none did, it waits. Then it returns possibly a fraction of it. (I am assuming that c is a TCP socket, not a UDP one.) Note that you should not assume that recv(N) returns N bytes; you should write a function like this:

def recvall(c, n):
    data = []
    while n > 0:
        s = c.recv(n)
        if not s: raise EOFError
        data.append(s)
        n -= len(s)
    return ''.join(data)

Anyway, to the point. The issue is not the speed of recv(). If I understood correctly, there are four operations:

• the server renders (1/25th sec)

• the server sends something on the socket, received by the client;

• the client renters (1/4th sec);

• the client send something back on the socket.

This takes almost (0.3 + 2 * network_delay) seconds. Nothing occurs in parallel. If you want more frames-per-second, you need to parallelize some of these four operations. For example, let's assume reasonably that the operation 3 is by far the slowest. Here's how we can make 3 run in parallel with the three other operations. You should change the client so that it receives data, process it, and immediately sends an answer to the server; and only then it proceeds to render it. This should be enough in this case, as it takes 1/4th seconds to do this rendering, which should be enough time for the answer to reach the server, the server to render, and the next packet to be sent again.