A while back i had a question about why my socket sometimes received only 653 octets ( for example ) when i sent 1024 octets and thanks to Rakis i understood: The OS allows reception to occur in arbitrarily sized chunks.
This time i need a confirmation :)
On any OS ( Well GNU/Linux and Windows at least ), In any Language ( I'm using Python here ), if i send a packet of a random number of bytes, can be 2 bytes, can be 12000 bytes, let's say X, when i write socket.send(X), am i absolutely guaranteed that X will be FULLY received ( regardless of any chunks the receiving OS divides it into ) on the other end of the socket BEFORE i do another socket.send(any string) ?
Or in other words if i have the code :
socket.send(X)
socket.send(Y)
Even if X > MTU so it will be obliged to send multiple packets, does it wait until every packet is sent and acknowledged by the endpoint of the socket before sending Y ? Well writing that makes me believe that the answer is yes it is guaranteed and that this is exactly the purpose of setting a socket in blocking mode but i want to be sure :D
Thanks in advance, Nolhian
You are guaranteed that X will be received (at the application level) before Y, if it's a stream socket. If it's a datagram socket, no guarantees.
Depending on the networking implementation, it's possible that at a lower level, X will be sent, lost in transmission, then Y will be sent, then X will be re-sent because no acknowledgement was received.
Even in blocking mode, the socket.send(Y) can execute before X even makes it "onto the wire", because the OS will buffer network traffic.