Home/ Questions/Q 7692789
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-31T20:56:01+00:00

I’m trying to write a basic client/server echo program, to test the usage of

  • 0

I’m trying to write a basic client/server echo program, to test the usage of timers for retransmission based on select() (though I had to comment out that bit to simplify debugging when it wasn’t working as intended). Here are snippets of the relevant code:

Server:

from socket import *
import sys
import select
address = ('localhost', 6005)
server_socket = socket(AF_INET, SOCK_DGRAM)
server_socket.bind(address)

while(1):
    print "Listening"
    recv_data, addr = server_socket.recvfrom(2048)
    print recv_data
    if recv_data == "Request 1" :
        print "Received request 1"
        server_socket.sendto("Response 1", address)
    elif recv_data == "Request 2" :
        print "Received request 2"
        data = "Response 2"
        server_socket.sendto(data, address)

Client:

from socket import *
import sys
import select

address = ('localhost', 6005)
client_socket = socket(AF_INET, SOCK_DGRAM)

num_retransmits = 0
while(num_retransmits < 60):
    num_retransmits = num_retransmits + 1


    data = "Request 1"
    client_socket.sendto(data, address)
    print "Sending request 1"

    recv_data, addr = client_socket.recvfrom(2048)

    print recv_data, "!!"

The output on the client is simply ‘Sending request 1’, and when a breakpoint is used on anything at or below the recvfrom call, it does not reach the breakpoint. So I figure the client is not receiving anything and is holding out til it does. On the other hand, the output on the server is:

  • Listening
  • Request 1
  • Received request 1
  • Listening
  • Response 1

and so on and so forth

After the first loop, the server loops again and prints RESPONSE 1. This means that what the server did was to receive request 1, send response 1 to the client, loop… but after it loops the second time, response 1 is STILL in its socket! That is why when it prints recv_data,it prints response 1. On the other hand, client is not printing recv_data, because client has not received it – it is still in the buffer of the server’s socket.

Please help – I have tried looking at other echo programs but they all seem to use TCP and are fairly straightforward (and I think I have pretty much followed the same steps). I have no idea why my UDP program is not working. I tried looking at the sendall() call but it only seems to work for TCP.

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    You have to send to addr instead of address.

    from socket import *
import sys
import select
address = ('localhost', 6005)
server_socket = socket(AF_INET, SOCK_DGRAM)
server_socket.bind(address)

while(1):
    print "Listening"
    recv_data, addr = server_socket.recvfrom(2048)
    print recv_data
    if recv_data == "Request 1" :
        print "Received request 1"
        server_socket.sendto("Response 1", addr)
    elif recv_data == "Request 2" :
        print "Received request 2"
        data = "Response 2"
        server_socket.sendto(data, addr)
    • 0