How does sending and receiving work in Python sockets?

Question

I have been working with python sockets for a while and I have written some simple simple programs.

The problem that I encounter every time concerns sending and receiving methods on sockets! Providing you with a simple and simple example:

This is the receiver (server!):

import socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind(('', 4001))
s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
s.listen(5)


while True:
  conn, addr = s.accept()
  print conn, addr

  data1 = conn.recv(64)
  data2 = conn.recv(64)

  print 'uname is %s , password is: %s' %(data1, data2, )
  conn.close()

And this is the sender (or client!):

import socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect(('', 4001))

uname = raw_input('enter username>')
passw = raw_input('enter password>')

s.send(uname)
s.send(passw)

print 'exiting ...'
s.close()

So, the problem is why the server receives both uname and passw in the first s.recv () method? This means that data2 is always empty!

I really don't know what happens when the client uses s.send (). I thought that each s.send () sends a "packet" to the destination (ip, port) !!

And can someone explain to me why this second code (another client) works?

import socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect(('', 4001))

uname = raw_input('enter username>')
s.send(uname)

passw = raw_input('enter password>')
s.send(passw)

print 'exiting ...'
s.close()

Answer 1

socket.SOCK_STREAM , TCP. , send, . send .

TCP, . , uname passw - .

data1 64 , uname passw.

, :

uname passw. (.. ).

, , . send : (, ).

, UDP:

socket.socket(socket.AF_INET, socket.SOCK_DGRAM)

Answer 2

. . , :