Heim > Backend-Entwicklung > Python-Tutorial > 在Python中使用异步Socket编程性能测试

在Python中使用异步Socket编程性能测试

WBOY
Freigeben: 2016-06-16 08:43:34
Original
1807 Leute haben es durchsucht

OK,首先写一个python socket的server段,对开放三个端口:10000,10001,10002.krondo的例子中是每个server绑定一个端口,测试的时候需要分别开3个shell,分别运行.这太麻烦了,就分别用三个Thread来运行这些services.

import optparse 
import os 
import socket 
import time 
from threading import Thread 
import StringIO 
  
txt = '''1111 
2222 
3333 
4444 
''' 
  
  def server(listen_socket): 
  while True: 
    buf = StringIO.StringIO(txt) 
    sock, addr = listen_socket.accept() 
    print 'Somebody at %s wants poetry!' % (addr,) 
    while True: 
        try: 
          line = buf.readline().strip() 
          if not line: 
            sock.close() 
            break 
          sock.sendall(line) # this is a blocking call 
          print 'send bytes to client:%s' % line 
          #sock.close() 
        except socket.error: 
          sock.close() 
          break 
        time.sleep(1) #server和client连接后,server会故意每发送一个单词后等待一秒钟后再发送另一个单词 
  
  def main(): 
  ports = [10000, 10001, 10002] 
  for port in ports: 
    listen_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM) 
    listen_socket.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1) 
    addres = (str('127.0.0.1'), port) 
    listen_socket.bind(addres) 
    listen_socket.listen(5) 
    print "start listen at:%s" % (port,) 
    worker = Thread(target = server, args = [listen_socket]) 
    worker.setDaemon(True) 
    worker.start() 
  
  if __name__ == '__main__': 
  main() 
  while True: 
    time.sleep(0.1) #如果不sleep的话,CPU会被Python完全占用了 
    pass 
Nach dem Login kopieren

下面是一个client,没有才用异步网络,连接这个三个端口的server:

import socket 
  
  
if __name__ == '__main__': 
  ports = [10000, 10001, 10002] 
  for port in ports: 
    address = (str('127.0.0.1'), port) 
    sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM) 
    sock.connect(address) 
    poem = '' 
    while True: 
      data = sock.recv(4) 
      if not data: 
        sock.close() 
        break 
      poem += data 
    print poem 
Nach dem Login kopieren

下面用异步的client来读取,代码如下:

import datetime, errno, optparse, select, socket 
  
def connect(port): 
  """Connect to the given server and return a non-blocking socket.""" 
  address = (str('127.0.0.1'), port) 
  sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM) 
  sock.connect(address) 
  sock.setblocking(0) 
  return sock 
  
def format_address(address): 
  host, port = address 
  return '%s:%s' % (host or '127.0.0.1', port) 
  
if __name__ == '__main__': 
  ports = [10000, 10001, 10002] 
  start = datetime.datetime.now() 
  
  sockets = map(connect, ports) 
  poems = dict.fromkeys(sockets, '') # socket -> accumulated poem  
  
  # socket -> task numbers 
  sock2task = dict([(s, i + 1) for i, s in enumerate(sockets)]) 
  sockets = list(sockets) # make a copy 
  
  while sockets: 
    #运用select来确保那些可读取的异步socket可以立即开始读取IO 
    #OS不停的搜索目前可以read的socket,有的话就返回rlist 
    rlist, _, _ = select.select(sockets, [], []) 
    for sock in rlist: 
      data = '' 
      while True: 
        try: 
          new_data = sock.recv(1024) 
        except socket.error, e: 
          if e.args[0] == errno.EWOULDBLOCK: 
            break 
          raise 
        else: 
          if not new_data: 
            break 
          else: 
            print new_data 
            data += new_data 
  
      task_num = sock2task[sock] 
      if not data: 
        sockets.remove(sock) 
        sock.close() 
        print 'Task %d finished' % task_num 
      else: 
        addr_fmt = format_address(sock.getpeername()) 
        msg = 'Task %d: got %d bytes of poetry from %s' 
        print msg % (task_num, len(data), addr_fmt) 
  
      poems[sock] += data 
  
  elapsed = datetime.datetime.now() - start 
  print 'Got poems in %s' % elapsed 
Nach dem Login kopieren

结果只需要4秒就完成了读取任务。效率是刚才同步socket的三倍。对客户端的异步改造主要有两点:

同步模式下,客户端分别创建socket;而在异步模式下,client开始就创建了所有的socket。
通过“sock.setblocking(0)”设置socket为异步模式。
通过Unix系统的select俩返回可读取IO
最为核心的是26行和29行。尤其是29行的select操作返回待读取socket的列表。

Verwandte Etiketten:
Quelle:php.cn
Erklärung dieser Website
Der Inhalt dieses Artikels wird freiwillig von Internetnutzern beigesteuert und das Urheberrecht liegt beim ursprünglichen Autor. Diese Website übernimmt keine entsprechende rechtliche Verantwortung. Wenn Sie Inhalte finden, bei denen der Verdacht eines Plagiats oder einer Rechtsverletzung besteht, wenden Sie sich bitte an admin@php.cn
Beliebte Tutorials
Mehr>
Neueste Downloads
Mehr>
Web-Effekte
Quellcode der Website
Website-Materialien
Frontend-Vorlage