use*_*752 3 python sockets multithreading python-2.7 python-3.x
我正在尝试将以下用 python 2 编写的代码转换为 python 3。此 python 代码执行 TCP 端口转发。它来自这个页面:http : //code.activestate.com/recipes/483730-port-forwarding/
import socket
import sys
import thread
def main(setup, error):
sys.stderr = file(error, 'a')
for settings in parse(setup):
thread.start_new_thread(server, settings)
lock = thread.allocate_lock()
lock.acquire()
lock.acquire()
def parse(setup):
settings = list()
for line in file(setup):
parts = line.split()
settings.append((parts[0], int(parts[1]), int(parts[2])))
return settings
def server(*settings):
try:
dock_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
dock_socket.bind(('', settings[2]))
dock_socket.listen(5)
while True:
client_socket = dock_socket.accept()[0]
server_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
server_socket.connect((settings[0], settings[1]))
thread.start_new_thread(forward, (client_socket, server_socket))
thread.start_new_thread(forward, (server_socket, client_socket))
finally:
thread.start_new_thread(server, settings)
def forward(source, destination):
string = ' '
while string:
string = source.recv(1024)
if string:
destination.sendall(string)
else:
source.shutdown(socket.SHUT_RD)
destination.shutdown(socket.SHUT_WR)
if __name__ == '__main__':
main('proxy.ini', 'error.log')
Run Code Online (Sandbox Code Playgroud)
这是我所拥有的:
import socket
import sys
import threading
import time
def main(setup, error):
# open file for error messages
sys.stderr = open(error, 'a')
# read settings for port forwarding
threads = []
for settings in parse(setup):
#thread.start_new_thread(server, settings)
t = threading.Thread(target=server, args=(settings))
t.start()
threads.append(t)
for t in threads:
t.join()
# wait for <ctrl-c>
while True:
time.sleep(60)
def parse(setup):
settings = list()
file = open(setup)
for line in iter(file):
parts = line.split()
settings.append((int(parts[0]), parts[1], int(parts[2])))
return settings
def server(*settings):
try:
dock_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
dock_socket.bind(('', settings[0]))
dock_socket.listen(5)
while True:
client_socket = dock_socket.accept()[0]
server_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
server_socket.connect((settings[1], settings[2]))
Thread1 = threading.Thread(target=forward, args=(client_socket, server_socket))
Thread1.start()
Thread2 = threading.Thread(target=forward, args=(server_socket, client_socket))
Thread2.start()
finally:
thread = threading.Thread(targer=server, args=settings)
thread.start()
def forward(source, destination):
string = ' '
while string:
string = source.recv(1024)
if string:
destination.sendall(string)
else:
source.shutdown(socket.SHUT_RD)
destination.shutdown(socket.SHUT_WR)
if __name__ == '__main__':
main('port-forward.config', 'error.log')
Run Code Online (Sandbox Code Playgroud)
python 3 版本似乎确实有效。但我不能完全确定它是否写得正确。我不清楚代码的线程部分。python 2 版本使用线程模块,而python 3 版本使用线程模块。查看python 2版本,它在主函数中使用了锁。我需要在 python 3 版本中使用锁吗?另一个问题是,我应该在 python 3 代码的服务器函数中加入两个线程(线程 1 和线程 2)吗?
无论 python 版本如何,我的另一个问题是服务器函数的参数。我知道“设置”变量是指一个列表。为什么在“设置”参数之前需要一个星号?我确实在 python 文档https://docs.python.org/3/tutorial/controlflow.html#arbitrary-argument-lists 中查看了以下页面
但我不明白为什么传递没有星号的设置列表不起作用。
我的另一个问题是关于 python 2 代码中的线程锁。为什么锁被获取两次?我尝试删除其中一个锁,程序在启动后立即结束。
通过使用命令行工具2to3将代码转换为 Python 3,您可以节省一些工作,如下所示(从bash或 Windows 命令行):
2to3 -w myscript.py
Run Code Online (Sandbox Code Playgroud)
但让我们回答你的实际问题:为什么定义中的明星
def server(*settings):
...
Run Code Online (Sandbox Code Playgroud)
该函数thread.start_new_thread实际上会server()像这样启动你的函数:
server(arg1, arg2, arg3)
Run Code Online (Sandbox Code Playgroud)
其中arg1,arg2等来自的第二个参数start_new_thread。该声明def server(*settings)将这些参数收集回单个列表 ,settings您的函数将继续使用该列表。如果你写def server(settings)(没有星号),你的函数被声明为接受一个参数,但会用几个参数调用。