多线程Python UDP客户端收回复仍抛超时异常的原因排查

客户端代码 #

from socket import *
import threading
import time


def ping_thread(num, clientSocket):
    try:
        clientSocket.settimeout(1)
        t1 = time.perf_counter()
        message = "Ping " + str(num) + " " + str(time.time())
        clientSocket.sendto(message.encode(), (serverName, serverPort))
        modifiedMessage, serverAddress = clientSocket.recvfrom(2048)
        t2 = time.perf_counter()
        lag = t2 - t1
        print(f"{modifiedMessage.decode()}: {lag:.8f}s")
    except TimeoutError:
        print(f"ping {num} time out!")


if __name__ == "__main__":
    threads = []
    clientSocket = socket(AF_INET, SOCK_DGRAM)
    serverName = "localhost"
    serverPort = 12000
    for i in range(1, 11):
        t = threading.Thread(target=ping_thread, args=(i, clientSocket))
        threads.append(t)
        t.start()

    for t in threads:
        t.join()

    clientSocket.close()

运行结果 #

PING 2 1710845145.1677558: 0.00191080s
PING 5 1710845145.169752: 0.00126800sPING 6 1710845145.169752: 0.00098610s

PING 1 1710845145.1658478: 0.00588400s
PING 3 1710845145.1677558: 0.00067630s
PING 8 1710845145.1719148: 0.00477900s
PING 10 1710845145.172912: 0.00083350s
ping 10 time out!
ping 4 time out!
ping 2 time out!

疑惑 #

为何部分线程（例如上述的Ping 2）明明收到了服务器回复，却仍抛出超时异常？

服务器代码 #

# We will need the following module to generate randomized lost packets
import random
from socket import *

# Create a UDP socket
# Notice the use of SOCK_DGRAM for UDP packets
serverSocket = socket(AF_INET, SOCK_DGRAM)
# Assign IP address and port number to socket
serverSocket.bind(('', 12000))

while True:
    # Generate random number in the range of 0 to 10
    rand = random.randint(0, 10)
    # Receive the client packet along with the address it is coming from
    message, address = serverSocket.recvfrom(1024)
    # Capitalize the message from the client
    message = message.upper()
    # If rand is less than 4, we consider the packet lost and do not respond
    if rand < 4:
        continue
    # Otherwise, the server responds
    serverSocket.sendto(message, address)

解答 #

问题的核心是所有线程共享同一个UDP套接字，而UDP是无连接协议，套接字的接收缓冲区是全局共享的：

• 当多个线程同时调用recvfrom()等待响应时，操作系统会把收到的任意一个UDP数据包分配给其中一个线程，不管这个数据包是不是该线程发送的请求对应的。
• 比如Ping 2的响应被线程5的recvfrom()抢到，线程5会打印Ping 2的回复信息，但线程5自己的Ping 5响应可能被其他线程抢走；而真正发送Ping 2请求的线程2，因为没拿到自己的响应，最终超时触发异常，打印ping 2 time out!。

修复方案 #

给每个线程创建独立的UDP套接字，这样每个套接字的接收缓冲区相互独立，服务器返回的响应只会被对应的发送线程接收：
修改客户端主函数部分代码：

if __name__ == "__main__":
    threads = []
    serverName = "localhost"
    serverPort = 12000
    for i in range(1, 11):
        # 每个线程创建自己的socket
        clientSocket = socket(AF_INET, SOCK_DGRAM)
        t = threading.Thread(target=ping_thread, args=(i, clientSocket))
        threads.append(t)
        t.start()

    for t in threads:
        t.join()

这样每个线程的请求和响应都会一一对应，不会出现响应被其他线程抢走的情况，也就不会出现“收到回复却超时”的矛盾现象。

内容的提问来源于stack exchange，提问作者echo_lin