CODEWITHSUNDEEP
pythonmultithreadingargs
Paul
Paul

Reputation: 205

socket threading.Thread args takes exactly 1 argument (2 given)

import threading
import socket

sock = socket.socket()
sock.bind(('0.0.0.0', 9999))
sock.listen(5)

def handle_client(sock_client):
        data = sock_client.recv(4096)
        while data:
                sock_client.send("this is server: " + data)
                data = sock_client.recv(4096)
        sock_client.close()

while True:
        client = sock.accept()
        t = threading.Thread(target=handle_client, args=client)
        t.start()

TypeError: handle_client() takes exactly 1 argument (2 given)

this Why, my thread args only give a parameter, why tell me to give two parameters.

how to understand this?

Upvotes: 0

Views: 2864

Answers (1)

Fejs
Fejs

Reputation: 2888

You're calling threading function in the wrong way. Currently, You use this:

t = threading.Thread(target=handle_client, args=client)

However, You should use this:

t = threading.Thread(target=handle_client, args=(client, ))

If You read the docs, You'll see that args parameter should be a tuple, and not a single variable.

Update:

Just tested my code, it works fine. Yes, program will still raise an error (AttributeError: 'tuple' object has no attribute 'recv'), but this is because You should use sock_client[0] to receive data (sock_client itself is a tuple consisting of connection object and address).

So, proper code is here:

import threading
import socket

sock = socket.socket()
sock.bind(('0.0.0.0', 9999))
sock.listen(5)

def handle_client(sock_client):
        conn = sock_client[0]
        address = sock_client[1]
        data = conn.recv(4096)
        while data:
                conn.send("this is server: " + data)
                data = conn.recv(4096)
        conn.close()

while True:
        client = sock.accept()
        t = threading.Thread(target=handle_client, args=(client, ))
        t.start()

Upvotes: 3

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