TypeError: send() argument after * must be an iterable, not socket in Python
Dung Do Tien
Sep 23 2021
159
Hello you guys, I am a newbie in Python and I'm also studying more about Python.
I have some lines of code, I have try to work with multi threading in Python and help me send a message with socket. Function as below:
import threading, socket
uri = ("localhost", 8080)
def send(sock):
sock.sendto("Message", uri)
print("Message sent")
s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
threading.Thread(target=send, args=(s)).start()
But I get an exception TypeError: send() argument after * must be an iterable, not socket when I run code above.
Exception in thread Thread-1:
Traceback (most recent call last):
File "/usr/lib/python3.8/threading.py", line 932, in _bootstrap_inner
self.run()
File "/usr/lib/python3.8/threading.py", line 870, in run
self._target(*self._args, **self._kwargs)
TypeError: send() argument after * must be an iterable, not socket
And I am using python 3.8.2
Anyone can explain it to me? How can I solve it?
Thanks for any response.
Have 1 answer(s) found.
-
E1
Eslam Ali Sep 23 2021
You need to add a comma
,after the variable s. Sending only s toargs= ()is an attempt to unpack multiple arguments instead of sending just one argument.Change:
threading.Thread(target=send, args=(s)).start()To
threading.Thread(target=send, args=(s,)).start()I hope it useful for you.
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