Python: TypeError: argument after * must be a sequence

Question

I have this piece of code in which I try to send an UDP datagram in a new thread

import threading, socket address = ("localhost", 9999) def send(sock): sock.sendto("Message", address) print "sent" s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM) threading.Thread(target=send, args=(s)).start()

But when I try to give the socket as an argument to the function, a TypeError exception is thrown:

Exception in thread Thread-1: Traceback (most recent call last): File "/usr/lib/python2.7/threading.py", line 810, in __bootstrap_inner self.run() File "/usr/lib/python2.7/threading.py", line 763, in self.__target(*self.__args, **self.__kwargs) TypeError: send() argument after * must be a sequence, not _socketobject

What that means?

Pythonista · Accepted Answer · 2018-10-09 20:25:32Z

You need to add a comma - , - after your variable s. Sending just s to args=() is trying to unpack a number of arguments instead of sending just that single arguement.

So you'd have threading.Thread(target=send, args=(s,)).start()

Also the splat - * - operator might be useful in this question explaining it's usage and unzipping arguments in general

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