Python: TypeError: argument after * must be a sequence

问题

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?

回答1:

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

来源：https://stackoverflow.com/questions/36387596/python-typeerror-argument-after-must-be-a-sequence