How to use Python thread with one argument function?

Question

I create a main function below with two argument function inside Thread .

if __name__ == "__main__":
    # Define ports
    ports_for_server_connection=[10003, 10004, 10005, 10006]
    for port_number in ports_for_server_connection:
        # Open multi thread sockets so that each will respond independently
        t = Thread(target=openServer, args=(port_number, 1))
        t.start()

However, I want to create that function with only one argument. When I tried to implement it with one argument( args=(port_number, 1) ) I got the following error.

Traceback (most recent call last):
  File "/usr/lib/python3.6/threading.py", line 916, in _bootstrap_inner
    self.run()
  File "/usr/lib/python3.6/threading.py", line 864, in run
    self._target(*self._args, **self._kwargs)
TypeError: openServer() argument after * must be an iterable, not int

How can I use thread with only one argument function?

Thanks,

Answer 1

Pass it only one argument:

t = Thread(target=openServer, args=(port_number,))

The trick bit is that (x,) is a tuple of length one those first item is x . If this is too hard just use a list:

t = Thread(target=openServer, args=[port_number])