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,

show openServer function definition

RomanPerekhrest
– RomanPerekhrest

2019-09-02 17:40:42 +00:00
Commented Sep 2, 2019 at 17:40 — RomanPerekhrest
– RomanPerekhrest, Commented Sep 2, 2019 at 17:40
'def openServer(port, listen_on):'

user10847524
– user10847524

2019-09-02 17:44:10 +00:00
Commented Sep 2, 2019 at 17:44 — user10847524
– user10847524, Commented Sep 2, 2019 at 17:44

Dan D. · Accepted Answer · 2019-09-02 17:42:03Z

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])

answered Sep 2, 2019 at 17:42

Dan D.

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1 Comment

user10847524 Over a year ago

I have tried what you suggest and it works well. Thanks.

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