Python multithreading argument error

Question

I've just started out with multi threading, so I thought I'd make myself a small and simple example:

import time
import threading

def count(who):
        count = 1
        while count <= 5:
                print who + " counted to: " + str(count)
                time.sleep(0.1)
                count += 1

thread1 = threading.Thread(target=count, args=('i'))
thread1.start()

Which works great and prints out the following:

>>> i counted to: 1
>>> i counted to: 2
>>> i counted to: 3
>>> i counted to: 4
>>> i counted to: 5

The odd thing is however, when I want to change the argument to another like: "john":

thread1 = threading.Thread(target=count, args=('john'))

Hoping it would produce:

>>> john counted to: 1
>>> john counted to: 2
>>> john counted to: 3
>>> john counted to: 4
>>> john counted to: 5

However, it procudes an error:

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 run
    self.__target(*self.__args, **self.__kwargs)
TypeError: count() takes exactly 1 argument (4 given)

I'm really not sure what is happening here... Does anyone know?

Reut Sharabani · Accepted Answer · 2017-02-10 17:18:42Z

5

Add a comma to make it explicit that you want a tuple:

thread1 = threading.Thread(target=count, args=('john', ))

Currently python thinks the parenthesis are redundant, so ("john") evaluates to "john" which are four characters, hence the message you're getting.

edited Feb 10, 2017 at 17:18

answered Feb 10, 2017 at 15:53

Reut Sharabani

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1 Comment

user5740843 Over a year ago

Oh okay, well I guess that makes sense. Thanks for that explanation!

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