multi thread issue in Python

Question

New to Python multi-thread and write such simple program, here is my code and error message, any ideas what is wrong? Thanks.

Using Python 2.7.

import time
import thread

def uploader(threadName):

    while True:
        time.sleep(5)
        print threadName

if __name__ == "__main__":

    numOfThreads = 5
    try:
        i = 0
        while i < numOfThreads:
            thread.start_new_thread(uploader, ('thread'+str(i)))
            i += 1

        print 'press any key to exit test'
        n=raw_input()

    except:
       print "Error: unable to start thread"

Unhandled exception in thread started by <pydev_monkey._NewThreadStartupWithTrace instance at 0x10e12c830>
Traceback (most recent call last):
  File "/Applications/PyCharm CE.app/Contents/helpers/pydev/pydev_monkey.py", line 521, in __call__
    return self.original_func(*self.args, **self.kwargs)
TypeError: uploader() takes exactly 1 argument (7 given)

thanks in advance, Lin

('thread'+str(i)) will give you a string, not a tuple. Try this: ('thread'+str(i),). — Justin O Barber
– Justin O Barber, Commented Mar 13, 2016 at 0:50
You mean you get this error: uploader() takes exactly 1 argument (7 given)? Are you sure you have saved it before re-running it? Incidentally, if you are new to python, you might find the threading module more helpful. — Justin O Barber
– Justin O Barber, Commented Mar 13, 2016 at 0:59
It seems to work for me when I change the line as @JustinBarber suggested. The threads start printing their names after sleeping for 5 secs. — jDo
– jDo, Commented Mar 13, 2016 at 1:05
@Lin Ma I have added the answer below. Glad you figured it out. Cheers. — Justin O Barber
– Justin O Barber, Commented Mar 13, 2016 at 1:06

Justin O Barber · Accepted Answer · 2016-03-13 01:05:17Z

3

The args of thread.start_new_thread need to be a tuple. Instead of this:

('thread' + str(i))  # results in a string

Try this for the args:

('thread' + str(i),)  # a tuple with a single element

Incidentally, you should check out the threading module, which is a higher-level interface than thread.

answered Mar 13, 2016 at 1:05

Justin O Barber

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2 Comments

Lin Ma Over a year ago

Thanks for the help, Justin. Mark your reply as an answer. :)

Justin O Barber Over a year ago

Thanks, @Lin Ma! Glad you got it figured out.

Jake Psimos · Accepted Answer · 2016-03-13 01:09:13Z

1

In the following, threadName is now a global variable defined towards the top of the program code, then the variable is initialized before the new thread is started with the target being the upload function.

Try this:

import time
import thread

threadName = ''

def uploader():

    while True:
        time.sleep(5)
        print threadName

if __name__ == "__main__":

    numOfThreads = 5
    try:
        i = 0
        while i < numOfThreads:
            threadName = 'thread' + str(i)
            newThread = threading.Thread(target=uploader)
            newThread.start()
            i += 1

        print 'press any key to exit test'
        n=raw_input()

    except:
       print "Error: unable to start thread"

edited Mar 13, 2016 at 1:09

answered Mar 13, 2016 at 1:04

Jake Psimos

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2 Comments

Alex Huszagh Over a year ago

Code only answers, although they may be part of a complete answer, are discouraged since they do not explain what was fixed or how. For people with similar issues in the future, this may not help them resolve their own questions without asking a subsequent question.

Jake Psimos Over a year ago

@AlexanderHuszagh Ok I will keep that in mind when answering. I just edited the answer with a quick explanation

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