CODEWITHSUNDEEP
pythonmultithreading
Lin Ma
Lin Ma

Reputation: 10159

multi thread issue in Python

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

Upvotes: 2

Views: 165

Answers (2)

Jake Psimos
Jake Psimos

Reputation: 640

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"

Upvotes: 1

Justin O Barber
Justin O Barber

Reputation: 11601

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.

Upvotes: 3

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