Python threads - number of arguments Error

Question

I am executing a command in a thread for almost 25k times like

if threaded is True:
                thread = Thread(target=threadedCommand, args=(cmd))
                thread.start()
                thread.join()  

def threadedCommand(command):
    if command is None:
        print 'can\'t execute threaded command'
        sys.exit(-1)
    print 'executing - %s'%(command)
    os.system(command)

and command is like

cp file dir

and what I see is

Traceback (most recent call last): File "/usr/lib64/python2.6/threading.py", line 525, in __bootstrap_inner self.run() File "/usr/lib64/python2.6/threading.py", line 477, in run self.__target(*self.__args, **self.__kwargs) TypeError: threadedCommand() takes exactly 1 argument (52 given)

^CException in thread Thread-9377: Traceback (most recent call last): File "/usr/lib64/python2.6/threading.py", line 525, in __bootstrap_inner self.run() File "/usr/lib64/python2.6/threading.py", line 477, in run self.__target(*self.__args, **self.__kwargs) TypeError: threadedCommand() takes exactly 1 argument (56 given)

if threaded is True is utterly pointless, fragile and verbose. Just use if threaded. — user395760
– user395760, Commented Sep 12, 2011 at 19:12

phihag · Accepted Answer · 2011-09-12 19:17:01Z

22

args must be a tuple. (cmd) is the same as cmd; you want a one-element tuple instead:

thread = Thread(target=threadedCommand, args=(cmd,))
#                                                ^

edited Sep 12, 2011 at 19:17

answered Sep 12, 2011 at 19:03

phihag

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1 Comment

daydreamer Over a year ago

wonderful, Thanks a lot, that clears it out! appreciate your help

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