Python Threading String Arguments

Question

I have a problem with Python threading and sending a string in the arguments.

def processLine(line) :
    print "hello";
    return;

.

dRecieved = connFile.readline();
processThread = threading.Thread(target=processLine, args=(dRecieved));
processThread.start();

Where dRecieved is the string of one line read by a connection. It calls a simple function which as of right now has only one job of printing "hello".

However I get the following error

Traceback (most recent call last):
File "C:\Python25\lib\threading.py", line 486, in __bootstrap_inner
self.run()
File "C:\Python25\lib\threading.py", line 446, in run
self.__target(*self.__args, **self.__kwargs)
TypeError: processLine() takes exactly 1 arguments (232 given)

232 is the length of the string that I am trying to pass, so I guess its breaking it up into each character and trying to pass the arguments like that. It works fine if I just call the function normally but I would really like to set it up as a separate thread.

Answer 1

You're trying to create a tuple, but you're just parenthesizing a string :)

Add an extra ',':

dRecieved = connFile.readline();
processThread = threading.Thread(target=processLine, args=(dRecieved,)); # <- note extra ','
processThread.start();

Or use brackets to make a list:

dRecieved = connFile.readline();
processThread = threading.Thread(target=processLine, args=[dRecieved])); # <- 1 element list
processThread.start();

---

If you notice, from the stack trace: self.__target(*self.__args, **self.__kwargs)

The *self.__args turns your string into a list of characters, passing them to the processLine function. If you pass it a one element list, it will pass that element as the first argument - in your case, the string.