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;

.

Answer 1

I hope to provide more background knowledge here.

First, constructor signature of the of method threading::Thread:

class threading.Thread(group=None, target=None, name=None, args=(), kwargs={}, *, daemon=None)

args is the argument tuple for the target invocation. Defaults to ().

Second, A quirk in Python about tuple:

Empty tuples are constructed by an empty pair of parentheses; a tuple with one item is constructed by following a value with a comma (it is not sufficient to enclose a single value in parentheses).

On the other hand, a string is a sequence of characters, like 'abc'[1] == 'b'. So if send a string to args, even in parentheses (still a sting), each character will be treated as a single parameter.

However, Python is so integrated and is not like JavaScript where extra arguments can be tolerated. Instead, it throws an TypeError to complain.

Answer 2

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.