通过线程将args传递给函数

Question

I am trying to pass two variable to a function via Thread as below:

from os import system
from sys import exc_info
from threading import Thread
import time


def play(file, priority):

    try:
        if priority == 0:
            system('ps aux | grep mpg321 | grep -v "grep mpg321" | awk \'{print $2}\' | xargs kill -9')

        statement = 'sudo -u pi mpg321 -g 1 -q -a bluetooth sound/' + file
        system(statement)

    except:
        print('There was an error playing the sound - ' + str(exc_info()[0]))

    finally:
        pass

t1Sound = 'presence.mp3'
t2Sound = 'ultra.mp3'

t1 = Thread(target=play(), args=(t1Sound, 0))
t2 = Thread(target=play(), args=(t2Sound, 1))

t1.start()
time.sleep(2)
t2.start()

But somehow I keep getting the error below:

sudo python test.py
Traceback (most recent call last):
  File "test.py", line 25, in <module>
    t1 = Thread(target=play(), args=(t1Sound, 0))
TypeError: play() takes exactly 2 arguments (0 given)

Do you guys know how to fix this? How I am supposed to pass those variables correctly?

Answer 1

You need to pass the function to the Thread , not call the function. Just remove the parenthesis at the end of play() .

t1 = Thread(target=play, args=(t1Sound, 0))
t2 = Thread(target=play, args=(t2Sound, 1))