Python In Linux: Put User Input Asynchronously Into Queue

I am trying to run a program that takes in input as a job is getting done. I have looked through several forms, and looked into the documentation. I'm running this in Debian, and

Solution 1:

I read more of this link, and there was an implementation of what I wanted at the bottom.

I used the select module for a Non-Blocking implementation on Linux. This times out in (5 seconds here) if no input is received. Particularly useful when used in a thread, so that the getch call is non-blocking and will allow the thread to exit cleanly

# This class gets a single character input from the keyboardclass_GetchUnix:
    def__init__(self):
        import tty, sys
        from select import select
    def__call__(self):
        import sys, tty, termios
        from select import select
        fd = sys.stdin.fileno()
        old_settings = termios.tcgetattr(fd)
        try:
                tty.setraw(sys.stdin.fileno())
                [i, o, e] = select([sys.stdin.fileno()], [], [], 2)
                if i: 
                ch=sys.stdin.read(1)
                else: 
                ch=''finally:
                    termios.tcsetattr(fd, termios.TCSADRAIN, old_settings)
            return ch
getch = _GetchUnix()
# End Class

Solution 2:

I've also used [i, o, e] = select([sys.stdin.fileno()], [], [], 2), but I've heard it might not work on windows. If anyone still needs a multi-threaded, non-blocking input example:

import threading
import sys
import time

bufferLock=threading.Lock()
inputBuffer=[]

classInputThread(threading.Thread):
    defrun(self):
        global inputBuffer
        print("starting input")
        whileTrue:
            line=sys.stdin.readline()
            bufferLock.acquire()
            inputBuffer.insert(0,line)
            bufferLock.release()

input_thread=InputThread()
input_thread.start()
whileTrue:
    time.sleep(4)
    bufferLock.acquire()
    iflen(inputBuffer)>0:
        print("Popping: "+inputBuffer.pop())
    bufferLock.release()