python- how to get the output of the function used in Timer

Question

I want to run a function for 10s then do other stuff. This is my code using Timer

from threading import Timer
import time

def timeout():
    b='true'
    return b

a='false'    
t = Timer(10,timeout)
t.start()

while(a=='false'):
    print '1'
    time.sleep(1)
print '2'

I know that using Timer I can print something at the end of the timer (print b instead of return b return true after 10s). What I want to know is : can I get the value returned by timeout() inside "a" to perform my while loop correctly?

Or is there another way to do it with another function ?

Answer 1

The return value of the function is just dropped by Timer, as we can see in the source. A way to go around this, is to pass a mutable argument and mutate it inside the function:

def work(container):
    container[0] = True

a = [False]
t = Timer(10, work, args=(a,))
t.start()

while not a[0]:
    print "Waiting, a[0]={0}...".format(a[0])
    time.sleep(1)

print "Done, result: {0}.".format(a[0])

Or, use global, but that's not the way a gentleman goes.