unexpected e32.ao_sleep behavior

[relbs]

Can anyone explain to me why executing this snippet of code:

import sys, e32

for i in range(5):
    e32.ao_sleep(0, lambda: sys.stdout.write("%d\n" % i))

e32.ao_sleep(4)

results in this output?

4
4
4
4
4

Regards,
Albert

[yemir]

The lambda closes over the variable i, which is assigned the values 0, 1, 2, 3 and then 4 successively by the for-loop. It turns out that all of your callbacks executed after i already had the value 4. (This behavior is different from Lisp, where each iteration of such a loop effictively defines a different variable i.)

Here's one solution:

for i in range(5):
    e32.ao_sleep(0, lambda i=i: sys.stdout.write("%d\n" % i))

Here's another:

for i in range(5):
    def callback(i=i):
        print i
    e32.ao_sleep(0, callback)

And here's another:

def async(i):
    e32.ao_sleep(0, lambda: sys.stdout.write("%d\n" % i))

for i in range(5):
    async(i)