Python Practice 20/22

Fix the bug: every handler reports the last channel

A deploy tool builds one notifier per alert channel up front, hands the list to the scheduler, and the scheduler calls them later when a build finishes. It ran for a month before anyone noticed every channel was reporting sms.

Nothing throws. The list holds three separate functions, exactly as intended, and each one hands back a clean string. They just all hand back the same string - the last one. Spot-check senders[-1]() and it looks perfect, which is why this walks through review.

The mechanism is not what most people assume. A lambda does not capture the value of name at the moment it is created. Its body does not run at all then. It captures the variable name - the same one the loop keeps reassigning - and reads it later, when it is finally called. All three lambdas share that one variable, and by the time the scheduler calls them the loop has long finished and left name set to "sms". These are not three functions that each remembered a different channel. They are three functions looking in the same box, and the box holds the last thing the loop put in it.

Your task: fix make_senders(channels) so senders[i]() returns "alert sent to <channels[i]>" at every index, not just the last one.

You'll practice:

Reading three identical outputs as a bug rather than a coincidence
Binding a loop variable's current value into a closure instead of the variable itself

# Each alert channel gets its own notifier. Run it - all three report the same channel.
def make_senders(channels):
    senders = []
    for name in channels:
        senders.append(lambda: f"alert sent to {name}")
    return senders

for send in make_senders(["email", "slack", "sms"]):
    print(send())