Execution order with try-finally in python generator functions

Question

def gen():
    try:
        yield 1
    finally:
        print("finally")


def main():
    print(next(gen()))

This code prints

finally
1

I don't understand the order of execution here. Why is "finally" printed before "1"?

Because `finally` block triggered on leaving `try` and value returned from generator function on leaving function. — Olvin Roght, Jul 19 '22 at 08:46
Yes @olvin, that's it. A subtle point about garbage collection. Thank you! — wiseacre, Jul 19 '22 at 09:16

score 0 · Answer 1 · answered Jul 19 '22 at 08:48

The reason why "finally" is printed before "1" is because the first thing Python does is resolve next(gen()), which involves returning 1 and printing "finally". Once Python is done executing next(gen()), it has printed finally, and has a value of 1. Now, it can resolve print(1), which prints "1".

score 0 · Answer 2 · answered Jul 19 '22 at 08:50

Here is an Example what is happening

next(gen()) is executed -> gen() prints "finally" -> next() gets the value
next() returns the value
print() prints the value

You have to look here inside first and then the outer shell and the outer outer shell etc.

score 0 · Accepted Answer · answered Jul 19 '22 at 09:14

You can rewrite the line print(next(gen())) as

g = gen()  # temporary variable
val = next(g)
#          # the generator is now deleted because the variable g is not referenced anymore
del(g)     # that calls the "finally" block `print("finally")`
print(val)

If you assign the generator to a variable, but don't delete it then it is recycled at the end of main function. That is what you probably expect.

A generator should be assigned to a variable if you use it by next(), otherwise there would be no way to use a generator with more items because it would be recycled after the first item!

Execution order with try-finally in python generator functions

3 Answers3