On my machine, an example like this results in a leak:

aut = spot.translate(spot.formula('(F(p & X(p & Xh)) & G((!h | Xp) & (!m | X!p))) | (F(h & X!p) & G((!m | X!p) & (!p | X(!p | X!h))))'))
for i in range(100):
newaut = spot.automaton(aut.to_str())
for i in range(0, 10):
for t in newaut.out(i):
pass

del newaut
print(resource.getrusage(resource.RUSAGE_SELF).ru_maxrss)

Indeed, the out_iteraser method is really a bit inconvenient to use ; )

On Fri, Jan 28, 2022 at 5:58 PM Alexandre Duret-Lutz <adl@lrde.epita.fr> wrote:

Yechuan Xia <xiaozi465@gmail.com> writes:

> Thank you for your reply!
>
> And I've found that it's not the memory of the automaton not being
> freed, but the containers acquired by 'aut.out(i)' not being freed in
> '...something'.
> This problem can be solved by using aut.out_iteraser(i).

Please provide a small example to reproduce the problem so it can be
fixed. What you describe sound unexpected.

Adding

for i in newaut.out(0):
pass

to my previous example does not leak.

> So I'm just wondering why there are two methods for getting out
> edges, and aut.out(i) doesn't allow erasure? And it seems that the
> aut.out(i) method is used by default in the spot's code examples.

out_iteraser does not follow C++'s and Python's convention for
iteration, so it is a bit cumbersome to use. There are very few cases
where out_iteraser() is needed.