On Tue, 2026-01-20 at 16:38 -0300, Wander Lairson Costa wrote: > On Tue, Jan 20, 2026 at 01:30:35PM +0100, Gabriele Monaco wrote: > > > You basically want i to be the length of the longest prefix common to at > > least > > another atom. > > > > You could assign i to some python trick doing the exact same thing the loop > > does, like: > > > > i = next((i for i in range(len(atom), -1, -1) > > if sum(a.startswith(atom[:i]) for a in atoms) > 1)) > > > > next() is basically doing the break at the first occurrence from the > > generator, > > just now your i doesn't live (only) inside the loop. > > > > So now you save 2 lines and get any C developer scratch their head when they > > look at the code, but hey, pyright is happy! > > > > Or just leave the assignment. > > > If you do find the trick with next() readable or have any better idea, feel > > free > > to try though. > > > > Definitely the next() trick is not worth to make pyright happy.
Alright, thinking on this again next() is the python way to do if(...) break , it looked a bit odd to me only because I didn't know about it, but if you're using python iterators, it kinda makes sense. Anyway I'm fine also with the dull assignment, there's no need to argue on this. Thanks, Gabriele > > > Thanks, > > Gabriele > > > > > I will modify it in v2. > > > > > > > > > > > Thanks, > > > > Gabriele > > > > > > > > [1] - https://github.com/microsoft/pyright/issues/844 > > > > > > > > > > > > > > Signed-off-by: Wander Lairson Costa <[email protected]> > > > > > --- > > > > > tools/verification/rvgen/rvgen/ltl2k.py | 1 + > > > > > 1 file changed, 1 insertion(+) > > > > > > > > > > diff --git a/tools/verification/rvgen/rvgen/ltl2k.py > > > > > b/tools/verification/rvgen/rvgen/ltl2k.py > > > > > index fa9ea6d597095..94dc64af1716d 100644 > > > > > --- a/tools/verification/rvgen/rvgen/ltl2k.py > > > > > +++ b/tools/verification/rvgen/rvgen/ltl2k.py > > > > > @@ -45,6 +45,7 @@ def abbreviate_atoms(atoms: list[str]) -> list[str]: > > > > > > > > > > abbrs = [] > > > > > for atom in atoms: > > > > > + i = 0 > > > > > for i in range(len(atom), -1, -1): > > > > > if sum(a.startswith(atom[:i]) for a in atoms) > 1: > > > > > break > > > > > >
