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[185.132.178.103]) by smtp.gmail.com with ESMTPSA id ffacd0b85a97d-43569921f6esm33540846f8f.4.2026.01.20.22.31.23 (version=TLS1_3 cipher=TLS_AES_256_GCM_SHA384 bits=256/256); Tue, 20 Jan 2026 22:31:24 -0800 (PST) Message-ID: <7e83ab8b607cdc9e1c677e7b118d46eabe35b7f8.camel@redhat.com> Subject: Re: [PATCH 17/26] rv/rvgen: fix possibly unbound variable in ltl2k From: Gabriele Monaco To: Wander Lairson Costa Cc: Steven Rostedt , Nam Cao , open list , "open list:RUNTIME VERIFICATION (RV)" Date: Wed, 21 Jan 2026 07:31:22 +0100 In-Reply-To: References: <20260119205601.105821-1-wander@redhat.com> <20260119205601.105821-18-wander@redhat.com> <3507c2f454ca66aa27dd3a24e33f5941fac96639.camel@redhat.com> Autocrypt: addr=gmonaco@redhat.com; prefer-encrypt=mutual; keydata=mDMEZuK5YxYJKwYBBAHaRw8BAQdAmJ3dM9Sz6/Hodu33Qrf8QH2bNeNbOikqYtxWFLVm0 1a0JEdhYnJpZWxlIE1vbmFjbyA8Z21vbmFjb0BrZXJuZWwub3JnPoiZBBMWCgBBFiEEysoR+AuB3R Zwp6j270psSVh4TfIFAmjKX2MCGwMFCQWjmoAFCwkIBwICIgIGFQoJCAsCBBYCAwECHgcCF4AACgk Q70psSVh4TfIQuAD+JulczTN6l7oJjyroySU55Fbjdvo52xiYYlMjPG7dCTsBAMFI7dSL5zg98I+8 cXY1J7kyNsY6/dcipqBM4RMaxXsOtCRHYWJyaWVsZSBNb25hY28gPGdtb25hY29AcmVkaGF0LmNvb T6InAQTFgoARAIbAwUJBaOagAULCQgHAgIiAgYVCgkICwIEFgIDAQIeBwIXgBYhBMrKEfgLgd0WcK eo9u9KbElYeE3yBQJoymCyAhkBAAoJEO9KbElYeE3yjX4BAJ/ETNnlHn8OjZPT77xGmal9kbT1bC1 7DfrYVISWV2Y1AP9HdAMhWNAvtCtN2S1beYjNybuK6IzWYcFfeOV+OBWRDQ== Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable User-Agent: Evolution 3.58.2 (3.58.2-1.fc43) Precedence: bulk X-Mailing-List: linux-kernel@vger.kernel.org List-Id: List-Subscribe: List-Unsubscribe: MIME-Version: 1.0 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: >=20 > > You basically want i to be the length of the longest prefix common to a= t > > least > > another atom. > >=20 > > You could assign i to some python trick doing the exact same thing the = loop > > does, like: > >=20 > > =C2=A0=C2=A0=C2=A0 i =3D next((i for i in range(len(atom), -1, -1) > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 if sum(a.startswith(atom[:i]= ) for a in atoms) > 1)) > >=20 > > next() is basically doing the break at the first occurrence from the > > generator, > > just now your i doesn't live (only) inside the loop. > >=20 > > 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! > >=20 >=20 > Or just leave the assignment. >=20 > > If you do find the trick with next() readable or have any better idea, = feel > > free > > to try though. > >=20 >=20 > Definitely the next() trick is not worth to make pyright happy. Alright, thinking on this again next() is the python way to do if(...) brea= k , it looked a bit odd to me only because I didn't know about it, but if you'r= e 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 >=20 > > Thanks, > > Gabriele > >=20 > > > I will modify it in v2. > > >=20 > > > >=20 > > > > Thanks, > > > > Gabriele > > > >=20 > > > > [1] - https://github.com/microsoft/pyright/issues/844 > > > >=20 > > > > >=20 > > > > > Signed-off-by: Wander Lairson Costa > > > > > --- > > > > > =C2=A0tools/verification/rvgen/rvgen/ltl2k.py | 1 + > > > > > =C2=A01 file changed, 1 insertion(+) > > > > >=20 > > > > > 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]: > > > > > =C2=A0 > > > > > =C2=A0=C2=A0=C2=A0=C2=A0 abbrs =3D [] > > > > > =C2=A0=C2=A0=C2=A0=C2=A0 for atom in atoms: > > > > > +=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 i =3D 0 > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 for i in range(l= en(atom), -1, -1): > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2= =A0=C2=A0 if sum(a.startswith(atom[:i]) for a in atoms) > 1: > > > > > =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2= =A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 break > > > >=20 > >=20