On Fri, May 24, 2024 at 12:33:11PM +0200, Gerion Entrup wrote:
> Am Dienstag, 7. Mai 2024, 19:46:28 MESZ schrieb Michael Niedermayer:
> > On Mon, May 06, 2024 at 12:30:39AM +0200, Gerion Entrup wrote:
> > > ---
> > > libavfilter/signature_lookup.c | 2 +-
> > > 1 file changed, 1 insertion(+), 1 deletion(-)
> > >
> > > diff --git a/libavfilter/signature_lookup.c
> > > b/libavfilter/signature_lookup.c
> > > index a0ca818a9b..b39a3e225b 100644
> > > --- a/libavfilter/signature_lookup.c
> > > +++ b/libavfilter/signature_lookup.c
> > > @@ -128,7 +128,7 @@ static int get_jaccarddist(SignatureContext *sc,
> > > CoarseSignature *first, CoarseS
> > > int jaccarddist, i, composdist = 0, cwthcount = 0;
> > > for (i = 0; i < 5; i++) {
> > > if ((jaccarddist = intersection_word(first->data[i],
> > > second->data[i])) > 0) {
> > > - jaccarddist /= union_word(first->data[i], second->data[i]);
> > > + jaccarddist /= FFMAX(union_word(first->data[i],
> > > second->data[i]), 1);
> > > }
> >
> > for which input data does this cause a division by 0 ?
>
> Sorry for the late answer. I missed your mail somehow.
> union_word counts the amount of one bits that are created when you are "or"ing
> the course signatures. So, when the underlying videos are so different that
> all
> bits of the created signatures are different, the "or"-operator will always
> return 0 and so also its sum (I have not tested this).the division only occurs if jaccarddist > 0 basically what iam asking is for which A and B do we have (A&B) != 0 && (A|B) == 0 or am i misreading the code ? thx [...] -- Michael GnuPG fingerprint: 9FF2128B147EF6730BADF133611EC787040B0FAB Awnsering whenever a program halts or runs forever is On a turing machine, in general impossible (turings halting problem). On any real computer, always possible as a real computer has a finite number of states N, and will either halt in less than N cycles or never halt.
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