On Wed, Jan 29, 2020 at 03:25:32PM -0800, Dale Curtis wrote: > On Wed, Jan 29, 2020 at 4:55 AM Michael Niedermayer <mich...@niedermayer.cc> > wrote: > > > simpler solution, and also behaves arithmetically more correct when the > > overflow happens in the othert direction: > > > > av_assert0(time_tolerance >= 0); > > > > if (e2_pts < e1_pts || e2_pts - (uint64_t)e1_pts < time_tolerance) > > > > Does that work? e1_pts is INT64_MIN in this case. So the (uint64_t)e1_pts > > e2_pts.
the original code was this: if (e2_pts - e1_pts < time_tolerance) time_tolerance is always positive e2_pts - e1_pts can be either positive or negative if its negative (e2_pts < e1_pts) then the original would be true and thats the first term of the new condition otherwise e2_pts - e1_pts is positive and can be represented as a uint64_t which is the 2nd term of the new condition so i think it works but maybe ive missed something, for which values of e2_pts do you see a problem with e1_pts = INT64_MIN? thx [...] -- Michael GnuPG fingerprint: 9FF2128B147EF6730BADF133611EC787040B0FAB Asymptotically faster algorithms should always be preferred if you have asymptotical amounts of data
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