On Sun, Dec 07, 2014 at 01:31:59PM +0100, Nicolas George wrote: > Only count the number of channels that are present on input. > Fix a case where input FL+FR+BL gives a better score to 6.1 > than quad because BL+BR→SL+SR counts as two remapped channels > while BL→BL+BR counts as one preserved and one extra. > > Signed-off-by: Nicolas George <geo...@nsup.org> > --- > libavfilter/avfiltergraph.c | 3 ++- > 1 file changed, 2 insertions(+), 1 deletion(-) > > > There are more issues in this score logic. > For example stereo+BR→stereo+SL is accepted because of the BL+BR→SL+SR > substitution, this is wrong. > > The obvious fix would be to change "(out_chlayout & cmp1)" into > "(out_chlayout & cmp1) == cmp1", i.e. only accept a substitution if the > target is entirely available in the output, but that would break the case > stereo+BR→stereo+SL. > > I do not have time for a more complete fix right now, but I believe the > present patch is valid and useful by itself.
LGTM though this is your code, you know it much better than i do [...] -- 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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