On Thu, Mar 12, 2015 at 07:14:54AM -0400, Ronald S. Bultje wrote: > Hi, > > On Wed, Mar 11, 2015 at 9:00 PM, Michael Niedermayer <michae...@gmx.at> > wrote: > > > Found-by: Clang -fsanitize=shift > > Reported-by: Thierry Foucu <tfo...@google.com> > > Signed-off-by: Michael Niedermayer <michae...@gmx.at> > > --- > > libavcodec/h264_mb.c | 14 +++++++------- > > 1 file changed, 7 insertions(+), 7 deletions(-) > > > > diff --git a/libavcodec/h264_mb.c b/libavcodec/h264_mb.c > > index dd406c7..a4653aa 100644 > > --- a/libavcodec/h264_mb.c > > +++ b/libavcodec/h264_mb.c > > @@ -213,7 +213,7 @@ static av_always_inline void mc_dir_part(H264Context > > *h, H264Picture *pic, > > const int mx = h->mv_cache[list][scan8[n]][0] + src_x_offset * 8; > > int my = h->mv_cache[list][scan8[n]][1] + src_y_offset * 8; > > const int luma_xy = (mx & 3) + ((my & 3) << 2); > > - ptrdiff_t offset = ((mx >> 2) << pixel_shift) + (my >> 2) * > > h->mb_linesize; > > + ptrdiff_t offset = (mx >> 2) * (1 << pixel_shift) + (my >> 2) * > > h->mb_linesize; > > > Why is this undefined?
Because C sucks 6.5.7 Bitwise shift operators ... 4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 * 2^E2 , reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 * 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined. [...] -- Michael GnuPG fingerprint: 9FF2128B147EF6730BADF133611EC787040B0FAB it is not once nor twice but times without number that the same ideas make their appearance in the world. -- Aristotle
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