The problematic function (ffmpeg/libavutil/rational.h):

/**

• Rational number num/den.
  */
  typedef struct AVRational{
  int num; ///< numerator
  int den; ///< denominator
  } AVRational;

/**

• Compare two rationals.

• @​param a first rational

• @​param b second rational

• @​return 0 if a==b, 1 if a>b and -1 if a<b.
  */
  static inline int av_cmp_q(AVRational a, AVRational b){
  const int64_t tmp= a.num * (int64_t)b.den - b.num * (int64_t)a.den;

  if(tmp) return (tmp>>63)Fixing Rust build #1;
  else return 0;
  }

I think the comment is wrong. This function returns 0 if a==b, 1 if a>b and
-0x7FFFFFFF if a<b.

fix
Arithmetic left shift is not the inverse operation of arithmetic right shift.
The arithmetic left shift can result in overflow. So, we should preserve the
sign bit beyond making the arithmetic left shift.

The patch is close, but not quite right. The sign bit definitely does need to be demanded in some
cases, but only in those cases where one of the shifted in bits is used. For example, consider:

%Y = ashr i8 %X, 3

If we demand (f.e.) the low two bits from Y, then the input sign bit is not needed. However, if any of
the top three bits are demanded, then the input sign bit is demanded.

So basically, in that code, if any of the "high bits" are demanded, you should set the sign bit as
demanded. If none of the high bits are demanded, the sign bit should not be demanded.

Thanks again for investigating this Lauro!

-Chris

Fixed.

http://lists.cs.uiuc.edu/pipermail/llvm-commits/Week-of-Mon-20070604/050287.html
http://lists.cs.uiuc.edu/pipermail/llvm-commits/Week-of-Mon-20070604/050288.html