libavutil/eval.c "double" error comparison is equal to

[张文兵]

libavutil/eval.c
func eval_expr

        case e_between: {
            double d = eval_expr(p, e->param[0]);
            return e->value * (d >= eval_expr(p, e->param[1]) &&
                               d <= eval_expr(p, e->param[2]));
        }

......
                case e_eq:  return e->value * (d == d2 ? 1.0 : 0.0);
                case e_gte: return e->value * (d >= d2 ? 1.0 : 0.0);
                case e_lte: return e->value * (d <= d2 ? 1.0 : 0.0);

[Gyan]

Can you elaborate what the bug is?

[张文兵]

Replying to Gyan:

Can you elaborate what the bug is?

Two "doubles" cannot be used directly as equals, because the precision problem will cause two values that look the same to be unequal.

ffmpeg -i <input> -vf "fps=fps=25,settb=AVTB,setpts=PTS-STARTPTS,delogo=x=432:y=1:w=173:h=463:enable='between(t,2376.760000,2381.400000)'" -s 1920x1080 -y -pix_fmt yuv420p -c:v libx264 -preset ultrafast -r 25.000000 -acodec aac -strict -2 -ac 2 -ar 48000 <output>

"between(t,2376.760000,2381.400000)'" cannot correctly determine 2376.760000

Here is how to reproduce.

#include <inttypes.h>
#include <math.h>
#include <stdio.h>

int main(void) {
    int64_t pts = 2376760000;
    double t = pts * (1 / (double)(1000000));

    double myt = 2376.76;

    printf("t:%.32f myt:%.32f pts:%lld %d\n",
           t,
           myt,
           pts,
           myt == t);
    return 0;
}

// int main(void) {
//     int64_t pts = 59419;
//     double t = pts * (1 / (double)(25));

//     double myt = 2376.76;

//     printf("t:%.32f myt:%.32f pts:%lld %d\n",
//            t,
//            myt,
//            pts,
//            myt == t);
//     return 0;
// }

[Elon Musk]

can be changed by adding comparisions using DBL_EPSILON