On Tue, 6 Jul 2021 at 23:39, Dean Rasheed <dean.a.rash...@gmail.com> wrote: > When I first read this: > > + /* half-round until we get down to unitBits */ > + while (rightshifts++ < unit->unitBits) > + size = half_rounded(size); > > it looked to me like it would be invoking half_rounded() multiple > times, which raised alarm bells because that would risk rounding the > wrong way. Eventually I realised that by the time it reaches that, > rightshifts will always equal unit->unitBits or unit->unitBits - 1, so > it'll never do more than one half-round, which is important.
It's true that based on how the units table is set up now, it'll only ever do it once for all but the first loop. I wrote the attached .c file just to try to see if it ever goes wrong and I didn't manage to find any inputs where it did. I always seem to get the half rounded value either the same as the shifted value or 1 higher towards positive infinity $ ./half_rounded -102 3 1. half_round(-102) == -51 :: -102 >> 1 = -51 2. half_round(-51) == -25 :: -51 >> 1 = -26 3. half_round(-25) == -12 :: -26 >> 1 = -13 $ ./half_rounded 6432 3 1. half_round(6432) == 3216 :: 6432 >> 1 = 3216 2. half_round(3216) == 1608 :: 3216 >> 1 = 1608 3. half_round(1608) == 804 :: 1608 >> 1 = 804 Can you give an example where calling half_rounded too many times will give the wrong value? Keeping in mind we call half_rounded the number of times that the passed in value would need to be left-shifted by to get the equivalent truncated value. David
#define half_rounded(x) (((x) + ((x) < 0 ? 0 : 1)) / 2) #include <stdio.h> #include <stdlib.h> int main(int argc, char **argv) { int num, times; int i; int r, shift; if (argc < 3) { printf("Usage: %s <num> <times>\n"); return -1; } num = atoi(argv[1]); times = atoi(argv[2]); r = num; shift = num; for (i = 0; i < times; i++) { int rr = half_rounded(r); printf("%d. half_round(%d) == %d :: %d >> 1 = %d\n", i+1, r, rr, shift, shift >> 1); r = rr; shift >>= 1; } return 0; }