in nearest rounding mode exact halfway cases were not following the
round to even rule if the rounding happened at a base
1000000000 digit
boundary of the internal representation and the previous digit was odd.
e.g. printf("%.0f", 1.5) printed 1 instead of 2.
if (x || d+1!=z) {
long double round = 2/LDBL_EPSILON;
long double small;
- if (*d/i & 1) round += 2;
+ if ((*d/i & 1) || (i==1000000000 && d>a && (d[-1]&1)))
+ round += 2;
if (x<i/2) small=0x0.8p0;
else if (x==i/2 && d+1==z) small=0x1.0p0;
else small=0x1.8p0;