return
}
- // TODO(rsc): Unless exp == minexp, if the number of digits in d
- // is less than 17, it seems likely that it would be
- // the shortest possible number already. So maybe we can
- // bail out without doing the extra multiprecision math here.
-
// Compute upper and lower such that any decimal number
// between upper and lower (possibly inclusive)
// will round to the original floating point number.
+ // We may see at once that the number is already shortest.
+ //
+ // Suppose d is not denormal, so that 2^exp <= d < 10^dp.
+ // The closest shorter number is at least 10^(dp-nd) away.
+ // The lower/upper bounds computed below are at distance
+ // at most 2^(exp-mantbits).
+ //
+ // So the number is already shortest if 10^(dp-nd) > 2^(exp-mantbits),
+ // or equivalently log2(10)*(dp-nd) > exp-mantbits.
+ // It is true if 332/100*(dp-nd) >= exp-mantbits (log2(10) > 3.32).
+ minexp := flt.bias + 1 // minimum possible exponent
+ if exp > minexp && 332*(d.dp-d.nd) >= 100*(exp-int(flt.mantbits)) {
+ // The number is already shortest.
+ return
+ }
+
// d = mant << (exp - mantbits)
// Next highest floating point number is mant+1 << exp-mantbits.
// Our upper bound is halfway inbetween, mant*2+1 << exp-mantbits-1.
// in which case the next lowest is mant*2-1 << exp-mantbits-1.
// Either way, call it mantlo << explo-mantbits.
// Our lower bound is halfway inbetween, mantlo*2+1 << explo-mantbits-1.
- minexp := flt.bias + 1 // minimum possible exponent
var mantlo uint64
var explo int
if mant > 1<<flt.mantbits || exp == minexp {