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doubleconv /
BignumDtoa.java
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// This file is available under and governed by the GNU General Public
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// However, the following notice accompanied the original version of this
// file:
//
// Copyright 2010 the V8 project authors. All rights reserved.
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//
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package jdk.nashorn.internal.runtime.doubleconv;
// Dtoa implementation based on our own Bignum implementation, supporting
// all conversion modes but slightly slower than the specialized implementations.
class BignumDtoa {
private static int normalizedExponent(long significand, int exponent) {
assert (significand != 0);
while ((significand & IeeeDouble.kHiddenBit) == 0) {
significand = significand << 1;
exponent = exponent - 1;
}
return exponent;
}
// Converts the given double 'v' to ascii.
// The result should be interpreted as buffer * 10^(point-length).
// The buffer will be null-terminated.
//
// The input v must be > 0 and different from NaN, and Infinity.
//
// The output depends on the given mode:
// - SHORTEST: produce the least amount of digits for which the internal
// identity requirement is still satisfied. If the digits are printed
// (together with the correct exponent) then reading this number will give
// 'v' again. The buffer will choose the representation that is closest to
// 'v'. If there are two at the same distance, than the number is round up.
// In this mode the 'requested_digits' parameter is ignored.
// - FIXED: produces digits necessary to print a given number with
// 'requested_digits' digits after the decimal point. The produced digits
// might be too short in which case the caller has to fill the gaps with '0's.
// Example: toFixed(0.001, 5) is allowed to return buffer="1", point=-2.
// Halfway cases are rounded up. The call toFixed(0.15, 2) thus returns
// buffer="2", point=0.
// Note: the length of the returned buffer has no meaning wrt the significance
// of its digits. That is, just because it contains '0's does not mean that
// any other digit would not satisfy the internal identity requirement.
// - PRECISION: produces 'requested_digits' where the first digit is not '0'.
// Even though the length of produced digits usually equals
// 'requested_digits', the function is allowed to return fewer digits, in
// which case the caller has to fill the missing digits with '0's.
// Halfway cases are again rounded up.
// 'BignumDtoa' expects the given buffer to be big enough to hold all digits
// and a terminating null-character.
static void bignumDtoa(final double v, final DtoaMode mode, final int requested_digits,
final DtoaBuffer buffer) {
assert (v > 0);
assert (!IeeeDouble.isSpecial(IeeeDouble.doubleToLong(v)));
final long significand;
final int exponent;
final boolean lower_boundary_is_closer;
final long l = IeeeDouble.doubleToLong(v);
significand = IeeeDouble.significand(l);
exponent = IeeeDouble.exponent(l);
lower_boundary_is_closer = IeeeDouble.lowerBoundaryIsCloser(l);
final boolean need_boundary_deltas = mode == DtoaMode.SHORTEST;
final boolean is_even = (significand & 1) == 0;
assert (significand != 0);
final int normalizedExponent = normalizedExponent(significand, exponent);
// estimated_power might be too low by 1.
final int estimated_power = estimatePower(normalizedExponent);
// Shortcut for Fixed.
// The requested digits correspond to the digits after the point. If the
// number is much too small, then there is no need in trying to get any
// digits.
if (mode == DtoaMode.FIXED && -estimated_power - 1 > requested_digits) {
buffer.reset();
// Set decimal-point to -requested_digits. This is what Gay does.
// Note that it should not have any effect anyways since the string is
// empty.
buffer.decimalPoint = -requested_digits;
return;
}
final Bignum numerator = new Bignum();
final Bignum denominator = new Bignum();
final Bignum delta_minus = new Bignum();
final Bignum delta_plus = new Bignum();
// Make sure the bignum can grow large enough. The smallest double equals
// 4e-324. In this case the denominator needs fewer than 324*4 binary digits.
// The maximum double is 1.7976931348623157e308 which needs fewer than
// 308*4 binary digits.
assert (Bignum.kMaxSignificantBits >= 324*4);
initialScaledStartValues(significand, exponent, lower_boundary_is_closer,
estimated_power, need_boundary_deltas,
numerator, denominator,
delta_minus, delta_plus);
// We now have v = (numerator / denominator) * 10^estimated_power.
buffer.decimalPoint = fixupMultiply10(estimated_power, is_even,
numerator, denominator,
delta_minus, delta_plus);
// We now have v = (numerator / denominator) * 10^(decimal_point-1), and
// 1 <= (numerator + delta_plus) / denominator < 10
switch (mode) {
case SHORTEST:
generateShortestDigits(numerator, denominator,
delta_minus, delta_plus,
is_even, buffer);
break;
case FIXED:
bignumToFixed(requested_digits,
numerator, denominator,
buffer);
break;
case PRECISION:
generateCountedDigits(requested_digits,
numerator, denominator,
buffer);
break;
default:
throw new RuntimeException();
}
}
// The procedure starts generating digits from the left to the right and stops
// when the generated digits yield the shortest decimal representation of v. A
// decimal representation of v is a number lying closer to v than to any other
// double, so it converts to v when read.
//
// This is true if d, the decimal representation, is between m- and m+, the
// upper and lower boundaries. d must be strictly between them if !is_even.
// m- := (numerator - delta_minus) / denominator
// m+ := (numerator + delta_plus) / denominator
//
// Precondition: 0 <= (numerator+delta_plus) / denominator < 10.
// If 1 <= (numerator+delta_plus) / denominator < 10 then no leading 0 digit
// will be produced. This should be the standard precondition.
static void generateShortestDigits(final Bignum numerator, final Bignum denominator,
final Bignum delta_minus, Bignum delta_plus,
final boolean is_even,
final DtoaBuffer buffer) {
// Small optimization: if delta_minus and delta_plus are the same just reuse
// one of the two bignums.
if (Bignum.equal(delta_minus, delta_plus)) {
delta_plus = delta_minus;
}
for (;;) {
final char digit;
digit = numerator.divideModuloIntBignum(denominator);
assert (digit <= 9); // digit is a uint16_t and therefore always positive.
// digit = numerator / denominator (integer division).
// numerator = numerator % denominator.
buffer.append((char) (digit + '0'));
// Can we stop already?
// If the remainder of the division is less than the distance to the lower
// boundary we can stop. In this case we simply round down (discarding the
// remainder).
// Similarly we test if we can round up (using the upper boundary).
final boolean in_delta_room_minus;
final boolean in_delta_room_plus;
if (is_even) {
in_delta_room_minus = Bignum.lessEqual(numerator, delta_minus);
} else {
in_delta_room_minus = Bignum.less(numerator, delta_minus);
}
if (is_even) {
in_delta_room_plus =
Bignum.plusCompare(numerator, delta_plus, denominator) >= 0;
} else {
in_delta_room_plus =
Bignum.plusCompare(numerator, delta_plus, denominator) > 0;
}
if (!in_delta_room_minus && !in_delta_room_plus) {
// Prepare for next iteration.
numerator.times10();
delta_minus.times10();
// We optimized delta_plus to be equal to delta_minus (if they share the
// same value). So don't multiply delta_plus if they point to the same
// object.
if (delta_minus != delta_plus) {
delta_plus.times10();
}
} else if (in_delta_room_minus && in_delta_room_plus) {
// Let's see if 2*numerator < denominator.
// If yes, then the next digit would be < 5 and we can round down.
final int compare = Bignum.plusCompare(numerator, numerator, denominator);
if (compare < 0) {
// Remaining digits are less than .5. -> Round down (== do nothing).
} else if (compare > 0) {
// Remaining digits are more than .5 of denominator. -> Round up.
// Note that the last digit could not be a '9' as otherwise the whole
// loop would have stopped earlier.
// We still have an assert here in case the preconditions were not
// satisfied.
assert (buffer.chars[buffer.length - 1] != '9');
buffer.chars[buffer.length - 1]++;
} else {
// Halfway case.
// TODO(floitsch): need a way to solve half-way cases.
// For now let's round towards even (since this is what Gay seems to
// do).
if ((buffer.chars[buffer.length - 1] - '0') % 2 == 0) {
// Round down => Do nothing.
} else {
assert (buffer.chars[buffer.length - 1] != '9');
buffer.chars[buffer.length - 1]++;
}
}
return;
} else if (in_delta_room_minus) {
// Round down (== do nothing).
return;
} else { // in_delta_room_plus
// Round up.
// Note again that the last digit could not be '9' since this would have
// stopped the loop earlier.
// We still have an ASSERT here, in case the preconditions were not
// satisfied.
assert (buffer.chars[buffer.length -1] != '9');
buffer.chars[buffer.length - 1]++;
return;
}
}
}
// Let v = numerator / denominator < 10.
// Then we generate 'count' digits of d = x.xxxxx... (without the decimal point)
// from left to right. Once 'count' digits have been produced we decide whether
// to round up or down. Remainders of exactly .5 round upwards. Numbers such
// as 9.999999 propagate a carry all the way, and change the
// exponent (decimal_point), when rounding upwards.
static void generateCountedDigits(final int count,
final Bignum numerator, final Bignum denominator,
final DtoaBuffer buffer) {
assert (count >= 0);
for (int i = 0; i < count - 1; ++i) {
final char digit;
digit = numerator.divideModuloIntBignum(denominator);
assert (digit <= 9); // digit is a uint16_t and therefore always positive.
// digit = numerator / denominator (integer division).
// numerator = numerator % denominator.
buffer.chars[i] = (char)(digit + '0');
// Prepare for next iteration.
numerator.times10();
}
// Generate the last digit.
char digit;
digit = numerator.divideModuloIntBignum(denominator);
if (Bignum.plusCompare(numerator, numerator, denominator) >= 0) {
digit++;
}
assert (digit <= 10);
buffer.chars[count - 1] = (char) (digit + '0');
// Correct bad digits (in case we had a sequence of '9's). Propagate the
// carry until we hat a non-'9' or til we reach the first digit.
for (int i = count - 1; i > 0; --i) {
if (buffer.chars[i] != '0' + 10) break;
buffer.chars[i] = '0';
buffer.chars[i - 1]++;
}
if (buffer.chars[0] == '0' + 10) {
// Propagate a carry past the top place.
buffer.chars[0] = '1';
buffer.decimalPoint++;
}
buffer.length = count;
}
// Generates 'requested_digits' after the decimal point. It might omit
// trailing '0's. If the input number is too small then no digits at all are
// generated (ex.: 2 fixed digits for 0.00001).
//
// Input verifies: 1 <= (numerator + delta) / denominator < 10.
static void bignumToFixed(final int requested_digits,
final Bignum numerator, final Bignum denominator,
final DtoaBuffer buffer) {
// Note that we have to look at more than just the requested_digits, since
// a number could be rounded up. Example: v=0.5 with requested_digits=0.
// Even though the power of v equals 0 we can't just stop here.
if (-buffer.decimalPoint > requested_digits) {
// The number is definitively too small.
// Ex: 0.001 with requested_digits == 1.
// Set decimal-decimalPoint to -requested_digits. This is what Gay does.
// Note that it should not have any effect anyways since the string is
// empty.
buffer.decimalPoint = -requested_digits;
buffer.length = 0;
// return;
} else if (-buffer.decimalPoint == requested_digits) {
// We only need to verify if the number rounds down or up.
// Ex: 0.04 and 0.06 with requested_digits == 1.
assert (buffer.decimalPoint == -requested_digits);
// Initially the fraction lies in range (1, 10]. Multiply the denominator
// by 10 so that we can compare more easily.
denominator.times10();
if (Bignum.plusCompare(numerator, numerator, denominator) >= 0) {
// If the fraction is >= 0.5 then we have to include the rounded
// digit.
buffer.chars[0] = '1';
buffer.length = 1;
buffer.decimalPoint++;
} else {
// Note that we caught most of similar cases earlier.
buffer.length = 0;
}
// return;
} else {
// The requested digits correspond to the digits after the point.
// The variable 'needed_digits' includes the digits before the point.
final int needed_digits = buffer.decimalPoint + requested_digits;
generateCountedDigits(needed_digits,
numerator, denominator,
buffer);
}
}
// Returns an estimation of k such that 10^(k-1) <= v < 10^k where
// v = f * 2^exponent and 2^52 <= f < 2^53.
// v is hence a normalized double with the given exponent. The output is an
// approximation for the exponent of the decimal approimation .digits * 10^k.
//
// The result might undershoot by 1 in which case 10^k <= v < 10^k+1.
// Note: this property holds for v's upper boundary m+ too.
// 10^k <= m+ < 10^k+1.
// (see explanation below).
//
// Examples:
// EstimatePower(0) => 16
// EstimatePower(-52) => 0
//
// Note: e >= 0 => EstimatedPower(e) > 0. No similar claim can be made for e<0.
static int estimatePower(final int exponent) {
// This function estimates log10 of v where v = f*2^e (with e == exponent).
// Note that 10^floor(log10(v)) <= v, but v <= 10^ceil(log10(v)).
// Note that f is bounded by its container size. Let p = 53 (the double's
// significand size). Then 2^(p-1) <= f < 2^p.
//
// Given that log10(v) == log2(v)/log2(10) and e+(len(f)-1) is quite close
// to log2(v) the function is simplified to (e+(len(f)-1)/log2(10)).
// The computed number undershoots by less than 0.631 (when we compute log3
// and not log10).
//
// Optimization: since we only need an approximated result this computation
// can be performed on 64 bit integers. On x86/x64 architecture the speedup is
// not really measurable, though.
//
// Since we want to avoid overshooting we decrement by 1e10 so that
// floating-point imprecisions don't affect us.
//
// Explanation for v's boundary m+: the computation takes advantage of
// the fact that 2^(p-1) <= f < 2^p. Boundaries still satisfy this requirement
// (even for denormals where the delta can be much more important).
final double k1Log10 = 0.30102999566398114; // 1/lg(10)
// For doubles len(f) == 53 (don't forget the hidden bit).
final int kSignificandSize = IeeeDouble.kSignificandSize;
final double estimate = Math.ceil((exponent + kSignificandSize - 1) * k1Log10 - 1e-10);
return (int) estimate;
}
// See comments for InitialScaledStartValues.
static void initialScaledStartValuesPositiveExponent(
final long significand, final int exponent,
final int estimated_power, final boolean need_boundary_deltas,
final Bignum numerator, final Bignum denominator,
final Bignum delta_minus, final Bignum delta_plus) {
// A positive exponent implies a positive power.
assert (estimated_power >= 0);
// Since the estimated_power is positive we simply multiply the denominator
// by 10^estimated_power.
// numerator = v.
numerator.assignUInt64(significand);
numerator.shiftLeft(exponent);
// denominator = 10^estimated_power.
denominator.assignPowerUInt16(10, estimated_power);
if (need_boundary_deltas) {
// Introduce a common denominator so that the deltas to the boundaries are
// integers.
denominator.shiftLeft(1);
numerator.shiftLeft(1);
// Let v = f * 2^e, then m+ - v = 1/2 * 2^e; With the common
// denominator (of 2) delta_plus equals 2^e.
delta_plus.assignUInt16((char) 1);
delta_plus.shiftLeft(exponent);
// Same for delta_minus. The adjustments if f == 2^p-1 are done later.
delta_minus.assignUInt16((char) 1);
delta_minus.shiftLeft(exponent);
}
}
// See comments for InitialScaledStartValues
static void initialScaledStartValuesNegativeExponentPositivePower(
final long significand, final int exponent,
final int estimated_power, final boolean need_boundary_deltas,
final Bignum numerator, final Bignum denominator,
final Bignum delta_minus, final Bignum delta_plus) {
// v = f * 2^e with e < 0, and with estimated_power >= 0.
// This means that e is close to 0 (have a look at how estimated_power is
// computed).
// numerator = significand
// since v = significand * 2^exponent this is equivalent to
// numerator = v * / 2^-exponent
numerator.assignUInt64(significand);
// denominator = 10^estimated_power * 2^-exponent (with exponent < 0)
denominator.assignPowerUInt16(10, estimated_power);
denominator.shiftLeft(-exponent);
if (need_boundary_deltas) {
// Introduce a common denominator so that the deltas to the boundaries are
// integers.
denominator.shiftLeft(1);
numerator.shiftLeft(1);
// Let v = f * 2^e, then m+ - v = 1/2 * 2^e; With the common
// denominator (of 2) delta_plus equals 2^e.
// Given that the denominator already includes v's exponent the distance
// to the boundaries is simply 1.
delta_plus.assignUInt16((char) 1);
// Same for delta_minus. The adjustments if f == 2^p-1 are done later.
delta_minus.assignUInt16((char) 1);
}
}
// See comments for InitialScaledStartValues
static void initialScaledStartValuesNegativeExponentNegativePower(
final long significand, final int exponent,
final int estimated_power, final boolean need_boundary_deltas,
final Bignum numerator, final Bignum denominator,
final Bignum delta_minus, final Bignum delta_plus) {
// Instead of multiplying the denominator with 10^estimated_power we
// multiply all values (numerator and deltas) by 10^-estimated_power.
// Use numerator as temporary container for power_ten.
final Bignum power_ten = numerator;
power_ten.assignPowerUInt16(10, -estimated_power);
if (need_boundary_deltas) {
// Since power_ten == numerator we must make a copy of 10^estimated_power
// before we complete the computation of the numerator.
// delta_plus = delta_minus = 10^estimated_power
delta_plus.assignBignum(power_ten);
delta_minus.assignBignum(power_ten);
}
// numerator = significand * 2 * 10^-estimated_power
// since v = significand * 2^exponent this is equivalent to
// numerator = v * 10^-estimated_power * 2 * 2^-exponent.
// Remember: numerator has been abused as power_ten. So no need to assign it
// to itself.
assert (numerator == power_ten);
numerator.multiplyByUInt64(significand);
// denominator = 2 * 2^-exponent with exponent < 0.
denominator.assignUInt16((char) 1);
denominator.shiftLeft(-exponent);
if (need_boundary_deltas) {
// Introduce a common denominator so that the deltas to the boundaries are
// integers.
numerator.shiftLeft(1);
denominator.shiftLeft(1);
// With this shift the boundaries have their correct value, since
// delta_plus = 10^-estimated_power, and
// delta_minus = 10^-estimated_power.
// These assignments have been done earlier.
// The adjustments if f == 2^p-1 (lower boundary is closer) are done later.
}
}
// Let v = significand * 2^exponent.
// Computes v / 10^estimated_power exactly, as a ratio of two bignums, numerator
// and denominator. The functions GenerateShortestDigits and
// GenerateCountedDigits will then convert this ratio to its decimal
// representation d, with the required accuracy.
// Then d * 10^estimated_power is the representation of v.
// (Note: the fraction and the estimated_power might get adjusted before
// generating the decimal representation.)
//
// The initial start values consist of:
// - a scaled numerator: s.t. numerator/denominator == v / 10^estimated_power.
// - a scaled (common) denominator.
// optionally (used by GenerateShortestDigits to decide if it has the shortest
// decimal converting back to v):
// - v - m-: the distance to the lower boundary.
// - m+ - v: the distance to the upper boundary.
//
// v, m+, m-, and therefore v - m- and m+ - v all share the same denominator.
//
// Let ep == estimated_power, then the returned values will satisfy:
// v / 10^ep = numerator / denominator.
// v's boundarys m- and m+:
// m- / 10^ep == v / 10^ep - delta_minus / denominator
// m+ / 10^ep == v / 10^ep + delta_plus / denominator
// Or in other words:
// m- == v - delta_minus * 10^ep / denominator;
// m+ == v + delta_plus * 10^ep / denominator;
//
// Since 10^(k-1) <= v < 10^k (with k == estimated_power)
// or 10^k <= v < 10^(k+1)
// we then have 0.1 <= numerator/denominator < 1
// or 1 <= numerator/denominator < 10
//
// It is then easy to kickstart the digit-generation routine.
//
// The boundary-deltas are only filled if the mode equals BIGNUM_DTOA_SHORTEST
// or BIGNUM_DTOA_SHORTEST_SINGLE.
static void initialScaledStartValues(final long significand,
final int exponent,
final boolean lower_boundary_is_closer,
final int estimated_power,
final boolean need_boundary_deltas,
final Bignum numerator,
final Bignum denominator,
final Bignum delta_minus,
final Bignum delta_plus) {
if (exponent >= 0) {
initialScaledStartValuesPositiveExponent(
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