/* -*- Mode: java; tab-width: 8; indent-tabs-mode: nil; c-basic-offset: 4 -*-
*
* ***** BEGIN LICENSE BLOCK *****
* Version: MPL 1.1/GPL 2.0
*
* The contents of this file are subject to the Mozilla Public License Version
* 1.1 (the "License"); you may not use this file except in compliance with
* the License. You may obtain a copy of the License at
* http://www.mozilla.org/MPL/
*
* Software distributed under the License is distributed on an "AS IS" basis,
* WITHOUT WARRANTY OF ANY KIND, either express or implied. See the License
* for the specific language governing rights and limitations under the
* License.
*
* The Original Code is Rhino code, released
* May 6, 1999.
*
* The Initial Developer of the Original Code is
* Netscape Communications Corporation.
* Portions created by the Initial Developer are Copyright (C) 1997-1999
* the Initial Developer. All Rights Reserved.
*
* Contributor(s):
* Waldemar Horwat
* Roger Lawrence
* Attila Szegedi
*
* Alternatively, the contents of this file may be used under the terms of
* the GNU General Public License Version 2 or later (the "GPL"), in which
* case the provisions of the GPL are applicable instead of those above. If
* you wish to allow use of your version of this file only under the terms of
* the GPL and not to allow others to use your version of this file under the
* MPL, indicate your decision by deleting the provisions above and replacing
* them with the notice and other provisions required by the GPL. If you do
* not delete the provisions above, a recipient may use your version of this
* file under either the MPL or the GPL.
*
* ***** END LICENSE BLOCK ***** */
/****************************************************************
*
* The author of this software is David M. Gay.
*
* Copyright (c) 1991, 2000, 2001 by Lucent Technologies.
*
* Permission to use, copy, modify, and distribute this software for any
* purpose without fee is hereby granted, provided that this entire notice
* is included in all copies of any software which is or includes a copy
* or modification of this software and in all copies of the supporting
* documentation for such software.
*
* THIS SOFTWARE IS BEING PROVIDED "AS IS", WITHOUT ANY EXPRESS OR IMPLIED
* WARRANTY. IN PARTICULAR, NEITHER THE AUTHOR NOR LUCENT MAKES ANY
* REPRESENTATION OR WARRANTY OF ANY KIND CONCERNING THE MERCHANTABILITY
* OF THIS SOFTWARE OR ITS FITNESS FOR ANY PARTICULAR PURPOSE.
*
***************************************************************/
package org.mozilla.javascript;
import java.math.BigInteger;
class DToA {
/* "-0.0000...(1073 zeros after decimal point)...0001\0" is the longest string that we could produce,
* which occurs when printing -5e-324 in binary. We could compute a better estimate of the size of
* the output string and malloc fewer bytes depending on d and base, but why bother? */
private static final int DTOBASESTR_BUFFER_SIZE = 1078;
private static char BASEDIGIT(int digit) {
return (char)((digit >= 10) ? 'a' - 10 + digit : '0' + digit);
}
static final int
DTOSTR_STANDARD = 0, /* Either fixed or exponential format; round-trip */
DTOSTR_STANDARD_EXPONENTIAL = 1, /* Always exponential format; round-trip */
DTOSTR_FIXED = 2, /* Round to <precision> digits after the decimal point; exponential if number is large */
DTOSTR_EXPONENTIAL = 3, /* Always exponential format; <precision> significant digits */
DTOSTR_PRECISION = 4; /* Either fixed or exponential format; <precision> significant digits */
private static final int Frac_mask = 0xfffff;
private static final int Exp_shift = 20;
private static final int Exp_msk1 = 0x100000;
private static final long Frac_maskL = 0xfffffffffffffL;
private static final int Exp_shiftL = 52;
private static final long Exp_msk1L = 0x10000000000000L;
private static final int Bias = 1023;
private static final int P = 53;
private static final int Exp_shift1 = 20;
private static final int Exp_mask = 0x7ff00000;
private static final int Exp_mask_shifted = 0x7ff;
private static final int Bndry_mask = 0xfffff;
private static final int Log2P = 1;
private static final int Sign_bit = 0x80000000;
private static final int Exp_11 = 0x3ff00000;
private static final int Ten_pmax = 22;
private static final int Quick_max = 14;
private static final int Bletch = 0x10;
private static final int Frac_mask1 = 0xfffff;
private static final int Int_max = 14;
private static final int n_bigtens = 5;
private static final double tens[] = {
1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9,
1e10, 1e11, 1e12, 1e13, 1e14, 1e15, 1e16, 1e17, 1e18, 1e19,
1e20, 1e21, 1e22
};
private static final double bigtens[] = { 1e16, 1e32, 1e64, 1e128, 1e256 };
private static int lo0bits(int y)
{
int k;
int x = y;
if ((x & 7) != 0) {
if ((x & 1) != 0)
return 0;
if ((x & 2) != 0) {
return 1;
}
return 2;
}
k = 0;
if ((x & 0xffff) == 0) {
k = 16;
x >>>= 16;
}
if ((x & 0xff) == 0) {
k += 8;
x >>>= 8;
}
if ((x & 0xf) == 0) {
k += 4;
x >>>= 4;
}
if ((x & 0x3) == 0) {
k += 2;
x >>>= 2;
}
if ((x & 1) == 0) {
k++;
x >>>= 1;
if ((x & 1) == 0)
return 32;
}
return k;
}
/* Return the number (0 through 32) of most significant zero bits in x. */
private static int hi0bits(int x)
{
int k = 0;
if ((x & 0xffff0000) == 0) {
k = 16;
x <<= 16;
}
if ((x & 0xff000000) == 0) {
k += 8;
x <<= 8;
}
if ((x & 0xf0000000) == 0) {
k += 4;
x <<= 4;
}
if ((x & 0xc0000000) == 0) {
k += 2;
x <<= 2;
}
if ((x & 0x80000000) == 0) {
k++;
if ((x & 0x40000000) == 0)
return 32;
}
return k;
}
private static void stuffBits(byte bits[], int offset, int val)
{
bits[offset] = (byte)(val >> 24);
bits[offset + 1] = (byte)(val >> 16);
bits[offset + 2] = (byte)(val >> 8);
bits[offset + 3] = (byte)(val);
}
/* Convert d into the form b*2^e, where b is an odd integer. b is the returned
* Bigint and e is the returned binary exponent. Return the number of significant
* bits in b in bits. d must be finite and nonzero. */
private static BigInteger d2b(double d, int[] e, int[] bits)
{
byte dbl_bits[];
int i, k, y, z, de;
long dBits = Double.doubleToLongBits(d);
int d0 = (int)(dBits >>> 32);
int d1 = (int)(dBits);
z = d0 & Frac_mask;
d0 &= 0x7fffffff; /* clear sign bit, which we ignore */
if ((de = (d0 >>> Exp_shift)) != 0)
z |= Exp_msk1;
if ((y = d1) != 0) {
dbl_bits = new byte[8];
k = lo0bits(y);
y >>>= k;
if (k != 0) {
stuffBits(dbl_bits, 4, y | z << (32 - k));
z >>= k;
}
else
stuffBits(dbl_bits, 4, y);
stuffBits(dbl_bits, 0, z);
i = (z != 0) ? 2 : 1;
}
else {
// JS_ASSERT(z);
dbl_bits = new byte[4];
k = lo0bits(z);
z >>>= k;
stuffBits(dbl_bits, 0, z);
k += 32;
i = 1;
}
if (de != 0) {
e[0] = de - Bias - (P-1) + k;
bits[0] = P - k;
}
else {
e[0] = de - Bias - (P-1) + 1 + k;
bits[0] = 32*i - hi0bits(z);
}
return new BigInteger(dbl_bits);
}
static String JS_dtobasestr(int base, double d)
{
if (!(2 <= base && base <= 36))
throw new IllegalArgumentException("Bad base: "+base);
/* Check for Infinity and NaN */
if (Double.isNaN(d)) {
return "NaN";
} else if (Double.isInfinite(d)) {
return (d > 0.0) ? "Infinity" : "-Infinity";
} else if (d == 0) {
// ALERT: should it distinguish -0.0 from +0.0 ?
return "0";
}
boolean negative;
if (d >= 0.0) {
negative = false;
} else {
negative = true;
d = -d;
}
/* Get the integer part of d including '-' sign. */
String intDigits;
double dfloor = Math.floor(d);
long lfloor = (long)dfloor;
if (lfloor == dfloor) {
// int part fits long
intDigits = Long.toString((negative) ? -lfloor : lfloor, base);
} else {
// BigInteger should be used
long floorBits = Double.doubleToLongBits(dfloor);
int exp = (int)(floorBits >> Exp_shiftL) & Exp_mask_shifted;
long mantissa;
if (exp == 0) {
mantissa = (floorBits & Frac_maskL) << 1;
} else {
mantissa = (floorBits & Frac_maskL) | Exp_msk1L;
}
if (negative) {
mantissa = -mantissa;
}
exp -= 1075;
BigInteger x = BigInteger.valueOf(mantissa);
if (exp > 0) {
x = x.shiftLeft(exp);
} else if (exp < 0) {
x = x.shiftRight(-exp);
}
intDigits = x.toString(base);
}
if (d == dfloor) {
// No fraction part
return intDigits;
} else {
/* We have a fraction. */
char[] buffer; /* The output string */
int p; /* index to current position in the buffer */
int digit;
double df; /* The fractional part of d */
BigInteger b;
buffer = new char[DTOBASESTR_BUFFER_SIZE];
p = 0;
df = d - dfloor;
long dBits = Double.doubleToLongBits(d);
int word0 = (int)(dBits >> 32);
int word1 = (int)(dBits);
int[] e = new int[1];
int[] bbits = new int[1];
b = d2b(df, e, bbits);
// JS_ASSERT(e < 0);
/* At this point df = b * 2^e. e must be less than zero because 0 < df < 1. */
int s2 = -(word0 >>> Exp_shift1 & Exp_mask >> Exp_shift1);
if (s2 == 0)
s2 = -1;
s2 += Bias + P;
/* 1/2^s2 = (nextDouble(d) - d)/2 */
// JS_ASSERT(-s2 < e);
BigInteger mlo = BigInteger.valueOf(1);
BigInteger mhi = mlo;
if ((word1 == 0) && ((word0 & Bndry_mask) == 0)
&& ((word0 & (Exp_mask & Exp_mask << 1)) != 0)) {
/* The special case. Here we want to be within a quarter of the last input
significant digit instead of one half of it when the output string's value is less than d. */
s2 += Log2P;
mhi = BigInteger.valueOf(1<<Log2P);
}
b = b.shiftLeft(e[0] + s2);
BigInteger s = BigInteger.valueOf(1);
s = s.shiftLeft(s2);
/* At this point we have the following:
* s = 2^s2;
* 1 > df = b/2^s2 > 0;
* (d - prevDouble(d))/2 = mlo/2^s2;
* (nextDouble(d) - d)/2 = mhi/2^s2. */
BigInteger bigBase = BigInteger.valueOf(base);
boolean done = false;
do {
b = b.multiply(bigBase);
BigInteger[] divResult = b.divideAndRemainder(s);
b = divResult[1];
digit = (char)(divResult[0].intValue());
if (mlo == mhi)
mlo = mhi = mlo.multiply(bigBase);
else {
mlo = mlo.multiply(bigBase);
mhi = mhi.multiply(bigBase);
}
/* Do we yet have the shortest string that will round to d? */
int j = b.compareTo(mlo);
/* j is b/2^s2 compared with mlo/2^s2. */
BigInteger delta = s.subtract(mhi);
int j1 = (delta.signum() <= 0) ? 1 : b.compareTo(delta);
/* j1 is b/2^s2 compared with 1 - mhi/2^s2. */
if (j1 == 0 && ((word1 & 1) == 0)) {
if (j > 0)
digit++;
done = true;
} else
if (j < 0 || (j == 0 && ((word1 & 1) == 0))) {
if (j1 > 0) {
/* Either dig or dig+1 would work here as the least significant digit.
Use whichever would produce an output value closer to d. */
b = b.shiftLeft(1);
j1 = b.compareTo(s);
if (j1 > 0) /* The even test (|| (j1 == 0 && (digit & 1))) is not here because it messes up odd base output
* such as 3.5 in base 3. */
digit++;
}
done = true;
} else if (j1 > 0) {
digit++;
done = true;
}
// JS_ASSERT(digit < (uint32)base);
buffer[p++] = BASEDIGIT(digit);
} while (!done);
StringBuffer sb = new StringBuffer(intDigits.length() + 1 + p);
sb.append(intDigits);
sb.append('.');
sb.append(buffer, 0, p);
return sb.toString();
}
}
/* dtoa for IEEE arithmetic (dmg): convert double to ASCII string.
*
* Inspired by "How to Print Floating-Point Numbers Accurately" by
* Guy L. Steele, Jr. and Jon L. White [Proc. ACM SIGPLAN '90, pp. 92-101].
*
* Modifications:
* 1. Rather than iterating, we use a simple numeric overestimate
* to determine k = floor(log10(d)). We scale relevant
* quantities using O(log2(k)) rather than O(k) multiplications.
* 2. For some modes > 2 (corresponding to ecvt and fcvt), we don't
* try to generate digits strictly left to right. Instead, we
* compute with fewer bits and propagate the carry if necessary
* when rounding the final digit up. This is often faster.
* 3. Under the assumption that input will be rounded nearest,
* mode 0 renders 1e23 as 1e23 rather than 9.999999999999999e22.
* That is, we allow equality in stopping tests when the
* round-nearest rule will give the same floating-point value
* as would satisfaction of the stopping test with strict
* inequality.
* 4. We remove common factors of powers of 2 from relevant
* quantities.
* 5. When converting floating-point integers less than 1e16,
* we use floating-point arithmetic rather than resorting
* to multiple-precision integers.
* 6. When asked to produce fewer than 15 digits, we first try
* to get by with floating-point arithmetic; we resort to
* multiple-precision integer arithmetic only if we cannot
* guarantee that the floating-point calculation has given
* the correctly rounded result. For k requested digits and
* "uniformly" distributed input, the probability is
* something like 10^(k-15) that we must resort to the Long
* calculation.
*/
static int word0(double d)
{
long dBits = Double.doubleToLongBits(d);
return (int)(dBits >> 32);
}
static double setWord0(double d, int i)
{
long dBits = Double.doubleToLongBits(d);
dBits = ((long)i << 32) | (dBits & 0x0FFFFFFFFL);
return Double.longBitsToDouble(dBits);
}
static int word1(double d)
{
long dBits = Double.doubleToLongBits(d);
return (int)(dBits);
}
/* Return b * 5^k. k must be nonnegative. */
// XXXX the C version built a cache of these
static BigInteger pow5mult(BigInteger b, int k)
{
return b.multiply(BigInteger.valueOf(5).pow(k));
}
static boolean roundOff(StringBuffer buf)
{
int i = buf.length();
while (i != 0) {
--i;
char c = buf.charAt(i);
if (c != '9') {
buf.setCharAt(i, (char)(c + 1));
buf.setLength(i + 1);
return false;
}
}
buf.setLength(0);
return true;
}
/* Always emits at least one digit. */
/* If biasUp is set, then rounding in modes 2 and 3 will round away from zero
* when the number is exactly halfway between two representable values. For example,
* rounding 2.5 to zero digits after the decimal point will return 3 and not 2.
* 2.49 will still round to 2, and 2.51 will still round to 3. */
/* bufsize should be at least 20 for modes 0 and 1. For the other modes,
* bufsize should be two greater than the maximum number of output characters expected. */
static int
JS_dtoa(double d, int mode, boolean biasUp, int ndigits,
boolean[] sign, StringBuffer buf)
{
/* Arguments ndigits, decpt, sign are similar to those
of ecvt and fcvt; trailing zeros are suppressed from
the returned string. If not null, *rve is set to point
to the end of the return value. If d is +-Infinity or NaN,
then *decpt is set to 9999.
mode:
0 ==> shortest string that yields d when read in
and rounded to nearest.
1 ==> like 0, but with Steele & White stopping rule;
e.g. with IEEE P754 arithmetic , mode 0 gives
1e23 whereas mode 1 gives 9.999999999999999e22.
2 ==> max(1,ndigits) significant digits. This gives a
return value similar to that of ecvt, except
that trailing zeros are suppressed.
3 ==> through ndigits past the decimal point. This
gives a return value similar to that from fcvt,
except that trailing zeros are suppressed, and
ndigits can be negative.
4-9 should give the same return values as 2-3, i.e.,
4 <= mode <= 9 ==> same return as mode
2 + (mode & 1). These modes are mainly for
debugging; often they run slower but sometimes
faster than modes 2-3.
4,5,8,9 ==> left-to-right digit generation.
6-9 ==> don't try fast floating-point estimate
(if applicable).
Values of mode other than 0-9 are treated as mode 0.
Sufficient space is allocated to the return value
to hold the suppressed trailing zeros.
*/
int b2, b5, i, ieps, ilim, ilim0, ilim1,
j, j1, k, k0, m2, m5, s2, s5;
char dig;
long L;
long x;
BigInteger b, b1, delta, mlo, mhi, S;
int[] be = new int[1];
int[] bbits = new int[1];
double d2, ds, eps;
boolean spec_case, denorm, k_check, try_quick, leftright;
if ((word0(d) & Sign_bit) != 0) {
/* set sign for everything, including 0's and NaNs */
sign[0] = true;
// word0(d) &= ~Sign_bit; /* clear sign bit */
d = setWord0(d, word0(d) & ~Sign_bit);
}
else
sign[0] = false;
if ((word0(d) & Exp_mask) == Exp_mask) {
/* Infinity or NaN */
buf.append(((word1(d) == 0) && ((word0(d) & Frac_mask) == 0)) ? "Infinity" : "NaN");
return 9999;
}
if (d == 0) {
// no_digits:
buf.setLength(0);
buf.append('0'); /* copy "0" to buffer */
return 1;
}
b = d2b(d, be, bbits);
if ((i = (word0(d) >>> Exp_shift1 & (Exp_mask>>Exp_shift1))) != 0) {
d2 = setWord0(d, (word0(d) & Frac_mask1) | Exp_11);
/* log(x) ~=~ log(1.5) + (x-1.5)/1.5
* log10(x) = log(x) / log(10)
* ~=~ log(1.5)/log(10) + (x-1.5)/(1.5*log(10))
* log10(d) = (i-Bias)*log(2)/log(10) + log10(d2)
*
* This suggests computing an approximation k to log10(d) by
*
* k = (i - Bias)*0.301029995663981
* + ( (d2-1.5)*0.289529654602168 + 0.176091259055681 );
*
* We want k to be too large rather than too small.
* The error in the first-order Taylor series approximation
* is in our favor, so we just round up the constant enough
* to compensate for any error in the multiplication of
* (i - Bias) by 0.301029995663981; since |i - Bias| <= 1077,
* and 1077 * 0.30103 * 2^-52 ~=~ 7.2e-14,
* adding 1e-13 to the constant term more than suffices.
* Hence we adjust the constant term to 0.1760912590558.
* (We could get a more accurate k by invoking log10,
* but this is probably not worthwhile.)
*/
i -= Bias;
denorm = false;
}
else {
/* d is denormalized */
i = bbits[0] + be[0] + (Bias + (P-1) - 1);
x = (i > 32) ? word0(d) << (64 - i) | word1(d) >>> (i - 32) : word1(d) << (32 - i);
// d2 = x;
// word0(d2) -= 31*Exp_msk1; /* adjust exponent */
d2 = setWord0(x, word0(x) - 31*Exp_msk1);
i -= (Bias + (P-1) - 1) + 1;
denorm = true;
}
/* At this point d = f*2^i, where 1 <= f < 2. d2 is an approximation of f. */
ds = (d2-1.5)*0.289529654602168 + 0.1760912590558 + i*0.301029995663981;
k = (int)ds;
if (ds < 0.0 && ds != k)
k--; /* want k = floor(ds) */
k_check = true;
if (k >= 0 && k <= Ten_pmax) {
if (d < tens[k])
k--;
k_check = false;
}
/* At this point floor(log10(d)) <= k <= floor(log10(d))+1.
If k_check is zero, we're guaranteed that k = floor(log10(d)). */
j = bbits[0] - i - 1;
/* At this point d = b/2^j, where b is an odd integer. */
if (j >= 0) {
b2 = 0;
s2 = j;
}
else {
b2 = -j;
s2 = 0;
}
if (k >= 0) {
b5 = 0;
s5 = k;
s2 += k;
}
else {
b2 -= k;
b5 = -k;
s5 = 0;
}
/* At this point d/10^k = (b * 2^b2 * 5^b5) / (2^s2 * 5^s5), where b is an odd integer,
b2 >= 0, b5 >= 0, s2 >= 0, and s5 >= 0. */
if (mode < 0 || mode > 9)
mode = 0;
try_quick = true;
if (mode > 5) {
mode -= 4;
try_quick = false;
}
leftright = true;
ilim = ilim1 = 0;
switch(mode) {
case 0:
case 1:
ilim = ilim1 = -1;
i = 18;
ndigits = 0;
break;
case 2:
leftright = false;
/* no break */
case 4:
if (ndigits <= 0)
ndigits = 1;
ilim = ilim1 = i = ndigits;
break;
case 3:
leftright = false;
/* no break */
case 5:
i = ndigits + k + 1;
ilim = i;
ilim1 = i - 1;
if (i <= 0)
i = 1;
}
/* ilim is the maximum number of significant digits we want, based on k and ndigits. */
/* ilim1 is the maximum number of significant digits we want, based on k and ndigits,
when it turns out that k was computed too high by one. */
boolean fast_failed = false;
if (ilim >= 0 && ilim <= Quick_max && try_quick) {
/* Try to get by with floating-point arithmetic. */
i = 0;
d2 = d;
k0 = k;
ilim0 = ilim;
ieps = 2; /* conservative */
/* Divide d by 10^k, keeping track of the roundoff error and avoiding overflows. */
if (k > 0) {
ds = tens[k&0xf];
j = k >> 4;
if ((j & Bletch) != 0) {
/* prevent overflows */
j &= Bletch - 1;
d /= bigtens[n_bigtens-1];
ieps++;
}
for(; (j != 0); j >>= 1, i++)
if ((j & 1) != 0) {
ieps++;
ds *= bigtens[i];
}
d /= ds;
}
else if ((j1 = -k) != 0) {
d *= tens[j1 & 0xf];
for(j = j1 >> 4; (j != 0); j >>= 1, i++)
if ((j & 1) != 0) {
ieps++;
d *= bigtens[i];
}
}
/* Check that k was computed correctly. */
if (k_check && d < 1.0 && ilim > 0) {
if (ilim1 <= 0)
fast_failed = true;
else {
ilim = ilim1;
k--;
d *= 10.;
ieps++;
}
}
/* eps bounds the cumulative error. */
// eps = ieps*d + 7.0;
// word0(eps) -= (P-1)*Exp_msk1;
eps = ieps*d + 7.0;
eps = setWord0(eps, word0(eps) - (P-1)*Exp_msk1);
if (ilim == 0) {
S = mhi = null;
d -= 5.0;
if (d > eps) {
buf.append('1');
k++;
return k + 1;
}
if (d < -eps) {
buf.setLength(0);
buf.append('0'); /* copy "0" to buffer */
return 1;
}
fast_failed = true;
}
if (!fast_failed) {
fast_failed = true;
if (leftright) {
/* Use Steele & White method of only
* generating digits needed.
*/
eps = 0.5/tens[ilim-1] - eps;
for(i = 0;;) {
L = (long)d;
d -= L;
buf.append((char)('0' + L));
if (d < eps) {
return k + 1;
}
if (1.0 - d < eps) {
// goto bump_up;
char lastCh;
while (true) {
lastCh = buf.charAt(buf.length() - 1);
buf.setLength(buf.length() - 1);
if (lastCh != '9') break;
if (buf.length() == 0) {
k++;
lastCh = '0';
break;
}
}
buf.append((char)(lastCh + 1));
return k + 1;
}
if (++i >= ilim)
break;
eps *= 10.0;
d *= 10.0;
}
}
else {
/* Generate ilim digits, then fix them up. */
eps *= tens[ilim-1];
for(i = 1;; i++, d *= 10.0) {
L = (long)d;
d -= L;
buf.append((char)('0' + L));
if (i == ilim) {
if (d > 0.5 + eps) {
// goto bump_up;
char lastCh;
while (true) {
lastCh = buf.charAt(buf.length() - 1);
buf.setLength(buf.length() - 1);
if (lastCh != '9') break;
if (buf.length() == 0) {
k++;
lastCh = '0';
break;
}
}
buf.append((char)(lastCh + 1));
return k + 1;
}
else
if (d < 0.5 - eps) {
stripTrailingZeroes(buf);
// while(*--s == '0') ;
// s++;
return k + 1;
}
break;
}
}
}
}
if (fast_failed) {
buf.setLength(0);
d = d2;
k = k0;
ilim = ilim0;
}
}
/* Do we have a "small" integer? */
if (be[0] >= 0 && k <= Int_max) {
/* Yes. */
ds = tens[k];
if (ndigits < 0 && ilim <= 0) {
S = mhi = null;
if (ilim < 0 || d < 5*ds || (!biasUp && d == 5*ds)) {
buf.setLength(0);
buf.append('0'); /* copy "0" to buffer */
return 1;
}
buf.append('1');
k++;
return k + 1;
}
for(i = 1;; i++) {
L = (long) (d / ds);
d -= L*ds;
buf.append((char)('0' + L));
if (i == ilim) {
d += d;
if ((d > ds) || (d == ds && (((L & 1) != 0) || biasUp))) {
// bump_up:
// while(*--s == '9')
// if (s == buf) {
// k++;
// *s = '0';
// break;
// }
// ++*s++;
char lastCh;
while (true) {
lastCh = buf.charAt(buf.length() - 1);
buf.setLength(buf.length() - 1);
if (lastCh != '9') break;
if (buf.length() == 0) {
k++;
lastCh = '0';
break;
}
}
buf.append((char)(lastCh + 1));
}
break;
}
d *= 10.0;
if (d == 0)
break;
}
return k + 1;
}
m2 = b2;
m5 = b5;
mhi = mlo = null;
if (leftright) {
if (mode < 2) {
i = (denorm) ? be[0] + (Bias + (P-1) - 1 + 1) : 1 + P - bbits[0];
/* i is 1 plus the number of trailing zero bits in d's significand. Thus,
(2^m2 * 5^m5) / (2^(s2+i) * 5^s5) = (1/2 lsb of d)/10^k. */
}
else {
j = ilim - 1;
if (m5 >= j)
m5 -= j;
else {
s5 += j -= m5;
b5 += j;
m5 = 0;
}
if ((i = ilim) < 0) {
m2 -= i;
i = 0;
}
/* (2^m2 * 5^m5) / (2^(s2+i) * 5^s5) = (1/2 * 10^(1-ilim))/10^k. */
}
b2 += i;
s2 += i;
mhi = BigInteger.valueOf(1);
/* (mhi * 2^m2 * 5^m5) / (2^s2 * 5^s5) = one-half of last printed (when mode >= 2) or
input (when mode < 2) significant digit, divided by 10^k. */
}
/* We still have d/10^k = (b * 2^b2 * 5^b5) / (2^s2 * 5^s5). Reduce common factors in
b2, m2, and s2 without changing the equalities. */
if (m2 > 0 && s2 > 0) {
i = (m2 < s2) ? m2 : s2;
b2 -= i;
m2 -= i;
s2 -= i;
}
/* Fold b5 into b and m5 into mhi. */
if (b5 > 0) {
if (leftright) {
if (m5 > 0) {
mhi = pow5mult(mhi, m5);
b1 = mhi.multiply(b);
b = b1;
}
if ((j = b5 - m5) != 0)
b = pow5mult(b, j);
}
else
b = pow5mult(b, b5);
}
/* Now we have d/10^k = (b * 2^b2) / (2^s2 * 5^s5) and
(mhi * 2^m2) / (2^s2 * 5^s5) = one-half of last printed or input significant digit, divided by 10^k. */
S = BigInteger.valueOf(1);
if (s5 > 0)
S = pow5mult(S, s5);
/* Now we have d/10^k = (b * 2^b2) / (S * 2^s2) and
(mhi * 2^m2) / (S * 2^s2) = one-half of last printed or input significant digit, divided by 10^k. */
/* Check for special case that d is a normalized power of 2. */
spec_case = false;
if (mode < 2) {
if ( (word1(d) == 0) && ((word0(d) & Bndry_mask) == 0)
&& ((word0(d) & (Exp_mask & Exp_mask << 1)) != 0)
) {
/* The special case. Here we want to be within a quarter of the last input
significant digit instead of one half of it when the decimal output string's value is less than d. */
b2 += Log2P;
s2 += Log2P;
spec_case = true;
}
}
/* Arrange for convenient computation of quotients:
* shift left if necessary so divisor has 4 leading 0 bits.
*
* Perhaps we should just compute leading 28 bits of S once
* and for all and pass them and a shift to quorem, so it
* can do shifts and ors to compute the numerator for q.
*/
byte [] S_bytes = S.toByteArray();
int S_hiWord = 0;
for (int idx = 0; idx < 4; idx++) {
S_hiWord = (S_hiWord << 8);
if (idx < S_bytes.length)
S_hiWord |= (S_bytes[idx] & 0xFF);
}
if ((i = (((s5 != 0) ? 32 - hi0bits(S_hiWord) : 1) + s2) & 0x1f) != 0)
i = 32 - i;
/* i is the number of leading zero bits in the most significant word of S*2^s2. */
if (i > 4) {
i -= 4;
b2 += i;
m2 += i;
s2 += i;
}
else if (i < 4) {
i += 28;
b2 += i;
m2 += i;
s2 += i;
}
/* Now S*2^s2 has exactly four leading zero bits in its most significant word. */
if (b2 > 0)
b = b.shiftLeft(b2);
if (s2 > 0)
S = S.shiftLeft(s2);
/* Now we have d/10^k = b/S and
(mhi * 2^m2) / S = maximum acceptable error, divided by 10^k. */
if (k_check) {
if (b.compareTo(S) < 0) {
k--;
b = b.multiply(BigInteger.valueOf(10)); /* we botched the k estimate */
if (leftright)
mhi = mhi.multiply(BigInteger.valueOf(10));
ilim = ilim1;
}
}
/* At this point 1 <= d/10^k = b/S < 10. */
if (ilim <= 0 && mode > 2) {
/* We're doing fixed-mode output and d is less than the minimum nonzero output in this mode.
Output either zero or the minimum nonzero output depending on which is closer to d. */
if ((ilim < 0 )
|| ((i = b.compareTo(S = S.multiply(BigInteger.valueOf(5)))) < 0)
|| ((i == 0 && !biasUp))) {
/* Always emit at least one digit. If the number appears to be zero
using the current mode, then emit one '0' digit and set decpt to 1. */
/*no_digits:
k = -1 - ndigits;
goto ret; */
buf.setLength(0);
buf.append('0'); /* copy "0" to buffer */
return 1;
// goto no_digits;
}
// one_digit:
buf.append('1');
k++;
return k + 1;
}
if (leftright) {
if (m2 > 0)
mhi = mhi.shiftLeft(m2);
/* Compute mlo -- check for special case
* that d is a normalized power of 2.
*/
mlo = mhi;
if (spec_case) {
mhi = mlo;
mhi = mhi.shiftLeft(Log2P);
}
/* mlo/S = maximum acceptable error, divided by 10^k, if the output is less than d. */
/* mhi/S = maximum acceptable error, divided by 10^k, if the output is greater than d. */
for(i = 1;;i++) {
BigInteger[] divResult = b.divideAndRemainder(S);
b = divResult[1];
dig = (char)(divResult[0].intValue() + '0');
/* Do we yet have the shortest decimal string
* that will round to d?
*/
j = b.compareTo(mlo);
/* j is b/S compared with mlo/S. */
delta = S.subtract(mhi);
j1 = (delta.signum() <= 0) ? 1 : b.compareTo(delta);
/* j1 is b/S compared with 1 - mhi/S. */
if ((j1 == 0) && (mode == 0) && ((word1(d) & 1) == 0)) {
if (dig == '9') {
buf.append('9');
if (roundOff(buf)) {
k++;
buf.append('1');
}
return k + 1;
// goto round_9_up;
}
if (j > 0)
dig++;
buf.append(dig);
return k + 1;
}
if ((j < 0)
|| ((j == 0)
&& (mode == 0)
&& ((word1(d) & 1) == 0)
)) {
if (j1 > 0) {
/* Either dig or dig+1 would work here as the least significant decimal digit.
Use whichever would produce a decimal value closer to d. */
b = b.shiftLeft(1);
j1 = b.compareTo(S);
if (((j1 > 0) || (j1 == 0 && (((dig & 1) == 1) || biasUp)))
&& (dig++ == '9')) {
buf.append('9');
if (roundOff(buf)) {
k++;
buf.append('1');
}
return k + 1;
// goto round_9_up;
}
}
buf.append(dig);
return k + 1;
}
if (j1 > 0) {
if (dig == '9') { /* possible if i == 1 */
// round_9_up:
// *s++ = '9';
// goto roundoff;
buf.append('9');
if (roundOff(buf)) {
k++;
buf.append('1');
}
return k + 1;
}
buf.append((char)(dig + 1));
return k + 1;
}
buf.append(dig);
if (i == ilim)
break;
b = b.multiply(BigInteger.valueOf(10));
if (mlo == mhi)
mlo = mhi = mhi.multiply(BigInteger.valueOf(10));
else {
mlo = mlo.multiply(BigInteger.valueOf(10));
mhi = mhi.multiply(BigInteger.valueOf(10));
}
}
}
else
for(i = 1;; i++) {
// (char)(dig = quorem(b,S) + '0');
BigInteger[] divResult = b.divideAndRemainder(S);
b = divResult[1];
dig = (char)(divResult[0].intValue() + '0');
buf.append(dig);
if (i >= ilim)
break;
b = b.multiply(BigInteger.valueOf(10));
}
/* Round off last digit */
b = b.shiftLeft(1);
j = b.compareTo(S);
if ((j > 0) || (j == 0 && (((dig & 1) == 1) || biasUp))) {
// roundoff:
// while(*--s == '9')
// if (s == buf) {
// k++;
// *s++ = '1';
// goto ret;
// }
// ++*s++;
if (roundOff(buf)) {
k++;
buf.append('1');
return k + 1;
}
}
else {
stripTrailingZeroes(buf);
// while(*--s == '0') ;
// s++;
}
// ret:
// Bfree(S);
// if (mhi) {
// if (mlo && mlo != mhi)
// Bfree(mlo);
// Bfree(mhi);
// }
// ret1:
// Bfree(b);
// JS_ASSERT(s < buf + bufsize);
return k + 1;
}
private static void
stripTrailingZeroes(StringBuffer buf)
{
// while(*--s == '0') ;
// s++;
int bl = buf.length();
while(bl-->0 && buf.charAt(bl) == '0') {
// empty
}
buf.setLength(bl + 1);
}
/* Mapping of JSDToStrMode -> JS_dtoa mode */
private static final int dtoaModes[] = {
0, /* DTOSTR_STANDARD */
0, /* DTOSTR_STANDARD_EXPONENTIAL, */
3, /* DTOSTR_FIXED, */
2, /* DTOSTR_EXPONENTIAL, */
2}; /* DTOSTR_PRECISION */
static void
JS_dtostr(StringBuffer buffer, int mode, int precision, double d)
{
int decPt; /* Position of decimal point relative to first digit returned by JS_dtoa */
boolean[] sign = new boolean[1]; /* true if the sign bit was set in d */
int nDigits; /* Number of significand digits returned by JS_dtoa */
// JS_ASSERT(bufferSize >= (size_t)(mode <= DTOSTR_STANDARD_EXPONENTIAL ? DTOSTR_STANDARD_BUFFER_SIZE :
// DTOSTR_VARIABLE_BUFFER_SIZE(precision)));
if (mode == DTOSTR_FIXED && (d >= 1e21 || d <= -1e21))
mode = DTOSTR_STANDARD; /* Change mode here rather than below because the buffer may not be large enough to hold a large integer. */
decPt = JS_dtoa(d, dtoaModes[mode], mode >= DTOSTR_FIXED, precision, sign, buffer);
nDigits = buffer.length();
/* If Infinity, -Infinity, or NaN, return the string regardless of the mode. */
if (decPt != 9999) {
boolean exponentialNotation = false;
int minNDigits = 0; /* Minimum number of significand digits required by mode and precision */
int p;
switch (mode) {
case DTOSTR_STANDARD:
if (decPt < -5 || decPt > 21)
exponentialNotation = true;
else
minNDigits = decPt;
break;
case DTOSTR_FIXED:
if (precision >= 0)
minNDigits = decPt + precision;
else
minNDigits = decPt;
break;
case DTOSTR_EXPONENTIAL:
// JS_ASSERT(precision > 0);
minNDigits = precision;
/* Fall through */
case DTOSTR_STANDARD_EXPONENTIAL:
exponentialNotation = true;
break;
case DTOSTR_PRECISION:
// JS_ASSERT(precision > 0);
minNDigits = precision;
if (decPt < -5 || decPt > precision)
exponentialNotation = true;
break;
}
/* If the number has fewer than minNDigits, pad it with zeros at the end */
if (nDigits < minNDigits) {
p = minNDigits;
nDigits = minNDigits;
do {
buffer.append('0');
} while (buffer.length() != p);
}
if (exponentialNotation) {
/* Insert a decimal point if more than one significand digit */
if (nDigits != 1) {
buffer.insert(1, '.');
}
buffer.append('e');
if ((decPt - 1) >= 0)
buffer.append('+');
buffer.append(decPt - 1);
// JS_snprintf(numEnd, bufferSize - (numEnd - buffer), "e%+d", decPt-1);
} else if (decPt != nDigits) {
/* Some kind of a fraction in fixed notation */
// JS_ASSERT(decPt <= nDigits);
if (decPt > 0) {
/* dd...dd . dd...dd */
buffer.insert(decPt, '.');
} else {
/* 0 . 00...00dd...dd */
for (int i = 0; i < 1 - decPt; i++)
buffer.insert(0, '0');
buffer.insert(1, '.');
}
}
}
/* If negative and neither -0.0 nor NaN, output a leading '-'. */
if (sign[0] &&
!(word0(d) == Sign_bit && word1(d) == 0) &&
!((word0(d) & Exp_mask) == Exp_mask &&
((word1(d) != 0) || ((word0(d) & Frac_mask) != 0)))) {
buffer.insert(0, '-');
}
}
}
