org.mozilla.javascript.DToA Maven / Gradle / Ivy
Go to download
Show more of this group Show more artifacts with this name
Show all versions of rhino Show documentation
Show all versions of rhino Show documentation
Rhino is an open-source implementation of JavaScript written entirely in Java. It is typically
embedded into Java applications to provide scripting to end users.
/* -*- Mode: java; tab-width: 8; indent-tabs-mode: nil; c-basic-offset: 4 -*-
*
* This Source Code Form is subject to the terms of the Mozilla Public
* License, v. 2.0. If a copy of the MPL was not distributed with this
* file, You can obtain one at http://mozilla.org/MPL/2.0/. */
/****************************************************************
*
* 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 {
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 digits after the decimal point; exponential if number is large */
DTOSTR_EXPONENTIAL = 3, /* Always exponential format; significant digits */
DTOSTR_PRECISION = 4; /* Either fixed or exponential format; 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;
}
/* We have a fraction. */
StringBuilder buffer; /* The output string */
int digit;
double df; /* The fractional part of d */
BigInteger b;
buffer = new StringBuilder();
buffer.append(intDigits).append('.');
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< 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.append(BASEDIGIT(digit));
} while (!done);
return buffer.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(StringBuilder 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, StringBuilder 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)
? ((long) word0(d)) << (64 - i) | word1(d) >>> (i - 32)
: ((long) 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;
/* fall through */
case 4:
if (ndigits <= 0)
ndigits = 1;
ilim = ilim1 = i = ndigits;
break;
case 3:
leftright = false;
/* fall through */
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(StringBuilder 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(StringBuilder 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, '-');
}
}
}
© 2015 - 2024 Weber Informatics LLC | Privacy Policy