reverted folder structure change for better mergeing with upstream

1 files changed, 0 insertions, 1271 deletions

diff --git a/infrastructure/rhino1_7R1/src/org/mozilla/javascript/DToA.java b/infrastructure/rhino1_7R1/src/org/mozilla/javascript/DToA.java
deleted file mode 100644
index ad2a68a..0000000
--- a/infrastructure/rhino1_7R1/src/org/mozilla/javascript/DToA.java
+++ /dev/null
@@ -1,1271 +0,0 @@
-/* -*- 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, '-');
-        }
-    }
-
-}
-