From 98e2821b38a775737e42a2479a6bc65107210859 Mon Sep 17 00:00:00 2001 From: Elliot Kroo Date: Thu, 11 Mar 2010 15:21:30 -0800 Subject: reorganizing the first level of folders (trunk/branch folders are not the git way :) --- .../src/org/mozilla/javascript/DToA.java | 1271 -------------------- 1 file changed, 1271 deletions(-) delete mode 100644 trunk/infrastructure/rhino1_7R1/src/org/mozilla/javascript/DToA.java (limited to 'trunk/infrastructure/rhino1_7R1/src/org/mozilla/javascript/DToA.java') diff --git a/trunk/infrastructure/rhino1_7R1/src/org/mozilla/javascript/DToA.java b/trunk/infrastructure/rhino1_7R1/src/org/mozilla/javascript/DToA.java deleted file mode 100644 index ad2a68a..0000000 --- a/trunk/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 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; - } 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< 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, '-'); - } - } - -} - -- cgit v1.2.3-1-g7c22