dtoa.java

java中比较著名的js引擎当属mozilla开源的rhino

            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 = (int)(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;