dtoa.java

javascript语言的解释器源码

    }

    /* 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;
        }
        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) {
                                    while (buf.charAt(buf.length() - 1) == '0')
                                        buf.setLength(buf.length() - 1);
//                                    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;
                }