dtoa.java

java中比较著名的js引擎当属mozilla开源的rhino

        }
        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);