dtoa.java

javascript语言的解释器源码

package org.mozilla.javascript;

import java.math.BigInteger;

class DToA {

	public static void main(String[] args) {
		DToA dt=new DToA();
		for(int i=0;i<18;i++)
		{
			System.out.println(DToA.BASEDIGIT(i));
			
		}
		
	}
/* "-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 = (int)(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 q;
            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);