// not a denorm, just a 0!                decExponent = 0;                digits = zero;                nDigits = 1;                return;            }            while ( (fractBits&singleFractHOB) == 0 ){                fractBits <<= 1;                binExp -= 1;            }            nSignificantBits = singleExpShift + binExp +1; // recall binExp is  - shift count.            binExp += 1;        } else {            fractBits |= singleFractHOB;            nSignificantBits = singleExpShift+1;        }        binExp -= singleExpBias;        // call the routine that actually does all the hard work.        dtoa( binExp, ((long)fractBits)<<(expShift-singleExpShift), nSignificantBits );    }     private void    dtoa( int binExp, long fractBits, int nSignificantBits )    {        int     nFractBits; // number of significant bits of fractBits;        int     nTinyBits;  // number of these to the right of the point.        int     decExp;         // Examine number. Determine if it is an easy case,        // which we can do pretty trivially using float/long conversion,        // or whether we must do real work.        nFractBits = countBits( fractBits );        nTinyBits = Math.max( 0, nFractBits - binExp - 1 );        if ( binExp <= maxSmallBinExp && binExp >= minSmallBinExp ){            // Look more closely at the number to decide if,            // with scaling by 10^nTinyBits, the result will fit in            // a long.            if ( (nTinyBits < long5pow.length) && ((nFractBits + n5bits[nTinyBits]) < 64 ) ){                /*                 * We can do this:                 * take the fraction bits, which are normalized.                 * (a) nTinyBits == 0: Shift left or right appropriately                 *     to align the binary point at the extreme right, i.e.                 *     where a long int point is expected to be. The integer                 *     result is easily converted to a string.                 * (b) nTinyBits > 0: Shift right by expShift-nFractBits,                 *     which effectively converts to long and scales by                 *     2^nTinyBits. Then multiply by 5^nTinyBits to                 *     complete the scaling. We know this won't overflow                 *     because we just counted the number of bits necessary                 *     in the result. The integer you get from this can                 *     then be converted to a string pretty easily.                 */                long halfULP;                if ( nTinyBits == 0 ) {                    if ( binExp > nSignificantBits ){                        halfULP = 1L << ( binExp-nSignificantBits-1);                    } else {                        halfULP = 0L;                    }                    if ( binExp >= expShift ){                        fractBits <<= (binExp-expShift);                    } else {                        fractBits >>>= (expShift-binExp) ;                    }                    developLongDigits( 0, fractBits, halfULP );                    return;                }                /*                 * The following causes excess digits to be printed                 * out in the single-float case. Our manipulation of                 * halfULP here is apparently not correct. If we                 * better understand how this works, perhaps we can                 * use this special case again. But for the time being,                 * we do not.                 * else {                 *     fractBits >>>= expShift+1-nFractBits;                 *     fractBits *= long5pow[ nTinyBits ];                 *     halfULP = long5pow[ nTinyBits ] >> (1+nSignificantBits-nFractBits);                 *     developLongDigits( -nTinyBits, fractBits, halfULP );                 *     return;                 * }                 */            }        }        /*         * This is the hard case. We are going to compute large positive         * integers B and S and integer decExp, s.t.         *      d = ( B / S ) * 10^decExp         *      1 <= B / S < 10         * Obvious choices are:         *      decExp = floor( log10(d) )         *      B      = d * 2^nTinyBits * 10^max( 0, -decExp )         *      S      = 10^max( 0, decExp) * 2^nTinyBits         * (noting that nTinyBits has already been forced to non-negative)         * I am also going to compute a large positive integer         *      M      = (1/2^nSignificantBits) * 2^nTinyBits * 10^max( 0, -decExp )         * i.e. M is (1/2) of the ULP of d, scaled like B.         * When we iterate through dividing B/S and picking off the         * quotient bits, we will know when to stop when the remainder         * is <= M.         *         * We keep track of powers of 2 and powers of 5.         */         /*         * Estimate decimal exponent. (If it is small-ish,         * we could double-check.)         *         * First, scale the mantissa bits such that 1 <= d2 < 2.         * We are then going to estimate         *          log10(d2) ~=~  (d2-1.5)/1.5 + log(1.5)         * and so we can estimate         *      log10(d) ~=~ log10(d2) + binExp * log10(2)         * take the floor and call it decExp.         * FIXME -- use more precise constants here. It costs no more.         */        double d2 = Double.longBitsToDouble(            expOne | ( fractBits &~ fractHOB ) );        decExp = (int)Math.floor(            (d2-1.5D)*0.289529654D + 0.176091259 + (double)binExp * 0.301029995663981 );        int B2, B5; // powers of 2 and powers of 5, respectively, in B        int S2, S5; // powers of 2 and powers of 5, respectively, in S        int M2, M5; // powers of 2 and powers of 5, respectively, in M        int Bbits; // binary digits needed to represent B, approx.        int tenSbits; // binary digits needed to represent 10*S, approx.        FDBigInt Sval, Bval, Mval;         B5 = Math.max( 0, -decExp );        B2 = B5 + nTinyBits + binExp;         S5 = Math.max( 0, decExp );        S2 = S5 + nTinyBits;         M5 = B5;        M2 = B2 - nSignificantBits;         /*         * the long integer fractBits contains the (nFractBits) interesting         * bits from the mantissa of d ( hidden 1 added if necessary) followed         * by (expShift+1-nFractBits) zeros. In the interest of compactness,         * I will shift out those zeros before turning fractBits into a         * FDBigInt. The resulting whole number will be         *      d * 2^(nFractBits-1-binExp).         */        fractBits >>>= (expShift+1-nFractBits);        B2 -= nFractBits-1;        int common2factor = Math.min( B2, S2 );        B2 -= common2factor;        S2 -= common2factor;        M2 -= common2factor;         /*         * HACK!! For exact powers of two, the next smallest number         * is only half as far away as we think (because the meaning of         * ULP changes at power-of-two bounds) for this reason, we         * hack M2. Hope this works.         */        if ( nFractBits == 1 )            M2 -= 1;         if ( M2 < 0 ){            // oops.            // since we cannot scale M down far enough,            // we must scale the other values up.            B2 -= M2;            S2 -= M2;            M2 =  0;        }        /*         * Construct, Scale, iterate.         * Some day, we'll write a stopping test that takes         * account of the asymmetry of the spacing of floating-point         * numbers below perfect powers of 2         * 26 Sept 96 is not that day.         * So we use a symmetric test.         */        char digits[] = this.digits = new char[18];        int  ndigit = 0;        boolean low, high;        long lowDigitDifference;        int  q;         /*         * Detect the special cases where all the numbers we are about         * to compute will fit in int or long integers.         * In these cases, we will avoid doing FDBigInt arithmetic.         * We use the same algorithms, except that we "normalize"         * our FDBigInts before iterating. This is to make division easier,         * as it makes our fist guess (quotient of high-order words)         * more accurate!         *         * Some day, we'll write a stopping test that takes         * account of the asymmetry of the spacing of floating-point         * numbers below perfect powers of 2         * 26 Sept 96 is not that day.         * So we use a symmetric test.         */        Bbits = nFractBits + B2 + (( B5 < n5bits.length )? n5bits[B5] : ( B5*3 ));        tenSbits = S2+1 + (( (S5+1) < n5bits.length )? n5bits[(S5+1)] : ( (S5+1)*3 ));        if ( Bbits < 64 && tenSbits < 64){            if ( Bbits < 32 && tenSbits < 32){                // wa-hoo! They're all ints!                int b = ((int)fractBits * small5pow[B5] ) << B2;                int s = small5pow[S5] << S2;                int m = small5pow[M5] << M2;                int tens = s * 10;                /*                 * Unroll the first iteration. If our decExp estimate                 * was too high, our first quotient will be zero. In this                 * case, we discard it and decrement decExp.                 */                ndigit = 0;                q = b / s;                b = 10 * ( b % s );                m *= 10;                low  = (b <  m );                high = (b+m > tens );                assert q < 10 : q; // excessively large digit                if ( (q == 0) && ! high ){                    // oops. Usually ignore leading zero.                    decExp--;                } else {                    digits[ndigit++] = (char)('0' + q);                }                /*                 * HACK! Java spec sez that we always have at least                 * one digit after the . in either F- or E-form output.                 * Thus we will need more than one digit if we're using                 * E-form                 */                if ( decExp < -3 || decExp >= 8 ){                    high = low = false;                }                while( ! low && ! high ){                    q = b / s;                    b = 10 * ( b % s );                    m *= 10;                    assert q < 10 : q; // excessively large digit                    if ( m > 0L ){                        low  = (b <  m );                        high = (b+m > tens );                    } else {                        // hack -- m might overflow!                        // in this case, it is certainly > b,                        // which won't                        // and b+m > tens, too, since that has overflowed                        // either!                        low = true;                        high = true;                    }                    digits[ndigit++] = (char)('0' + q);                }                lowDigitDifference = (b<<1) - tens;            } else {                // still good! they're all longs!                long b = (fractBits * long5pow[B5] ) << B2;                long s = long5pow[S5] << S2;                long m = long5pow[M5] << M2;                long tens = s * 10L;                /*                 * Unroll the first iteration. If our decExp estimate                 * was too high, our first quotient will be zero. In this                 * case, we discard it and decrement decExp.                 */                ndigit = 0;                q = (int) ( b / s );                b = 10L * ( b % s );                m *= 10L;                low  = (b <  m );                high = (b+m > tens );                assert q < 10 : q; // excessively large digit                if ( (q == 0) && ! high ){                    // oops. Usually ignore leading zero.                    decExp--;                } else {                    digits[ndigit++] = (char)('0' + q);                }                /*                 * HACK! Java spec sez that we always have at least                 * one digit after the . in either F- or E-form output.                 * Thus we will need more than one digit if we're using                 * E-form                 */                if ( decExp < -3 || decExp >= 8 ){                    high = low = false;                }                while( ! low && ! high ){                    q = (int) ( b / s );                    b = 10 * ( b % s );                    m *= 10;                    assert q < 10 : q;  // excessively large digit                    if ( m > 0L ){                        low  = (b <  m );                        high = (b+m > tens );                    } else {                        // hack -- m might overflow!                        // in this case, it is certainly > b,                        // which won't                        // and b+m > tens, too, since that has overflowed                        // either!                        low = true;                        high = true;                    }                    digits[ndigit++] = (char)('0' + q);                }                lowDigitDifference = (b<<1) - tens;            }        } else {            FDBigInt tenSval;            int  shiftBias;             /*             * We really must do FDBigInt arithmetic.             * Fist, construct our FDBigInt initial values.             */            Bval = multPow52( new FDBigInt( fractBits  ), B5, B2 );            Sval = constructPow52( S5, S2 );            Mval = constructPow52( M5, M2 );              // normalize so that division works better            Bval.lshiftMe( shiftBias = Sval.normalizeMe() );            Mval.lshiftMe( shiftBias );            tenSval = Sval.mult( 10 );            /*             * Unroll the first iteration. If our decExp estimate             * was too high, our first quotient will be zero. In this             * case, we discard it and decrement decExp.             */            ndigit = 0;            q = Bval.quoRemIteration( Sval );            Mval = Mval.mult( 10 );            low  = (Bval.cmp( Mval ) < 0);            high = (Bval.add( Mval ).cmp( tenSval ) > 0 );            assert q < 10 : q; // excessively large digit            if ( (q == 0) && ! high ){                // oops. Usually ignore leading zero.                decExp--;            } else {                digits[ndigit++] = (char)('0' + q);            }            /*             * HACK! Java spec sez that we always have at least             * one digit after the . in either F- or E-form output.             * Thus we will need more than one digit if we're using             * E-form             */            if ( decExp < -3 || decExp >= 8 ){                high = low = false;            }            while( ! low && ! high ){                q = Bval.quoRemIteration( Sval );                Mval = Mval.mult( 10 );                assert q < 10 : q;  // excessively large digit                low  = (Bval.cmp( Mval ) < 0);                high = (Bval.add( Mval ).cmp( tenSval ) > 0 );                digits[ndigit++] = (char)('0' + q);            }            if ( high && low ){                Bval.lshiftMe(1);                lowDigitDifference = Bval.cmp(tenSval);            } else                lowDigitDifference = 0L; // this here only for flow analysis!        }        this.decExponent = decExp+1;        this.digits = digits;        this.nDigits = ndigit;        /*         * Last digit gets rounded based on stopping condition.         */        if ( high ){            if ( low ){                if ( lowDigitDifference == 0L ){                    // it's a tie!                    // choose based on which digits we like.                    if ( (digits[nDigits-1]&1) != 0 ) roundup();                } else if ( lowDigitDifference > 0 ){                    roundup();                }            } else {                roundup();            }        }    }     public String    toString(){        // most brain-dead version        StringBuffer result = new StringBuffer( nDigits+8 );        if ( isNegative ){ result.append( '-' ); }        if ( isExceptional ){            result.append( digits, 0, nDigits );        } else {            result.append( "0.");            result.append( digits, 0, nDigits );            result.append('e');            result.append( decExponent );        }        return new String(result);    }     public String toJavaFormatString() {        char result[] = (char[])(perThreadBuffer.get());        int i = getChars(result);        return new String(result, 0, i);    }     private int getChars(char[] result) {        assert nDigits <= 19 : nDigits; // generous bound on size of nDigits        int i = 0;        if (isNegative) { result[0] = '-'; i = 1; }        if (isExceptional) {            System.arraycopy(digits, 0, result, i, nDigits);            i += nDigits;        } else {            if (decExponent > 0 && decExponent < 8) {                // print digits.digits.                int charLength = Math.min(nDigits, decExponent);                System.arraycopy(digits, 0, result, i, charLength);                i += charLength;                if (charLength < decExponent) {                    charLength = decExponent-charLength;                    System.arraycopy(zero, 0, result, i, charLength);                    i += charLength;                    result[i++] = '.';                    result[i++] = '0';                } else {                    result[i++] = '.';                    if (charLength < nDigits) {                        int t = nDigits - charLength;                        System.arraycopy(digits, charLength, result, i, t);                        i += t;                    } else {                        result[i++] = '0';                    }                }            } else if (decExponent <=0 && decExponent > -3) {                result[i++] = '0';                result[i++] = '.';                if (decExponent != 0) {                    System.arraycopy(zero, 0, result, i, -decExponent);                    i -= decExponent;                }                System.arraycopy(digits, 0, result, i, nDigits);                i += nDigits;            } else {                result[i++] = digits[0];                result[i++] = '.';                if (nDigits > 1) {                    System.arraycopy(digits, 1, result, i, nDigits-1);                    i += nDigits-1;                } else {                    result[i++] = '0';                }                result[i++] = 'E';                int e;                if (decExponent <= 0) {                    result[i++] = '-';                    e = -decExponent+1;                } else {                    e = decExponent-1;                }                // decExponent has 1, 2, or 3, digits                if (e <= 9) {                    result[i++] = (char)(e+'0');                } else if (e <= 99) {                    result[i++] = (char)(e/10 +'0');                    result[i++] = (char)(e%10 + '0');                } else {                    result[i++] = (char)(e/100+'0');                    e %= 100;                    result[i++] = (char)(e/10+'0');                    result[i++] = (char)(e%10 + '0');                }            }        }        return i;    }     // Per-thread buffer for string/stringbuffer conversion    private static ThreadLocal perThreadBuffer = new ThreadLocal() {            protected synchronized Object initialValue() {                return new char[26];            }        };     public void appendTo(Appendable buf) {          char result[] = (char[])(perThreadBuffer.get());          int i = getChars(result);        if (buf instanceof StringBuilder)            ((StringBuilder) buf).append(result, 0, i);        else if (buf instanceof StringBuffer)            ((StringBuffer) buf).append(result, 0, i);        else            assert false;    }     public static FloatingDecimal    readJavaFormatString( String in ) throws NumberFormatException {

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