JDK 1.1 Source Code Directory

JDK 1.1 source code directory contains Java source code for JDK 1.1 core classes: "C:\fyicenter\jdk-1.1.8\src".

Here is the list of Java classes of the JDK 1.1 source code:

✍: FYIcenter

java/lang/FloatingDecimal.java

/*
 * @(#)FloatingDecimal.java	1.11 01/12/10
 *
 * Copyright 2002 Sun Microsystems, Inc. All rights reserved.
 * SUN PROPRIETARY/CONFIDENTIAL. Use is subject to license terms.
 */

package java.lang;

class FloatingDecimal{
    boolean	isExceptional;
    boolean	isNegative;
    int		decExponent;
    char	digits[];
    int		nDigits;

    private	FloatingDecimal( boolean negSign, int decExponent, char []digits, int n,  boolean e )
    {
	isNegative = negSign;
	isExceptional = e;
	this.decExponent = decExponent;
	this.digits = digits;
	this.nDigits = n;
    }

    /*
     * Constants of the implementation
     * Most are IEEE-754 related.
     * (There are more really boring constants at the end.)
     */
    static final long	signMask = 0x8000000000000000L;
    static final long	expMask  = 0x7ff0000000000000L;
    static final long	fractMask= ~(signMask|expMask);
    static final int	expShift = 52;
    static final int	expBias  = 1023;
    static final long	fractHOB = ( 1L<<expShift ); // assumed High-Order bit
    static final long	expOne	 = ((long)expBias)<<expShift; // exponent of 1.0
    static final int	maxSmallBinExp = 62;
    static final int	minSmallBinExp = -( 63 / 3 );

    static final long	highbyte = 0xff00000000000000L;
    static final long	highbit  = 0x8000000000000000L;
    static final long	lowbytes = ~highbyte;

    static final int	singleSignMask =    0x80000000;
    static final int	singleExpMask  =    0x7f800000;
    static final int	singleFractMask =   ~(singleSignMask|singleExpMask);
    static final int	singleExpShift	=   23;
    static final int	singleFractHOB	=   1<<singleExpShift;
    static final int	singleExpBias	=   127;

    /*
     * count number of bits from high-order 1 bit to low-order 1 bit,
     * inclusive.
     */
    private static int
    countBits( long v ){
	// 
	// the strategy is to shift until we get a non-zero sign bit
	// then shift until we have no bits left, counting the difference.
	// we do byte shifting as a hack. Hope it helps.
	//
	if ( v == 0L ) return 0;

	while ( ( v & highbyte ) == 0L ){
	    v <<= 8;
	}
	while ( v > 0L ) { // i.e. while ((v&highbit) == 0L )
	    v <<= 1;
	}

	int n = 0;
	while (( v & lowbytes ) != 0L ){
	    v <<= 8;
	    n += 8;
	}
	while ( v != 0L ){
	    v <<= 1;
	    n += 1;
	}
	return n;
    }

    /*
     * Keep big powers of 5 handy for future reference.
     */
    private static FDBigInt b5p[];

    private static FDBigInt
    big5pow( int p ){
	if ( p < 0 )
	    throw new RuntimeException( "Assertion botch: negative power of 5");
	if ( b5p == null ){
	    b5p = new FDBigInt[ p+1 ];
	}else if (b5p.length <= p ){
	    FDBigInt t[] = new FDBigInt[ p+1 ];
	    System.arraycopy( b5p, 0, t, 0, b5p.length );
	    b5p = t;
	}
	if ( b5p[p] != null )
	    return b5p[p];
	else if ( p < small5pow.length )
	    return b5p[p] = new FDBigInt( small5pow[p] );
	else if ( p < long5pow.length )
	    return b5p[p] = new FDBigInt( long5pow[p] );
	else {
	    // construct the damn thing.
	    // recursively.
	    int q, r;
	    // in order to compute 5^p,
	    // compute its square root, 5^(p/2) and square.
	    // or, let q = p / 2, r = p -q, then
	    // 5^p = 5^(q+r) = 5^q * 5^r
	    q = p >> 1;
	    r = p - q;
	    FDBigInt bigq =  b5p[q];
	    if ( bigq == null ) 
		bigq = big5pow ( q );
	    if ( r < small5pow.length ){
		return (b5p[p] = bigq.mult( small5pow[r] ) );
	    }else{
		FDBigInt bigr = b5p[ r ];
		if ( bigr == null ) 
		    bigr = big5pow( r );
		return (b5p[p] = bigq.mult( bigr ) );
	    }
	}
    }

    /*
     * This is the easy subcase -- 
     * all the significant bits, after scaling, are held in lvalue.
     * negSign and decExponent tell us what processing and scaling
     * has already been done. Exceptional cases have already been
     * stripped out. 
     * In particular:
     * lvalue is a finite number (not Inf, nor NaN)
     * lvalue > 0L (not zero, nor negative).
     *
     * The only reason that we develop the digits here, rather than
     * calling on Long.toString() is that we can do it a little faster,
     * and besides want to treat trailing 0s specially. If Long.toString
     * changes, we should re-evaluate this strategy!
     */
    private void
    developLongDigits( int decExponent, long lvalue, long insignificant ){
	char digits[];
	int  ndigits;
	int  digitno;
	int  c;
	//
	// Discard non-significant low-order bits, while rounding,
	// up to insignificant value.
	int i;
	for ( i = 0; insignificant >= 10L; i++ )
	    insignificant /= 10L;
	if ( i != 0 ){
	    long pow10 = long5pow[i] << i; // 10^i == 5^i * 2^i;
	    long residue = lvalue % pow10;
	    lvalue /= pow10;
	    decExponent += i;
	    if ( residue >= (pow10>>1) ){
		// round up based on the low-order bits we're discarding
		lvalue++;
	    }
	}
	if ( lvalue <= Integer.MAX_VALUE ){
	    if ( lvalue <= 0L )
		throw new RuntimeException("Assertion botch: value "+lvalue+" <= 0");

	    // even easier subcase!
	    // can do int arithmetic rather than long!
	    int  ivalue = (int)lvalue;
	    digits = new char[ ndigits=10 ];
	    digitno = ndigits-1;
	    c = ivalue%10;
	    ivalue /= 10;
	    while ( c == 0 ){
		decExponent++;
		c = ivalue%10;
		ivalue /= 10;
	    }
	    while ( ivalue != 0){
		digits[digitno--] = (char)(c+'0');
		decExponent++;
		c = ivalue%10;
		ivalue /= 10;
	    }
	    digits[digitno] = (char)(c+'0');
	} else {
	    // same algorithm as above (same bugs, too )
	    // but using long arithmetic.
	    digits = new char[ ndigits=20 ];
	    digitno = ndigits-1;
	    c = (int)(lvalue%10L);
	    lvalue /= 10L;
	    while ( c == 0 ){
		decExponent++;
		c = (int)(lvalue%10L);
		lvalue /= 10L;
	    }
	    while ( lvalue != 0L ){
		digits[digitno--] = (char)(c+'0');
		decExponent++;
		c = (int)(lvalue%10L);
		lvalue /= 10;
	    }
	    digits[digitno] = (char)(c+'0');
	}
	char result [];
	ndigits -= digitno;
	if ( digitno == 0 )
	    result = digits;
	else {
	    result = new char[ ndigits ];
	    System.arraycopy( digits, digitno, result, 0, ndigits );
	}
	this.digits = result;
	this.decExponent = decExponent+1;
	this.nDigits = ndigits;
    }

    //
    // add one to the least significant digit.
    // in the unlikely event there is a carry out,
    // deal with it.
    // assert that this will only happen where there
    // is only one digit, e.g. (float)1e-44 seems to do it.
    //
    private void
    roundup(){
	int i;
	int q = digits[ i = (nDigits-1)];
	if ( q == '9' ){
	    while ( q == '9' && i > 0 ){
		digits[i] = '0';
		q = digits[--i];
	    }
	    if ( q == '9' ){
		// carryout! High-order 1, rest 0s, larger exp.
		decExponent += 1;
		digits[0] = '1';
		return;
	    }
	    // else fall through.
	}
	digits[i] = (char)(q+1);
    }

    /*
     * FIRST IMPORTANT CONSTRUCTOR: DOUBLE
     */
    public FloatingDecimal( double d )
    {
	long	dBits = Double.doubleToLongBits( d );
	long	fractBits;
	int	binExp;
	int	nSignificantBits;

	// discover and delete sign
	if ( (dBits&signMask) != 0 ){
	    isNegative = true;
	    dBits ^= signMask;
	} else {
	    isNegative = false;
	}
	// Begin to unpack
	// Discover obvious special cases of NaN and Infinity.
	binExp = (int)( (dBits&expMask) >> expShift );
	fractBits = dBits&fractMask;
	if ( binExp == (int)(expMask>>expShift) ) {
	    isExceptional = true;
	    if ( fractBits == 0L ){
		digits =  infinity;
	    } else {
		digits = notANumber;
		isNegative = false; // NaN has no sign!
	    }
	    nDigits = digits.length;
	    return;
	}
	isExceptional = false;
	// Finish unpacking
	// Normalize denormalized numbers.
	// Insert assumed high-order bit for normalized numbers.
	// Subtract exponent bias.
	if ( binExp == 0 ){
	    if ( fractBits == 0L ){
		// not a denorm, just a 0!
		decExponent = 0;
		digits = zero;
		nDigits = 1;
		return;
	    }
	    while ( (fractBits&fractHOB) == 0L ){
		fractBits <<= 1;
		binExp -= 1;
	    }
	    nSignificantBits = expShift + binExp; // recall binExp is  - shift count.
	    binExp += 1;
	} else {
	    fractBits |= fractHOB;
	    nSignificantBits = expShift+1;
	}
	binExp -= expBias;
	// call the routine that actually does all the hard work.
	dtoa( binExp, fractBits, nSignificantBits );
    }

    /*
     * SECOND IMPORTANT CONSTRUCTOR: SINGLE
     */
    public FloatingDecimal( float f )
    {
	int	fBits = Float.floatToIntBits( f );
	int	fractBits;
	int	binExp;
	int	nSignificantBits;

	// discover and delete sign
	if ( (fBits&singleSignMask) != 0 ){
	    isNegative = true;
	    fBits ^= singleSignMask;
	} else {
	    isNegative = false;
	}
	// Begin to unpack
	// Discover obvious special cases of NaN and Infinity.
	binExp = (int)( (fBits&singleExpMask) >> singleExpShift );
	fractBits = fBits&singleFractMask;
	if ( binExp == (int)(singleExpMask>>singleExpShift) ) {
	    isExceptional = true;
	    if ( fractBits == 0L ){
		digits =  infinity;
	    } else {
		digits = notANumber;
		isNegative = false; // NaN has no sign!
	    }
	    nDigits = digits.length;
	    return;
	}
	isExceptional = false;
	// Finish unpacking
	// Normalize denormalized numbers.
	// Insert assumed high-order bit for normalized numbers.
	// Subtract exponent bias.
	if ( binExp == 0 ){
	    if ( fractBits == 0 ){
		// not a denorm, just a 0!
		decExponent = 0;
		digits = zero;
		nDigits = 1;
		return;
	    }
	    while ( (fractBits&singleFractHOB) == 0 ){
		fractBits <<= 1;
		binExp -= 1;
	    }
	    nSignificantBits = singleExpShift + binExp; // 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 assymetry 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 assymetry 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 = (int) ( b / s );
		b = 10 * ( b % s );
		m *= 10;
		low  = (b <  m );
		high = (b+m > tens );
		if ( q >= 10 ){
		    // bummer, dude
		    throw new RuntimeException( "Assertion botch: excessivly large digit "+q);
		} else 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;
		    if ( q >= 10 ){
			// bummer, dude
			throw new RuntimeException( "Assertion botch: excessivly large digit "+q);
		    }
		    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 );
		if ( q >= 10 ){
		    // bummer, dude
		    throw new RuntimeException( "Assertion botch: excessivly large digit "+q);
		} else 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;
		    if ( q >= 10 ){
			// bummer, dude
			throw new RuntimeException( "Assertion botch: excessivly large digit "+q);
		    }
		    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 = new FDBigInt( fractBits  );
	    if ( B5 != 0 ){
		if ( B5 < small5pow.length ){
		    Bval = Bval.mult( small5pow[B5] );
		} else {
		    Bval = Bval.mult( big5pow( B5 ) );
		}
	    }
	    if ( B2 != 0 ){
		Bval.lshiftMe( B2 );
	    }
	    Sval = new FDBigInt( big5pow( S5 ) );
	    if ( S2 != 0 ){
		Sval.lshiftMe( S2 );
	    }
	    Mval = new FDBigInt( big5pow( M5 ) );
	    if ( M2 != 0 ){
		Mval.lshiftMe( 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 );
	    if ( q >= 10 ){
		// bummer, dude
		throw new RuntimeException( "Assertion botch: excessivly large digit "+q);
	    } else 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 );
		if ( q >= 10 ){
		    // bummer, dude
		    throw new RuntimeException( "Assertion botch: excessivly large digit "+q);
		}
		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[] = new char[ nDigits + 10 ];
	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 new String(result, 0, i);
    }

    private static final int small5pow[] = {
	1,
	5,
	5*5,
	5*5*5,
	5*5*5*5,
	5*5*5*5*5,
	5*5*5*5*5*5,
	5*5*5*5*5*5*5,
	5*5*5*5*5*5*5*5,
	5*5*5*5*5*5*5*5*5,
	5*5*5*5*5*5*5*5*5*5,
	5*5*5*5*5*5*5*5*5*5*5,
	5*5*5*5*5*5*5*5*5*5*5*5,
	5*5*5*5*5*5*5*5*5*5*5*5*5
    };

    private static final long long5pow[] = {
	1L,
	5L,
	5L*5,
	5L*5*5,
	5L*5*5*5,
	5L*5*5*5*5,
	5L*5*5*5*5*5,
	5L*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
	5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
    };

    // approximately ceil( log2( long5pow[i] ) )
    private static final int n5bits[] = {
	0,
	3,
	5,
	7,
	10,
	12,
	14,
	17,
	19,
	21,
	24,
	26,
	28,
	31,
	33,
	35,
	38,
	40,
	42,
	45,
	47,
	49,
	52,
	54,
	56,
	59,
	61,
    };

    private static final char infinity[] = { 'I', 'n', 'f', 'i', 'n', 'i', 't', 'y' };
    private static final char notANumber[] = { 'N', 'a', 'N' };
    private static final char zero[] = { '0', '0', '0', '0', '0', '0', '0', '0' };
}

/*
 * A really, really simple bigint package
 * tailored to the needs of floating base conversion.
 */
class FDBigInt {
    int	nWords; // number of words used
    int data[]; // value: data[0] is least significant

    private static boolean debugging = false;

    public static void setDebugging( boolean d ) { debugging = d; }

    public FDBigInt( int v ){
	nWords = 1;
	data = new int[1];
	data[0] = v;
    }

    public FDBigInt( long v ){
	data = new int[2];
	data[0] = (int)v;
	data[1] = (int)(v>>>32);
	nWords = (data[1]==0) ? 1 : 2;
    }

    public FDBigInt( FDBigInt other ){
	data = new int[nWords = other.nWords];
	System.arraycopy( other.data, 0, data, 0, nWords );
    }

    private FDBigInt( int [] d, int n ){
	data = d;
	nWords = n;
    }

    /*
     * Left shift by c bits.
     * Shifts this in place.
     */
    public void
    lshiftMe( int c )throws IllegalArgumentException {
	if ( c <= 0 ){
	    if ( c == 0 )
		return; // silly.
	    else
		throw new IllegalArgumentException("negative shift count");
	}
	int wordcount = c>>5;
	int bitcount  = c & 0x1f;
	int anticount = 32-bitcount;
	int t[] = data;
	int s[] = data;
	if ( nWords+wordcount+1 > t.length ){
	    // reallocate.
	    t = new int[ nWords+wordcount+1 ];
	}
	int target = nWords+wordcount;
	int src    = nWords-1;
	if ( bitcount == 0 ){
	    // special hack, since an anticount of 32 won't go!
	    System.arraycopy( s, 0, t, wordcount, nWords );
	    target = wordcount-1;
	} else {
	    t[target--] = s[src]>>>anticount;
	    while ( src >= 1 ){
		t[target--] = (s[src]<<bitcount) | (s[--src]>>>anticount);
	    }
	    t[target--] = s[src]<<bitcount;
	}
	while( target >= 0 ){
	    t[target--] = 0;
	}
	data = t;
	nWords += wordcount + 1;
	// may have constructed high-order word of 0.
	// if so, trim it
	while ( nWords > 1 && data[nWords-1] == 0 )
	    nWords--;
    }

    /*
     * normalize this number by shifting until
     * the MSB of the number is at 0x08000000.
     * This is in preparation for quoRemIteration, below.
     * The idea is that, to make division easier, we want the
     * divisor to be "normalized" -- usually this means shifting
     * the MSB into the high words sign bit. But because we know that
     * the quotient will be 0 < q < 10, we would like to arrange that
     * the dividend not span up into another word of precision.
     * (This needs to be explained more clearly!)
     */
    public int
    normalizeMe() throws IllegalArgumentException {
	int src;
	int wordcount = 0;
	int bitcount  = 0;
	int v = 0;
	for ( src= nWords-1 ; src >= 0 && (v=data[src]) == 0 ; src--){
	    wordcount += 1;
	}
	if ( src < 0 ){
	    // oops. Value is zero. Cannot normalize it!
	    throw new IllegalArgumentException("zero value");
	}
	/*
	 * In most cases, we assume that wordcount is zero. This only
	 * makes sense, as we try not to maintain any high-order
	 * words full of zeros. In fact, if there are zeros, we will
	 * simply SHORTEN our number at this point. Watch closely...
	 */
	nWords -= wordcount;
	/*
	 * Compute how far left we have to shift v s.t. its highest-
	 * order bit is in the right place. Then call lshiftMe to
	 * do the work.
	 */
	if ( (v & 0xf0000000) != 0 ){
	    // will have to shift up into the next word.
	    // too bad.
	    for( bitcount = 32 ; (v & 0xf0000000) != 0 ; bitcount-- )
		v >>>= 1;
	} else {
	    while ( v <= 0x000fffff ){
		// hack: byte-at-a-time shifting
		v <<= 8;
		bitcount += 8;
	    }
	    while ( v <= 0x07ffffff ){
		v <<= 1;
		bitcount += 1;
	    }
	}
	if ( bitcount != 0 )
	    lshiftMe( bitcount );
	return bitcount;
    }

    /*
     * Multiply a FDBigInt by an int.
     * Result is a new FDBigInt.
     */
    public FDBigInt
    mult( int iv ) {
	long v = iv;
	int r[];
	long p;

	// guess adequate size of r.
	r = new int[ ( v * ((long)data[nWords-1]&0xffffffffL) > 0xfffffffL ) ? nWords+1 : nWords ];
	p = 0L;
	for( int i=0; i < nWords; i++ ) {
	    p += v * ((long)data[i]&0xffffffffL);
	    r[i] = (int)p;
	    p >>>= 32;
	}
	if ( p == 0L){
	    return new FDBigInt( r, nWords );
	} else {
	    r[nWords] = (int)p;
	    return new FDBigInt( r, nWords+1 );
	}
    }

    /*
     * Multiply a FDBigInt by another FDBigInt.
     * Result is a new FDBigInt.
     */
    public FDBigInt
    mult( FDBigInt other ){
	// crudely guess adequate size for r
	int r[] = new int[ nWords + other.nWords ];
	int i;
	// I think I am promised zeros...

	for( i = 0; i < this.nWords; i++ ){
	    long v = (long)this.data[i] & 0xffffffffL; // UNSIGNED CONVERSION
	    long p = 0L;
	    int j;
	    for( j = 0; j < other.nWords; j++ ){
		p += ((long)r[i+j]&0xffffffffL) + v*((long)other.data[j]&0xffffffffL); // UNSIGNED CONVERSIONS ALL 'ROUND.
		r[i+j] = (int)p;
		p >>>= 32;
	    }
	    r[i+j] = (int)p;
	}
	// compute how much of r we actually needed for all that.
	for ( i = r.length-1; i> 0; i--)
	    if ( r[i] != 0 )
		break;
	return new FDBigInt( r, i+1 );
    }

    /*
     * Add one FDBigInt to another. Return a FDBigInt
     */
    public FDBigInt
    add( FDBigInt other ){
	int i;
	int a[], b[];
	int n, m;
	long c = 0L;
	// arrange such that a.nWords >= b.nWords;
	// n = a.nWords, m = b.nWords
	if ( this.nWords >= other.nWords ){
	    a = this.data;
	    n = this.nWords;
	    b = other.data;
	    m = other.nWords;
	} else {
	    a = other.data;
	    n = other.nWords;
	    b = this.data;
	    m = this.nWords;
	}
	int r[] = new int[ n ];
	for ( i = 0; i < n; i++ ){
	    c += (long)a[i] & 0xffffffffL;
	    if ( i < m ){
		c += (long)b[i] & 0xffffffffL;
	    }
	    r[i] = (int) c;
	    c >>= 32; // signed shift.
	}
	if ( c != 0L ){
	    // oops -- carry out -- need longer result.
	    int s[] = new int[ r.length+1 ];
	    System.arraycopy( r, 0, s, 0, r.length );
	    s[i++] = (int)c;
	    return new FDBigInt( s, i );
	}
	return new FDBigInt( r, i );
    }

    /*
     * Subtract one FDBigInt from another. Return a FDBigInt
     * Assert that the result is positive.
     */
    public FDBigInt
    sub( FDBigInt other ){
	int r[] = new int[ this.nWords ];
	int i;
	int n = this.nWords;
	int m = other.nWords;
	int nzeros = 0;
	long c = 0L;
	for ( i = 0; i < n; i++ ){
	    c += (long)this.data[i] & 0xffffffffL;
	    if ( i < m ){
		c -= (long)other.data[i] & 0xffffffffL;
	    }
	    if ( ( r[i] = (int) c ) == 0 )
		nzeros++;
	    else
		nzeros = 0;
	    c >>= 32; // signed shift.
	}
	if ( c != 0L )
	    throw new RuntimeException("Assertion botch: borrow out of subtract");
	while ( i < m )
	    if ( other.data[i++] != 0 )
		throw new RuntimeException("Assertion botch: negative result of subtract");
	return new FDBigInt( r, n-nzeros );
    }

    /*
     * Compare FDBigInt with another FDBigInt. Return an integer
     * >0: this > other
     *  0: this == other
     * <0: this < other
     */
    public int
    cmp( FDBigInt other ){
        int i;
	if ( this.nWords > other.nWords ){
	    // if any of my high-order words is non-zero,
	    // then the answer is evident
	    int j = other.nWords-1;
	    for ( i = this.nWords-1; i > j ; i-- )
		if ( this.data[i] != 0 ) return 1;
	}else if ( this.nWords < other.nWords ){
	    // if any of other's high-order words is non-zero,
	    // then the answer is evident
	    int j = this.nWords-1;
	    for ( i = other.nWords-1; i > j ; i-- )
		if ( other.data[i] != 0 ) return -1;
	} else{
	    i = this.nWords-1;
	}
	for ( ; i > 0 ; i-- )
	    if ( this.data[i] != other.data[i] )
		break;
	// careful! want unsigned compare!
	// use brute force here.
	int a = this.data[i];
	int b = other.data[i];
	if ( a < 0 ){
	    // a is really big, unsigned
	    if ( b < 0 ){
		return a-b; // both big, negative
	    } else {
		return 1; // b not big, answer is obvious;
	    }
	} else {
	    // a is not really big
	    if ( b < 0 ) {
		// but b is really big
		return -1;
	    } else {
		return a - b;
	    }
	}
    }

    /*
     * Compute
     * q = (int)( this / S )
     * this = 10 * ( this mod S )
     * Return q.
     * This is the iteration step of digit development for output.
     * We assume that S has been normalized, as above, and that
     * "this" has been lshift'ed accordingly.
     * Also assume, of course, that the result, q, can be expressed
     * as an integer, 0 <= q < 10.
     */
    public int
    quoRemIteration( FDBigInt S )throws IllegalArgumentException {
	// ensure that this and S have the same number of
	// digits. If S is properly normalized and q < 10 then
	// this must be so.
	if ( nWords != S.nWords ){
	    throw new IllegalArgumentException("disparate values");
	}
	// estimate q the obvious way. We will usually be
	// right. If not, then we're only off by a little and
	// will re-add.
	int n = nWords-1;
	long q = ((long)data[n]&0xffffffffL) / (long)S.data[n];
	long diff = 0L;
	for ( int i = 0; i <= n ; i++ ){
	    diff += ((long)data[i]&0xffffffffL) -  q*((long)S.data[i]&0xffffffffL);
	    data[i] = (int)diff;
	    diff >>= 32; // N.B. SIGNED shift.
	}
	if ( diff != 0L ) {
	    // damn, damn, damn. q is too big.
	    // add S back in until this turns +. This should
	    // not be very many times!
	    long sum = 0L;
	    while ( sum ==  0L ){
		sum = 0L;
		for ( int i = 0; i <= n; i++ ){
		    sum += ((long)data[i]&0xffffffffL) +  ((long)S.data[i]&0xffffffffL);
		    data[i] = (int) sum;
		    sum >>= 32; // Signed or unsigned, answer is 0 or 1
		}
		/*
		 * Originally the following line read
		 * "if ( sum !=0 && sum != -1 )"
		 * but that would be wrong, because of the 
		 * treatment of the two values as entirely unsigned,
		 * it would be impossible for a carry-out to be interpreted
		 * as -1 -- it would have to be a single-bit carry-out, or
		 * +1.
		 */
		if ( sum !=0 && sum != 1 )
		    throw new RuntimeException("Assertion botch: "+sum+" carry out of division correction");
		q -= 1;
	    }
	}
	// finally, we can multiply this by 10.
	// it cannot overflow, right, as the high-order word has
	// at least 4 high-order zeros!
	long p = 0L;
	for ( int i = 0; i <= n; i++ ){
	    p += 10*((long)data[i]&0xffffffffL);
	    data[i] = (int)p;
	    p >>= 32; // SIGNED shift.
	}
	if ( p != 0L )
	    throw new RuntimeException("Assertion botch: carry out of *10");

	return (int)q;
    }

    public long
    longValue(){
	// if this can be represented as a long,
	// return the value
	int i;
	for ( i = this.nWords-1; i > 1 ; i-- ){
	    if ( data[i] != 0 ){
		throw new RuntimeException("Assertion botch: value too big");
	    }
	}
	switch(i){
	case 1:
	    if ( data[1] < 0 )
		throw new RuntimeException("Assertion botch: value too big");
	    return ((long)(data[1]) << 32) | ((long)data[0]&0xffffffffL);
	case 0:
	    return ((long)data[0]&0xffffffffL);
	default:
	    throw new RuntimeException("Assertion botch: longValue confused");
	}
    }

    public String
    toString() {
	StringBuffer r = new StringBuffer(30);
	r.append('[');
	int i = Math.min( nWords-1, data.length-1) ;
	if ( nWords > data.length ){
	    r.append( "("+data.length+"<"+nWords+"!)" );
	}
	for( ; i> 0 ; i-- ){
	    r.append( Integer.toHexString( data[i] ) );
	    r.append( (char) ' ' );
	}
	r.append( Integer.toHexString( data[0] ) );
	r.append( (char) ']' );
	return new String( r );
    }
}

java/lang/FloatingDecimal.java

 

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