--- a/jdk/test/sun/misc/FloatingDecimal/OldFloatingDecimalForTest.java Thu Dec 24 10:33:21 2015 -0800
+++ /dev/null Thu Jan 01 00:00:00 1970 +0000
@@ -1,2434 +0,0 @@
-/*
- * Copyright (c) 1996, 2013, Oracle and/or its affiliates. All rights reserved.
- * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
- *
- * This code is free software; you can redistribute it and/or modify it
- * under the terms of the GNU General Public License version 2 only, as
- * published by the Free Software Foundation.
- *
- * This code is distributed in the hope that it will be useful, but WITHOUT
- * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
- * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
- * version 2 for more details (a copy is included in the LICENSE file that
- * accompanied this code).
- *
- * You should have received a copy of the GNU General Public License version
- * 2 along with this work; if not, write to the Free Software Foundation,
- * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
- *
- * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
- * or visit www.oracle.com if you need additional information or have any
- * questions.
- */
-
-//package sun.misc;
-
-import java.util.regex.*;
-
-public class OldFloatingDecimalForTest{
- boolean isExceptional;
- boolean isNegative;
- int decExponent;
- char digits[];
- int nDigits;
- int bigIntExp;
- int bigIntNBits;
- boolean mustSetRoundDir = false;
- boolean fromHex = false;
- int roundDir = 0; // set by doubleValue
-
- /*
- * The fields below provides additional information about the result of
- * the binary to decimal digits conversion done in dtoa() and roundup()
- * methods. They are changed if needed by those two methods.
- */
-
- // True if the dtoa() binary to decimal conversion was exact.
- boolean exactDecimalConversion = false;
-
- // True if the result of the binary to decimal conversion was rounded-up
- // at the end of the conversion process, i.e. roundUp() method was called.
- boolean decimalDigitsRoundedUp = false;
-
- private OldFloatingDecimalForTest( 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 int maxDecimalDigits = 15;
- static final int maxDecimalExponent = 308;
- static final int minDecimalExponent = -324;
- static final int bigDecimalExponent = 324; // i.e. abs(minDecimalExponent)
-
- 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;
- static final int singleMaxDecimalDigits = 7;
- static final int singleMaxDecimalExponent = 38;
- static final int singleMinDecimalExponent = -45;
-
- static final int intDecimalDigits = 9;
-
-
- /*
- * 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 OldFDBigIntForTest b5p[];
-
- private static synchronized OldFDBigIntForTest
- big5pow( int p ){
- assert p >= 0 : p; // negative power of 5
- if ( b5p == null ){
- b5p = new OldFDBigIntForTest[ p+1 ];
- }else if (b5p.length <= p ){
- OldFDBigIntForTest t[] = new OldFDBigIntForTest[ 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 OldFDBigIntForTest( small5pow[p] );
- else if ( p < long5pow.length )
- return b5p[p] = new OldFDBigIntForTest( long5pow[p] );
- else {
- // construct the value.
- // 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;
- OldFDBigIntForTest bigq = b5p[q];
- if ( bigq == null )
- bigq = big5pow ( q );
- if ( r < small5pow.length ){
- return (b5p[p] = bigq.mult( small5pow[r] ) );
- }else{
- OldFDBigIntForTest bigr = b5p[ r ];
- if ( bigr == null )
- bigr = big5pow( r );
- return (b5p[p] = bigq.mult( bigr ) );
- }
- }
- }
-
- //
- // a common operation
- //
- private static OldFDBigIntForTest
- multPow52( OldFDBigIntForTest v, int p5, int p2 ){
- if ( p5 != 0 ){
- if ( p5 < small5pow.length ){
- v = v.mult( small5pow[p5] );
- } else {
- v = v.mult( big5pow( p5 ) );
- }
- }
- if ( p2 != 0 ){
- v.lshiftMe( p2 );
- }
- return v;
- }
-
- //
- // another common operation
- //
- private static OldFDBigIntForTest
- constructPow52( int p5, int p2 ){
- OldFDBigIntForTest v = new OldFDBigIntForTest( big5pow( p5 ) );
- if ( p2 != 0 ){
- v.lshiftMe( p2 );
- }
- return v;
- }
-
- /*
- * Make a floating double into a OldFDBigIntForTest.
- * This could also be structured as a OldFDBigIntForTest
- * constructor, but we'd have to build a lot of knowledge
- * about floating-point representation into it, and we don't want to.
- *
- * AS A SIDE EFFECT, THIS METHOD WILL SET THE INSTANCE VARIABLES
- * bigIntExp and bigIntNBits
- *
- */
- private OldFDBigIntForTest
- doubleToBigInt( double dval ){
- long lbits = Double.doubleToLongBits( dval ) & ~signMask;
- int binexp = (int)(lbits >>> expShift);
- lbits &= fractMask;
- if ( binexp > 0 ){
- lbits |= fractHOB;
- } else {
- assert lbits != 0L : lbits; // doubleToBigInt(0.0)
- binexp +=1;
- while ( (lbits & fractHOB ) == 0L){
- lbits <<= 1;
- binexp -= 1;
- }
- }
- binexp -= expBias;
- int nbits = countBits( lbits );
- /*
- * We now know where the high-order 1 bit is,
- * and we know how many there are.
- */
- int lowOrderZeros = expShift+1-nbits;
- lbits >>>= lowOrderZeros;
-
- bigIntExp = binexp+1-nbits;
- bigIntNBits = nbits;
- return new OldFDBigIntForTest( lbits );
- }
-
- /*
- * Compute a number that is the ULP of the given value,
- * for purposes of addition/subtraction. Generally easy.
- * More difficult if subtracting and the argument
- * is a normalized a power of 2, as the ULP changes at these points.
- */
- private static double ulp( double dval, boolean subtracting ){
- long lbits = Double.doubleToLongBits( dval ) & ~signMask;
- int binexp = (int)(lbits >>> expShift);
- double ulpval;
- if ( subtracting && ( binexp >= expShift ) && ((lbits&fractMask) == 0L) ){
- // for subtraction from normalized, powers of 2,
- // use next-smaller exponent
- binexp -= 1;
- }
- if ( binexp > expShift ){
- ulpval = Double.longBitsToDouble( ((long)(binexp-expShift))<<expShift );
- } else if ( binexp == 0 ){
- ulpval = Double.MIN_VALUE;
- } else {
- ulpval = Double.longBitsToDouble( 1L<<(binexp-1) );
- }
- if ( subtracting ) ulpval = - ulpval;
-
- return ulpval;
- }
-
- /*
- * Round a double to a float.
- * In addition to the fraction bits of the double,
- * look at the class instance variable roundDir,
- * which should help us avoid double-rounding error.
- * roundDir was set in hardValueOf if the estimate was
- * close enough, but not exact. It tells us which direction
- * of rounding is preferred.
- */
- float
- stickyRound( double dval ){
- long lbits = Double.doubleToLongBits( dval );
- long binexp = lbits & expMask;
- if ( binexp == 0L || binexp == expMask ){
- // what we have here is special.
- // don't worry, the right thing will happen.
- return (float) dval;
- }
- lbits += (long)roundDir; // hack-o-matic.
- return (float)Double.longBitsToDouble( lbits );
- }
-
-
- /*
- * 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 ){
- assert lvalue > 0L : lvalue; // lvalue <= 0
- // even easier subcase!
- // can do int arithmetic rather than long!
- int ivalue = (int)lvalue;
- ndigits = 10;
- digits = perThreadBuffer.get();
- 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.
- ndigits = 20;
- digits = perThreadBuffer.get();
- 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;
- 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);
- decimalDigitsRoundedUp = true;
- }
-
- public boolean digitsRoundedUp() {
- return decimalDigitsRoundedUp;
- }
-
- /*
- * FIRST IMPORTANT CONSTRUCTOR: DOUBLE
- */
- public OldFloatingDecimalForTest( 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 +1; // 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 OldFloatingDecimalForTest( 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 = (fBits&singleExpMask) >> singleExpShift;
- fractBits = fBits&singleFractMask;
- if ( binExp == (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 +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.
- OldFDBigIntForTest 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
- * OldFDBigIntForTest. 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 OldFDBigIntForTest arithmetic.
- * We use the same algorithms, except that we "normalize"
- * our OldFDBigIntForTests 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;
- exactDecimalConversion = (b == 0);
- } 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;
- exactDecimalConversion = (b == 0);
- }
- } else {
- OldFDBigIntForTest ZeroVal = new OldFDBigIntForTest(0);
- OldFDBigIntForTest tenSval;
- int shiftBias;
-
- /*
- * We really must do OldFDBigIntForTest arithmetic.
- * Fist, construct our OldFDBigIntForTest initial values.
- */
- Bval = multPow52( new OldFDBigIntForTest( 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!
- }
- exactDecimalConversion = (Bval.cmp( ZeroVal ) == 0);
- }
- 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 boolean decimalDigitsExact() {
- return exactDecimalConversion;
- }
-
- 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[] = 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<char[]> perThreadBuffer = new ThreadLocal<char[]>() {
- protected synchronized char[] initialValue() {
- return new char[26];
- }
- };
-
- public void appendTo(Appendable buf) {
- char result[] = 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;
- }
-
- @SuppressWarnings("fallthrough")
- public static OldFloatingDecimalForTest
- readJavaFormatString( String in ) throws NumberFormatException {
- boolean isNegative = false;
- boolean signSeen = false;
- int decExp;
- char c;
-
- parseNumber:
- try{
- in = in.trim(); // don't fool around with white space.
- // throws NullPointerException if null
- int l = in.length();
- if ( l == 0 ) throw new NumberFormatException("empty String");
- int i = 0;
- switch ( c = in.charAt( i ) ){
- case '-':
- isNegative = true;
- //FALLTHROUGH
- case '+':
- i++;
- signSeen = true;
- }
-
- // Check for NaN and Infinity strings
- c = in.charAt(i);
- if(c == 'N' || c == 'I') { // possible NaN or infinity
- boolean potentialNaN = false;
- char targetChars[] = null; // char array of "NaN" or "Infinity"
-
- if(c == 'N') {
- targetChars = notANumber;
- potentialNaN = true;
- } else {
- targetChars = infinity;
- }
-
- // compare Input string to "NaN" or "Infinity"
- int j = 0;
- while(i < l && j < targetChars.length) {
- if(in.charAt(i) == targetChars[j]) {
- i++; j++;
- }
- else // something is amiss, throw exception
- break parseNumber;
- }
-
- // For the candidate string to be a NaN or infinity,
- // all characters in input string and target char[]
- // must be matched ==> j must equal targetChars.length
- // and i must equal l
- if( (j == targetChars.length) && (i == l) ) { // return NaN or infinity
- return (potentialNaN ? new OldFloatingDecimalForTest(Double.NaN) // NaN has no sign
- : new OldFloatingDecimalForTest(isNegative?
- Double.NEGATIVE_INFINITY:
- Double.POSITIVE_INFINITY)) ;
- }
- else { // something went wrong, throw exception
- break parseNumber;
- }
-
- } else if (c == '0') { // check for hexadecimal floating-point number
- if (l > i+1 ) {
- char ch = in.charAt(i+1);
- if (ch == 'x' || ch == 'X' ) // possible hex string
- return parseHexString(in);
- }
- } // look for and process decimal floating-point string
-
- char[] digits = new char[ l ];
- int nDigits= 0;
- boolean decSeen = false;
- int decPt = 0;
- int nLeadZero = 0;
- int nTrailZero= 0;
- digitLoop:
- while ( i < l ){
- switch ( c = in.charAt( i ) ){
- case '0':
- if ( nDigits > 0 ){
- nTrailZero += 1;
- } else {
- nLeadZero += 1;
- }
- break; // out of switch.
- case '1':
- case '2':
- case '3':
- case '4':
- case '5':
- case '6':
- case '7':
- case '8':
- case '9':
- while ( nTrailZero > 0 ){
- digits[nDigits++] = '0';
- nTrailZero -= 1;
- }
- digits[nDigits++] = c;
- break; // out of switch.
- case '.':
- if ( decSeen ){
- // already saw one ., this is the 2nd.
- throw new NumberFormatException("multiple points");
- }
- decPt = i;
- if ( signSeen ){
- decPt -= 1;
- }
- decSeen = true;
- break; // out of switch.
- default:
- break digitLoop;
- }
- i++;
- }
- /*
- * At this point, we've scanned all the digits and decimal
- * point we're going to see. Trim off leading and trailing
- * zeros, which will just confuse us later, and adjust
- * our initial decimal exponent accordingly.
- * To review:
- * we have seen i total characters.
- * nLeadZero of them were zeros before any other digits.
- * nTrailZero of them were zeros after any other digits.
- * if ( decSeen ), then a . was seen after decPt characters
- * ( including leading zeros which have been discarded )
- * nDigits characters were neither lead nor trailing
- * zeros, nor point
- */
- /*
- * special hack: if we saw no non-zero digits, then the
- * answer is zero!
- * Unfortunately, we feel honor-bound to keep parsing!
- */
- if ( nDigits == 0 ){
- digits = zero;
- nDigits = 1;
- if ( nLeadZero == 0 ){
- // we saw NO DIGITS AT ALL,
- // not even a crummy 0!
- // this is not allowed.
- break parseNumber; // go throw exception
- }
-
- }
-
- /* Our initial exponent is decPt, adjusted by the number of
- * discarded zeros. Or, if there was no decPt,
- * then its just nDigits adjusted by discarded trailing zeros.
- */
- if ( decSeen ){
- decExp = decPt - nLeadZero;
- } else {
- decExp = nDigits+nTrailZero;
- }
-
- /*
- * Look for 'e' or 'E' and an optionally signed integer.
- */
- if ( (i < l) && (((c = in.charAt(i) )=='e') || (c == 'E') ) ){
- int expSign = 1;
- int expVal = 0;
- int reallyBig = Integer.MAX_VALUE / 10;
- boolean expOverflow = false;
- switch( in.charAt(++i) ){
- case '-':
- expSign = -1;
- //FALLTHROUGH
- case '+':
- i++;
- }
- int expAt = i;
- expLoop:
- while ( i < l ){
- if ( expVal >= reallyBig ){
- // the next character will cause integer
- // overflow.
- expOverflow = true;
- }
- switch ( c = in.charAt(i++) ){
- case '0':
- case '1':
- case '2':
- case '3':
- case '4':
- case '5':
- case '6':
- case '7':
- case '8':
- case '9':
- expVal = expVal*10 + ( (int)c - (int)'0' );
- continue;
- default:
- i--; // back up.
- break expLoop; // stop parsing exponent.
- }
- }
- int expLimit = bigDecimalExponent+nDigits+nTrailZero;
- if ( expOverflow || ( expVal > expLimit ) ){
- //
- // The intent here is to end up with
- // infinity or zero, as appropriate.
- // The reason for yielding such a small decExponent,
- // rather than something intuitive such as
- // expSign*Integer.MAX_VALUE, is that this value
- // is subject to further manipulation in
- // doubleValue() and floatValue(), and I don't want
- // it to be able to cause overflow there!
- // (The only way we can get into trouble here is for
- // really outrageous nDigits+nTrailZero, such as 2 billion. )
- //
- decExp = expSign*expLimit;
- } else {
- // this should not overflow, since we tested
- // for expVal > (MAX+N), where N >= abs(decExp)
- decExp = decExp + expSign*expVal;
- }
-
- // if we saw something not a digit ( or end of string )
- // after the [Ee][+-], without seeing any digits at all
- // this is certainly an error. If we saw some digits,
- // but then some trailing garbage, that might be ok.
- // so we just fall through in that case.
- // HUMBUG
- if ( i == expAt )
- break parseNumber; // certainly bad
- }
- /*
- * We parsed everything we could.
- * If there are leftovers, then this is not good input!
- */
- if ( i < l &&
- ((i != l - 1) ||
- (in.charAt(i) != 'f' &&
- in.charAt(i) != 'F' &&
- in.charAt(i) != 'd' &&
- in.charAt(i) != 'D'))) {
- break parseNumber; // go throw exception
- }
-
- return new OldFloatingDecimalForTest( isNegative, decExp, digits, nDigits, false );
- } catch ( StringIndexOutOfBoundsException e ){ }
- throw new NumberFormatException("For input string: \"" + in + "\"");
- }
-
- /*
- * Take a FloatingDecimal, which we presumably just scanned in,
- * and find out what its value is, as a double.
- *
- * AS A SIDE EFFECT, SET roundDir TO INDICATE PREFERRED
- * ROUNDING DIRECTION in case the result is really destined
- * for a single-precision float.
- */
-
- public strictfp double doubleValue(){
- int kDigits = Math.min( nDigits, maxDecimalDigits+1 );
- long lValue;
- double dValue;
- double rValue, tValue;
-
- // First, check for NaN and Infinity values
- if(digits == infinity || digits == notANumber) {
- if(digits == notANumber)
- return Double.NaN;
- else
- return (isNegative?Double.NEGATIVE_INFINITY:Double.POSITIVE_INFINITY);
- }
- else {
- if (mustSetRoundDir) {
- roundDir = 0;
- }
- /*
- * convert the lead kDigits to a long integer.
- */
- // (special performance hack: start to do it using int)
- int iValue = (int)digits[0]-(int)'0';
- int iDigits = Math.min( kDigits, intDecimalDigits );
- for ( int i=1; i < iDigits; i++ ){
- iValue = iValue*10 + (int)digits[i]-(int)'0';
- }
- lValue = (long)iValue;
- for ( int i=iDigits; i < kDigits; i++ ){
- lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
- }
- dValue = (double)lValue;
- int exp = decExponent-kDigits;
- /*
- * lValue now contains a long integer with the value of
- * the first kDigits digits of the number.
- * dValue contains the (double) of the same.
- */
-
- if ( nDigits <= maxDecimalDigits ){
- /*
- * possibly an easy case.
- * We know that the digits can be represented
- * exactly. And if the exponent isn't too outrageous,
- * the whole thing can be done with one operation,
- * thus one rounding error.
- * Note that all our constructors trim all leading and
- * trailing zeros, so simple values (including zero)
- * will always end up here
- */
- if (exp == 0 || dValue == 0.0)
- return (isNegative)? -dValue : dValue; // small floating integer
- else if ( exp >= 0 ){
- if ( exp <= maxSmallTen ){
- /*
- * Can get the answer with one operation,
- * thus one roundoff.
- */
- rValue = dValue * small10pow[exp];
- if ( mustSetRoundDir ){
- tValue = rValue / small10pow[exp];
- roundDir = ( tValue == dValue ) ? 0
- :( tValue < dValue ) ? 1
- : -1;
- }
- return (isNegative)? -rValue : rValue;
- }
- int slop = maxDecimalDigits - kDigits;
- if ( exp <= maxSmallTen+slop ){
- /*
- * We can multiply dValue by 10^(slop)
- * and it is still "small" and exact.
- * Then we can multiply by 10^(exp-slop)
- * with one rounding.
- */
- dValue *= small10pow[slop];
- rValue = dValue * small10pow[exp-slop];
-
- if ( mustSetRoundDir ){
- tValue = rValue / small10pow[exp-slop];
- roundDir = ( tValue == dValue ) ? 0
- :( tValue < dValue ) ? 1
- : -1;
- }
- return (isNegative)? -rValue : rValue;
- }
- /*
- * Else we have a hard case with a positive exp.
- */
- } else {
- if ( exp >= -maxSmallTen ){
- /*
- * Can get the answer in one division.
- */
- rValue = dValue / small10pow[-exp];
- tValue = rValue * small10pow[-exp];
- if ( mustSetRoundDir ){
- roundDir = ( tValue == dValue ) ? 0
- :( tValue < dValue ) ? 1
- : -1;
- }
- return (isNegative)? -rValue : rValue;
- }
- /*
- * Else we have a hard case with a negative exp.
- */
- }
- }
-
- /*
- * Harder cases:
- * The sum of digits plus exponent is greater than
- * what we think we can do with one error.
- *
- * Start by approximating the right answer by,
- * naively, scaling by powers of 10.
- */
- if ( exp > 0 ){
- if ( decExponent > maxDecimalExponent+1 ){
- /*
- * Lets face it. This is going to be
- * Infinity. Cut to the chase.
- */
- return (isNegative)? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
- }
- if ( (exp&15) != 0 ){
- dValue *= small10pow[exp&15];
- }
- if ( (exp>>=4) != 0 ){
- int j;
- for( j = 0; exp > 1; j++, exp>>=1 ){
- if ( (exp&1)!=0)
- dValue *= big10pow[j];
- }
- /*
- * The reason for the weird exp > 1 condition
- * in the above loop was so that the last multiply
- * would get unrolled. We handle it here.
- * It could overflow.
- */
- double t = dValue * big10pow[j];
- if ( Double.isInfinite( t ) ){
- /*
- * It did overflow.
- * Look more closely at the result.
- * If the exponent is just one too large,
- * then use the maximum finite as our estimate
- * value. Else call the result infinity
- * and punt it.
- * ( I presume this could happen because
- * rounding forces the result here to be
- * an ULP or two larger than
- * Double.MAX_VALUE ).
- */
- t = dValue / 2.0;
- t *= big10pow[j];
- if ( Double.isInfinite( t ) ){
- return (isNegative)? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
- }
- t = Double.MAX_VALUE;
- }
- dValue = t;
- }
- } else if ( exp < 0 ){
- exp = -exp;
- if ( decExponent < minDecimalExponent-1 ){
- /*
- * Lets face it. This is going to be
- * zero. Cut to the chase.
- */
- return (isNegative)? -0.0 : 0.0;
- }
- if ( (exp&15) != 0 ){
- dValue /= small10pow[exp&15];
- }
- if ( (exp>>=4) != 0 ){
- int j;
- for( j = 0; exp > 1; j++, exp>>=1 ){
- if ( (exp&1)!=0)
- dValue *= tiny10pow[j];
- }
- /*
- * The reason for the weird exp > 1 condition
- * in the above loop was so that the last multiply
- * would get unrolled. We handle it here.
- * It could underflow.
- */
- double t = dValue * tiny10pow[j];
- if ( t == 0.0 ){
- /*
- * It did underflow.
- * Look more closely at the result.
- * If the exponent is just one too small,
- * then use the minimum finite as our estimate
- * value. Else call the result 0.0
- * and punt it.
- * ( I presume this could happen because
- * rounding forces the result here to be
- * an ULP or two less than
- * Double.MIN_VALUE ).
- */
- t = dValue * 2.0;
- t *= tiny10pow[j];
- if ( t == 0.0 ){
- return (isNegative)? -0.0 : 0.0;
- }
- t = Double.MIN_VALUE;
- }
- dValue = t;
- }
- }
-
- /*
- * dValue is now approximately the result.
- * The hard part is adjusting it, by comparison
- * with OldFDBigIntForTest arithmetic.
- * Formulate the EXACT big-number result as
- * bigD0 * 10^exp
- */
- OldFDBigIntForTest bigD0 = new OldFDBigIntForTest( lValue, digits, kDigits, nDigits );
- exp = decExponent - nDigits;
-
- correctionLoop:
- while(true){
- /* AS A SIDE EFFECT, THIS METHOD WILL SET THE INSTANCE VARIABLES
- * bigIntExp and bigIntNBits
- */
- OldFDBigIntForTest bigB = doubleToBigInt( dValue );
-
- /*
- * Scale bigD, bigB appropriately for
- * big-integer operations.
- * Naively, we multiply by powers of ten
- * and powers of two. What we actually do
- * is keep track of the powers of 5 and
- * powers of 2 we would use, then factor out
- * common divisors before doing the work.
- */
- int B2, B5; // powers of 2, 5 in bigB
- int D2, D5; // powers of 2, 5 in bigD
- int Ulp2; // powers of 2 in halfUlp.
- if ( exp >= 0 ){
- B2 = B5 = 0;
- D2 = D5 = exp;
- } else {
- B2 = B5 = -exp;
- D2 = D5 = 0;
- }
- if ( bigIntExp >= 0 ){
- B2 += bigIntExp;
- } else {
- D2 -= bigIntExp;
- }
- Ulp2 = B2;
- // shift bigB and bigD left by a number s. t.
- // halfUlp is still an integer.
- int hulpbias;
- if ( bigIntExp+bigIntNBits <= -expBias+1 ){
- // This is going to be a denormalized number
- // (if not actually zero).
- // half an ULP is at 2^-(expBias+expShift+1)
- hulpbias = bigIntExp+ expBias + expShift;
- } else {
- hulpbias = expShift + 2 - bigIntNBits;
- }
- B2 += hulpbias;
- D2 += hulpbias;
- // if there are common factors of 2, we might just as well
- // factor them out, as they add nothing useful.
- int common2 = Math.min( B2, Math.min( D2, Ulp2 ) );
- B2 -= common2;
- D2 -= common2;
- Ulp2 -= common2;
- // do multiplications by powers of 5 and 2
- bigB = multPow52( bigB, B5, B2 );
- OldFDBigIntForTest bigD = multPow52( new OldFDBigIntForTest( bigD0 ), D5, D2 );
- //
- // to recap:
- // bigB is the scaled-big-int version of our floating-point
- // candidate.
- // bigD is the scaled-big-int version of the exact value
- // as we understand it.
- // halfUlp is 1/2 an ulp of bigB, except for special cases
- // of exact powers of 2
- //
- // the plan is to compare bigB with bigD, and if the difference
- // is less than halfUlp, then we're satisfied. Otherwise,
- // use the ratio of difference to halfUlp to calculate a fudge
- // factor to add to the floating value, then go 'round again.
- //
- OldFDBigIntForTest diff;
- int cmpResult;
- boolean overvalue;
- if ( (cmpResult = bigB.cmp( bigD ) ) > 0 ){
- overvalue = true; // our candidate is too big.
- diff = bigB.sub( bigD );
- if ( (bigIntNBits == 1) && (bigIntExp > -expBias+1) ){
- // candidate is a normalized exact power of 2 and
- // is too big. We will be subtracting.
- // For our purposes, ulp is the ulp of the
- // next smaller range.
- Ulp2 -= 1;
- if ( Ulp2 < 0 ){
- // rats. Cannot de-scale ulp this far.
- // must scale diff in other direction.
- Ulp2 = 0;
- diff.lshiftMe( 1 );
- }
- }
- } else if ( cmpResult < 0 ){
- overvalue = false; // our candidate is too small.
- diff = bigD.sub( bigB );
- } else {
- // the candidate is exactly right!
- // this happens with surprising frequency
- break correctionLoop;
- }
- OldFDBigIntForTest halfUlp = constructPow52( B5, Ulp2 );
- if ( (cmpResult = diff.cmp( halfUlp ) ) < 0 ){
- // difference is small.
- // this is close enough
- if (mustSetRoundDir) {
- roundDir = overvalue ? -1 : 1;
- }
- break correctionLoop;
- } else if ( cmpResult == 0 ){
- // difference is exactly half an ULP
- // round to some other value maybe, then finish
- dValue += 0.5*ulp( dValue, overvalue );
- // should check for bigIntNBits == 1 here??
- if (mustSetRoundDir) {
- roundDir = overvalue ? -1 : 1;
- }
- break correctionLoop;
- } else {
- // difference is non-trivial.
- // could scale addend by ratio of difference to
- // halfUlp here, if we bothered to compute that difference.
- // Most of the time ( I hope ) it is about 1 anyway.
- dValue += ulp( dValue, overvalue );
- if ( dValue == 0.0 || dValue == Double.POSITIVE_INFINITY )
- break correctionLoop; // oops. Fell off end of range.
- continue; // try again.
- }
-
- }
- return (isNegative)? -dValue : dValue;
- }
- }
-
- /*
- * Take a FloatingDecimal, which we presumably just scanned in,
- * and find out what its value is, as a float.
- * This is distinct from doubleValue() to avoid the extremely
- * unlikely case of a double rounding error, wherein the conversion
- * to double has one rounding error, and the conversion of that double
- * to a float has another rounding error, IN THE WRONG DIRECTION,
- * ( because of the preference to a zero low-order bit ).
- */
-
- public strictfp float floatValue(){
- int kDigits = Math.min( nDigits, singleMaxDecimalDigits+1 );
- int iValue;
- float fValue;
-
- // First, check for NaN and Infinity values
- if(digits == infinity || digits == notANumber) {
- if(digits == notANumber)
- return Float.NaN;
- else
- return (isNegative?Float.NEGATIVE_INFINITY:Float.POSITIVE_INFINITY);
- }
- else {
- /*
- * convert the lead kDigits to an integer.
- */
- iValue = (int)digits[0]-(int)'0';
- for ( int i=1; i < kDigits; i++ ){
- iValue = iValue*10 + (int)digits[i]-(int)'0';
- }
- fValue = (float)iValue;
- int exp = decExponent-kDigits;
- /*
- * iValue now contains an integer with the value of
- * the first kDigits digits of the number.
- * fValue contains the (float) of the same.
- */
-
- if ( nDigits <= singleMaxDecimalDigits ){
- /*
- * possibly an easy case.
- * We know that the digits can be represented
- * exactly. And if the exponent isn't too outrageous,
- * the whole thing can be done with one operation,
- * thus one rounding error.
- * Note that all our constructors trim all leading and
- * trailing zeros, so simple values (including zero)
- * will always end up here.
- */
- if (exp == 0 || fValue == 0.0f)
- return (isNegative)? -fValue : fValue; // small floating integer
- else if ( exp >= 0 ){
- if ( exp <= singleMaxSmallTen ){
- /*
- * Can get the answer with one operation,
- * thus one roundoff.
- */
- fValue *= singleSmall10pow[exp];
- return (isNegative)? -fValue : fValue;
- }
- int slop = singleMaxDecimalDigits - kDigits;
- if ( exp <= singleMaxSmallTen+slop ){
- /*
- * We can multiply dValue by 10^(slop)
- * and it is still "small" and exact.
- * Then we can multiply by 10^(exp-slop)
- * with one rounding.
- */
- fValue *= singleSmall10pow[slop];
- fValue *= singleSmall10pow[exp-slop];
- return (isNegative)? -fValue : fValue;
- }
- /*
- * Else we have a hard case with a positive exp.
- */
- } else {
- if ( exp >= -singleMaxSmallTen ){
- /*
- * Can get the answer in one division.
- */
- fValue /= singleSmall10pow[-exp];
- return (isNegative)? -fValue : fValue;
- }
- /*
- * Else we have a hard case with a negative exp.
- */
- }
- } else if ( (decExponent >= nDigits) && (nDigits+decExponent <= maxDecimalDigits) ){
- /*
- * In double-precision, this is an exact floating integer.
- * So we can compute to double, then shorten to float
- * with one round, and get the right answer.
- *
- * First, finish accumulating digits.
- * Then convert that integer to a double, multiply
- * by the appropriate power of ten, and convert to float.
- */
- long lValue = (long)iValue;
- for ( int i=kDigits; i < nDigits; i++ ){
- lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
- }
- double dValue = (double)lValue;
- exp = decExponent-nDigits;
- dValue *= small10pow[exp];
- fValue = (float)dValue;
- return (isNegative)? -fValue : fValue;
-
- }
- /*
- * Harder cases:
- * The sum of digits plus exponent is greater than
- * what we think we can do with one error.
- *
- * Start by weeding out obviously out-of-range
- * results, then convert to double and go to
- * common hard-case code.
- */
- if ( decExponent > singleMaxDecimalExponent+1 ){
- /*
- * Lets face it. This is going to be
- * Infinity. Cut to the chase.
- */
- return (isNegative)? Float.NEGATIVE_INFINITY : Float.POSITIVE_INFINITY;
- } else if ( decExponent < singleMinDecimalExponent-1 ){
- /*
- * Lets face it. This is going to be
- * zero. Cut to the chase.
- */
- return (isNegative)? -0.0f : 0.0f;
- }
-
- /*
- * Here, we do 'way too much work, but throwing away
- * our partial results, and going and doing the whole
- * thing as double, then throwing away half the bits that computes
- * when we convert back to float.
- *
- * The alternative is to reproduce the whole multiple-precision
- * algorithm for float precision, or to try to parameterize it
- * for common usage. The former will take about 400 lines of code,
- * and the latter I tried without success. Thus the semi-hack
- * answer here.
- */
- mustSetRoundDir = !fromHex;
- double dValue = doubleValue();
- return stickyRound( dValue );
- }
- }
-
-
- /*
- * All the positive powers of 10 that can be
- * represented exactly in double/float.
- */
- private static final double small10pow[] = {
- 1.0e0,
- 1.0e1, 1.0e2, 1.0e3, 1.0e4, 1.0e5,
- 1.0e6, 1.0e7, 1.0e8, 1.0e9, 1.0e10,
- 1.0e11, 1.0e12, 1.0e13, 1.0e14, 1.0e15,
- 1.0e16, 1.0e17, 1.0e18, 1.0e19, 1.0e20,
- 1.0e21, 1.0e22
- };
-
- private static final float singleSmall10pow[] = {
- 1.0e0f,
- 1.0e1f, 1.0e2f, 1.0e3f, 1.0e4f, 1.0e5f,
- 1.0e6f, 1.0e7f, 1.0e8f, 1.0e9f, 1.0e10f
- };
-
- private static final double big10pow[] = {
- 1e16, 1e32, 1e64, 1e128, 1e256 };
- private static final double tiny10pow[] = {
- 1e-16, 1e-32, 1e-64, 1e-128, 1e-256 };
-
- private static final int maxSmallTen = small10pow.length-1;
- private static final int singleMaxSmallTen = singleSmall10pow.length-1;
-
- 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' };
-
-
- /*
- * Grammar is compatible with hexadecimal floating-point constants
- * described in section 6.4.4.2 of the C99 specification.
- */
- private static Pattern hexFloatPattern = null;
- private static synchronized Pattern getHexFloatPattern() {
- if (hexFloatPattern == null) {
- hexFloatPattern = Pattern.compile(
- //1 234 56 7 8 9
- "([-+])?0[xX](((\\p{XDigit}+)\\.?)|((\\p{XDigit}*)\\.(\\p{XDigit}+)))[pP]([-+])?(\\p{Digit}+)[fFdD]?"
- );
- }
- return hexFloatPattern;
- }
-
- /*
- * Convert string s to a suitable floating decimal; uses the
- * double constructor and set the roundDir variable appropriately
- * in case the value is later converted to a float.
- */
- static OldFloatingDecimalForTest parseHexString(String s) {
- // Verify string is a member of the hexadecimal floating-point
- // string language.
- Matcher m = getHexFloatPattern().matcher(s);
- boolean validInput = m.matches();
-
- if (!validInput) {
- // Input does not match pattern
- throw new NumberFormatException("For input string: \"" + s + "\"");
- } else { // validInput
- /*
- * We must isolate the sign, significand, and exponent
- * fields. The sign value is straightforward. Since
- * floating-point numbers are stored with a normalized
- * representation, the significand and exponent are
- * interrelated.
- *
- * After extracting the sign, we normalized the
- * significand as a hexadecimal value, calculating an
- * exponent adjust for any shifts made during
- * normalization. If the significand is zero, the
- * exponent doesn't need to be examined since the output
- * will be zero.
- *
- * Next the exponent in the input string is extracted.
- * Afterwards, the significand is normalized as a *binary*
- * value and the input value's normalized exponent can be
- * computed. The significand bits are copied into a
- * double significand; if the string has more logical bits
- * than can fit in a double, the extra bits affect the
- * round and sticky bits which are used to round the final
- * value.
- */
-
- // Extract significand sign
- String group1 = m.group(1);
- double sign = (( group1 == null ) || group1.equals("+"))? 1.0 : -1.0;
-
-
- // Extract Significand magnitude
- /*
- * Based on the form of the significand, calculate how the
- * binary exponent needs to be adjusted to create a
- * normalized *hexadecimal* floating-point number; that
- * is, a number where there is one nonzero hex digit to
- * the left of the (hexa)decimal point. Since we are
- * adjusting a binary, not hexadecimal exponent, the
- * exponent is adjusted by a multiple of 4.
- *
- * There are a number of significand scenarios to consider;
- * letters are used in indicate nonzero digits:
- *
- * 1. 000xxxx => x.xxx normalized
- * increase exponent by (number of x's - 1)*4
- *
- * 2. 000xxx.yyyy => x.xxyyyy normalized
- * increase exponent by (number of x's - 1)*4
- *
- * 3. .000yyy => y.yy normalized
- * decrease exponent by (number of zeros + 1)*4
- *
- * 4. 000.00000yyy => y.yy normalized
- * decrease exponent by (number of zeros to right of point + 1)*4
- *
- * If the significand is exactly zero, return a properly
- * signed zero.
- */
-
- String significandString =null;
- int signifLength = 0;
- int exponentAdjust = 0;
- {
- int leftDigits = 0; // number of meaningful digits to
- // left of "decimal" point
- // (leading zeros stripped)
- int rightDigits = 0; // number of digits to right of
- // "decimal" point; leading zeros
- // must always be accounted for
- /*
- * The significand is made up of either
- *
- * 1. group 4 entirely (integer portion only)
- *
- * OR
- *
- * 2. the fractional portion from group 7 plus any
- * (optional) integer portions from group 6.
- */
- String group4;
- if( (group4 = m.group(4)) != null) { // Integer-only significand
- // Leading zeros never matter on the integer portion
- significandString = stripLeadingZeros(group4);
- leftDigits = significandString.length();
- }
- else {
- // Group 6 is the optional integer; leading zeros
- // never matter on the integer portion
- String group6 = stripLeadingZeros(m.group(6));
- leftDigits = group6.length();
-
- // fraction
- String group7 = m.group(7);
- rightDigits = group7.length();
-
- // Turn "integer.fraction" into "integer"+"fraction"
- significandString =
- ((group6 == null)?"":group6) + // is the null
- // check necessary?
- group7;
- }
-
- significandString = stripLeadingZeros(significandString);
- signifLength = significandString.length();
-
- /*
- * Adjust exponent as described above
- */
- if (leftDigits >= 1) { // Cases 1 and 2
- exponentAdjust = 4*(leftDigits - 1);
- } else { // Cases 3 and 4
- exponentAdjust = -4*( rightDigits - signifLength + 1);
- }
-
- // If the significand is zero, the exponent doesn't
- // matter; return a properly signed zero.
-
- if (signifLength == 0) { // Only zeros in input
- return new OldFloatingDecimalForTest(sign * 0.0);
- }
- }
-
- // Extract Exponent
- /*
- * Use an int to read in the exponent value; this should
- * provide more than sufficient range for non-contrived
- * inputs. If reading the exponent in as an int does
- * overflow, examine the sign of the exponent and
- * significand to determine what to do.
- */
- String group8 = m.group(8);
- boolean positiveExponent = ( group8 == null ) || group8.equals("+");
- long unsignedRawExponent;
- try {
- unsignedRawExponent = Integer.parseInt(m.group(9));
- }
- catch (NumberFormatException e) {
- // At this point, we know the exponent is
- // syntactically well-formed as a sequence of
- // digits. Therefore, if an NumberFormatException
- // is thrown, it must be due to overflowing int's
- // range. Also, at this point, we have already
- // checked for a zero significand. Thus the signs
- // of the exponent and significand determine the
- // final result:
- //
- // significand
- // + -
- // exponent + +infinity -infinity
- // - +0.0 -0.0
- return new OldFloatingDecimalForTest(sign * (positiveExponent ?
- Double.POSITIVE_INFINITY : 0.0));
- }
-
- long rawExponent =
- (positiveExponent ? 1L : -1L) * // exponent sign
- unsignedRawExponent; // exponent magnitude
-
- // Calculate partially adjusted exponent
- long exponent = rawExponent + exponentAdjust ;
-
- // Starting copying non-zero bits into proper position in
- // a long; copy explicit bit too; this will be masked
- // later for normal values.
-
- boolean round = false;
- boolean sticky = false;
- int bitsCopied=0;
- int nextShift=0;
- long significand=0L;
- // First iteration is different, since we only copy
- // from the leading significand bit; one more exponent
- // adjust will be needed...
-
- // IMPORTANT: make leadingDigit a long to avoid
- // surprising shift semantics!
- long leadingDigit = getHexDigit(significandString, 0);
-
- /*
- * Left shift the leading digit (53 - (bit position of
- * leading 1 in digit)); this sets the top bit of the
- * significand to 1. The nextShift value is adjusted
- * to take into account the number of bit positions of
- * the leadingDigit actually used. Finally, the
- * exponent is adjusted to normalize the significand
- * as a binary value, not just a hex value.
- */
- if (leadingDigit == 1) {
- significand |= leadingDigit << 52;
- nextShift = 52 - 4;
- /* exponent += 0 */ }
- else if (leadingDigit <= 3) { // [2, 3]
- significand |= leadingDigit << 51;
- nextShift = 52 - 5;
- exponent += 1;
- }
- else if (leadingDigit <= 7) { // [4, 7]
- significand |= leadingDigit << 50;
- nextShift = 52 - 6;
- exponent += 2;
- }
- else if (leadingDigit <= 15) { // [8, f]
- significand |= leadingDigit << 49;
- nextShift = 52 - 7;
- exponent += 3;
- } else {
- throw new AssertionError("Result from digit conversion too large!");
- }
- // The preceding if-else could be replaced by a single
- // code block based on the high-order bit set in
- // leadingDigit. Given leadingOnePosition,
-
- // significand |= leadingDigit << (SIGNIFICAND_WIDTH - leadingOnePosition);
- // nextShift = 52 - (3 + leadingOnePosition);
- // exponent += (leadingOnePosition-1);
-
-
- /*
- * Now the exponent variable is equal to the normalized
- * binary exponent. Code below will make representation
- * adjustments if the exponent is incremented after
- * rounding (includes overflows to infinity) or if the
- * result is subnormal.
- */
-
- // Copy digit into significand until the significand can't
- // hold another full hex digit or there are no more input
- // hex digits.
- int i = 0;
- for(i = 1;
- i < signifLength && nextShift >= 0;
- i++) {
- long currentDigit = getHexDigit(significandString, i);
- significand |= (currentDigit << nextShift);
- nextShift-=4;
- }
-
- // After the above loop, the bulk of the string is copied.
- // Now, we must copy any partial hex digits into the
- // significand AND compute the round bit and start computing
- // sticky bit.
-
- if ( i < signifLength ) { // at least one hex input digit exists
- long currentDigit = getHexDigit(significandString, i);
-
- // from nextShift, figure out how many bits need
- // to be copied, if any
- switch(nextShift) { // must be negative
- case -1:
- // three bits need to be copied in; can
- // set round bit
- significand |= ((currentDigit & 0xEL) >> 1);
- round = (currentDigit & 0x1L) != 0L;
- break;
-
- case -2:
- // two bits need to be copied in; can
- // set round and start sticky
- significand |= ((currentDigit & 0xCL) >> 2);
- round = (currentDigit &0x2L) != 0L;
- sticky = (currentDigit & 0x1L) != 0;
- break;
-
- case -3:
- // one bit needs to be copied in
- significand |= ((currentDigit & 0x8L)>>3);
- // Now set round and start sticky, if possible
- round = (currentDigit &0x4L) != 0L;
- sticky = (currentDigit & 0x3L) != 0;
- break;
-
- case -4:
- // all bits copied into significand; set
- // round and start sticky
- round = ((currentDigit & 0x8L) != 0); // is top bit set?
- // nonzeros in three low order bits?
- sticky = (currentDigit & 0x7L) != 0;
- break;
-
- default:
- throw new AssertionError("Unexpected shift distance remainder.");
- // break;
- }
-
- // Round is set; sticky might be set.
-
- // For the sticky bit, it suffices to check the
- // current digit and test for any nonzero digits in
- // the remaining unprocessed input.
- i++;
- while(i < signifLength && !sticky) {
- currentDigit = getHexDigit(significandString,i);
- sticky = sticky || (currentDigit != 0);
- i++;
- }
-
- }
- // else all of string was seen, round and sticky are
- // correct as false.
-
-
- // Check for overflow and update exponent accordingly.
-
- if (exponent > Double.MAX_EXPONENT) { // Infinite result
- // overflow to properly signed infinity
- return new OldFloatingDecimalForTest(sign * Double.POSITIVE_INFINITY);
- } else { // Finite return value
- if (exponent <= Double.MAX_EXPONENT && // (Usually) normal result
- exponent >= Double.MIN_EXPONENT) {
-
- // The result returned in this block cannot be a
- // zero or subnormal; however after the
- // significand is adjusted from rounding, we could
- // still overflow in infinity.
-
- // AND exponent bits into significand; if the
- // significand is incremented and overflows from
- // rounding, this combination will update the
- // exponent correctly, even in the case of
- // Double.MAX_VALUE overflowing to infinity.
-
- significand = (( (exponent +
- (long)DoubleConsts.EXP_BIAS) <<
- (DoubleConsts.SIGNIFICAND_WIDTH-1))
- & DoubleConsts.EXP_BIT_MASK) |
- (DoubleConsts.SIGNIF_BIT_MASK & significand);
-
- } else { // Subnormal or zero
- // (exponent < Double.MIN_EXPONENT)
-
- if (exponent < (DoubleConsts.MIN_SUB_EXPONENT -1 )) {
- // No way to round back to nonzero value
- // regardless of significand if the exponent is
- // less than -1075.
- return new OldFloatingDecimalForTest(sign * 0.0);
- } else { // -1075 <= exponent <= MIN_EXPONENT -1 = -1023
- /*
- * Find bit position to round to; recompute
- * round and sticky bits, and shift
- * significand right appropriately.
- */
-
- sticky = sticky || round;
- round = false;
-
- // Number of bits of significand to preserve is
- // exponent - abs_min_exp +1
- // check:
- // -1075 +1074 + 1 = 0
- // -1023 +1074 + 1 = 52
-
- int bitsDiscarded = 53 -
- ((int)exponent - DoubleConsts.MIN_SUB_EXPONENT + 1);
- assert bitsDiscarded >= 1 && bitsDiscarded <= 53;
-
- // What to do here:
- // First, isolate the new round bit
- round = (significand & (1L << (bitsDiscarded -1))) != 0L;
- if (bitsDiscarded > 1) {
- // create mask to update sticky bits; low
- // order bitsDiscarded bits should be 1
- long mask = ~((~0L) << (bitsDiscarded -1));
- sticky = sticky || ((significand & mask) != 0L ) ;
- }
-
- // Now, discard the bits
- significand = significand >> bitsDiscarded;
-
- significand = (( ((long)(Double.MIN_EXPONENT -1) + // subnorm exp.
- (long)DoubleConsts.EXP_BIAS) <<
- (DoubleConsts.SIGNIFICAND_WIDTH-1))
- & DoubleConsts.EXP_BIT_MASK) |
- (DoubleConsts.SIGNIF_BIT_MASK & significand);
- }
- }
-
- // The significand variable now contains the currently
- // appropriate exponent bits too.
-
- /*
- * Determine if significand should be incremented;
- * making this determination depends on the least
- * significant bit and the round and sticky bits.
- *
- * Round to nearest even rounding table, adapted from
- * table 4.7 in "Computer Arithmetic" by IsraelKoren.
- * The digit to the left of the "decimal" point is the
- * least significant bit, the digits to the right of
- * the point are the round and sticky bits
- *
- * Number Round(x)
- * x0.00 x0.
- * x0.01 x0.
- * x0.10 x0.
- * x0.11 x1. = x0. +1
- * x1.00 x1.
- * x1.01 x1.
- * x1.10 x1. + 1
- * x1.11 x1. + 1
- */
- boolean incremented = false;
- boolean leastZero = ((significand & 1L) == 0L);
- if( ( leastZero && round && sticky ) ||
- ((!leastZero) && round )) {
- incremented = true;
- significand++;
- }
-
- OldFloatingDecimalForTest fd = new OldFloatingDecimalForTest(Math.copySign(
- Double.longBitsToDouble(significand),
- sign));
-
- /*
- * Set roundingDir variable field of fd properly so
- * that the input string can be properly rounded to a
- * float value. There are two cases to consider:
- *
- * 1. rounding to double discards sticky bit
- * information that would change the result of a float
- * rounding (near halfway case between two floats)
- *
- * 2. rounding to double rounds up when rounding up
- * would not occur when rounding to float.
- *
- * For former case only needs to be considered when
- * the bits rounded away when casting to float are all
- * zero; otherwise, float round bit is properly set
- * and sticky will already be true.
- *
- * The lower exponent bound for the code below is the
- * minimum (normalized) subnormal exponent - 1 since a
- * value with that exponent can round up to the
- * minimum subnormal value and the sticky bit
- * information must be preserved (i.e. case 1).
- */
- if ((exponent >= FloatConsts.MIN_SUB_EXPONENT-1) &&
- (exponent <= Float.MAX_EXPONENT ) ){
- // Outside above exponent range, the float value
- // will be zero or infinity.
-
- /*
- * If the low-order 28 bits of a rounded double
- * significand are 0, the double could be a
- * half-way case for a rounding to float. If the
- * double value is a half-way case, the double
- * significand may have to be modified to round
- * the the right float value (see the stickyRound
- * method). If the rounding to double has lost
- * what would be float sticky bit information, the
- * double significand must be incremented. If the
- * double value's significand was itself
- * incremented, the float value may end up too
- * large so the increment should be undone.
- */
- if ((significand & 0xfffffffL) == 0x0L) {
- // For negative values, the sign of the
- // roundDir is the same as for positive values
- // since adding 1 increasing the significand's
- // magnitude and subtracting 1 decreases the
- // significand's magnitude. If neither round
- // nor sticky is true, the double value is
- // exact and no adjustment is required for a
- // proper float rounding.
- if( round || sticky) {
- if (leastZero) { // prerounding lsb is 0
- // If round and sticky were both true,
- // and the least significant
- // significand bit were 0, the rounded
- // significand would not have its
- // low-order bits be zero. Therefore,
- // we only need to adjust the
- // significand if round XOR sticky is
- // true.
- if (round ^ sticky) {
- fd.roundDir = 1;
- }
- }
- else { // prerounding lsb is 1
- // If the prerounding lsb is 1 and the
- // resulting significand has its
- // low-order bits zero, the significand
- // was incremented. Here, we undo the
- // increment, which will ensure the
- // right guard and sticky bits for the
- // float rounding.
- if (round)
- fd.roundDir = -1;
- }
- }
- }
- }
-
- fd.fromHex = true;
- return fd;
- }
- }
- }
-
- /**
- * Return <code>s</code> with any leading zeros removed.
- */
- static String stripLeadingZeros(String s) {
- return s.replaceFirst("^0+", "");
- }
-
- /**
- * Extract a hexadecimal digit from position <code>position</code>
- * of string <code>s</code>.
- */
- static int getHexDigit(String s, int position) {
- int value = Character.digit(s.charAt(position), 16);
- if (value <= -1 || value >= 16) {
- throw new AssertionError("Unexpected failure of digit conversion of " +
- s.charAt(position));
- }
- return value;
- }
-
-
-}