/java-1.7.0-openjdk/openjdk/jdk/src/share/classes/sun/misc/FloatingDecimal.java
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Possible License(s): GPL-2.0, BSD-3-Clause-No-Nuclear-License-2014, LGPL-3.0, LGPL-2.0
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- /*
- * Copyright (c) 1996, 2011, 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. Oracle designates this
- * particular file as subject to the "Classpath" exception as provided
- * by Oracle in the LICENSE file that accompanied this code.
- *
- * 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 sun.misc.FpUtils;
- import sun.misc.DoubleConsts;
- import sun.misc.FloatConsts;
- import java.util.regex.*;
- public class FloatingDecimal{
- 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
- 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 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 FDBigInt b5p[];
- private static synchronized FDBigInt
- big5pow( int p ){
- assert p >= 0 : p; // 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 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;
- 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 ) );
- }
- }
- }
- //
- // a common operation
- //
- private static FDBigInt
- multPow52( FDBigInt 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 FDBigInt
- constructPow52( int p5, int p2 ){
- FDBigInt v = new FDBigInt( big5pow( p5 ) );
- if ( p2 != 0 ){
- v.lshiftMe( p2 );
- }
- return v;
- }
- /*
- * Make a floating double into a FDBigInt.
- * This could also be structured as a FDBigInt
- * 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 FDBigInt
- 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 FDBigInt( 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 = (char[])(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 = (char[])(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);
- }
- /*
- * 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 +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 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 +1; // recall binExp is - shift count.
- binExp += 1;
- } else {
- fractBits |= singleFractHOB;
- nSignificantBits = singleExpShift+1;
- }
- binExp -= singleExpBias;
- // call the routine that actually does all the hard work.
- dtoa( binExp, ((long)fractBits)<<(expShift-singleExpShift), nSignificantBits );
- }
- private void
- dtoa( int binExp, long fractBits, int nSignificantBits )
- {
- int nFractBits; // number of significant bits of fractBits;
- int nTinyBits; // number of these to the right of the point.
- int decExp;
- // Examine number. Determine if it is an easy case,
- // which we can do pretty trivially using float/long conversion,
- // or whether we must do real work.
- nFractBits = countBits( fractBits );
- nTinyBits = Math.max( 0, nFractBits - binExp - 1 );
- if ( binExp <= maxSmallBinExp && binExp >= minSmallBinExp ){
- // Look more closely at the number to decide if,
- // with scaling by 10^nTinyBits, the result will fit in
- // a long.
- if ( (nTinyBits < long5pow.length) && ((nFractBits + n5bits[nTinyBits]) < 64 ) ){
- /*
- * We can do this:
- * take the fraction bits, which are normalized.
- * (a) nTinyBits == 0: Shift left or right appropriately
- * to align the binary point at the extreme right, i.e.
- * where a long int point is expected to be. The integer
- * result is easily converted to a string.
- * (b) nTinyBits > 0: Shift right by expShift-nFractBits,
- * which effectively converts to long and scales by
- * 2^nTinyBits. Then multiply by 5^nTinyBits to
- * complete the scaling. We know this won't overflow
- * because we just counted the number of bits necessary
- * in the result. The integer you get from this can
- * then be converted to a string pretty easily.
- */
- long halfULP;
- if ( nTinyBits == 0 ) {
- if ( binExp > nSignificantBits ){
- halfULP = 1L << ( binExp-nSignificantBits-1);
- } else {
- halfULP = 0L;
- }
- if ( binExp >= expShift ){
- fractBits <<= (binExp-expShift);
- } else {
- fractBits >>>= (expShift-binExp) ;
- }
- developLongDigits( 0, fractBits, halfULP );
- return;
- }
- /*
- * The following causes excess digits to be printed
- * out in the single-float case. Our manipulation of
- * halfULP here is apparently not correct. If we
- * better understand how this works, perhaps we can
- * use this special case again. But for the time being,
- * we do not.
- * else {
- * fractBits >>>= expShift+1-nFractBits;
- * fractBits *= long5pow[ nTinyBits ];
- * halfULP = long5pow[ nTinyBits ] >> (1+nSignificantBits-nFractBits);
- * developLongDigits( -nTinyBits, fractBits, halfULP );
- * return;
- * }
- */
- }
- }
- /*
- * This is the hard case. We are going to compute large positive
- * integers B and S and integer decExp, s.t.
- * d = ( B / S ) * 10^decExp
- * 1 <= B / S < 10
- * Obvious choices are:
- * decExp = floor( log10(d) )
- * B = d * 2^nTinyBits * 10^max( 0, -decExp )
- * S = 10^max( 0, decExp) * 2^nTinyBits
- * (noting that nTinyBits has already been forced to non-negative)
- * I am also going to compute a large positive integer
- * M = (1/2^nSignificantBits) * 2^nTinyBits * 10^max( 0, -decExp )
- * i.e. M is (1/2) of the ULP of d, scaled like B.
- * When we iterate through dividing B/S and picking off the
- * quotient bits, we will know when to stop when the remainder
- * is <= M.
- *
- * We keep track of powers of 2 and powers of 5.
- */
- /*
- * Estimate decimal exponent. (If it is small-ish,
- * we could double-check.)
- *
- * First, scale the mantissa bits such that 1 <= d2 < 2.
- * We are then going to estimate
- * log10(d2) ~=~ (d2-1.5)/1.5 + log(1.5)
- * and so we can estimate
- * log10(d) ~=~ log10(d2) + binExp * log10(2)
- * take the floor and call it decExp.
- * FIXME -- use more precise constants here. It costs no more.
- */
- double d2 = Double.longBitsToDouble(
- expOne | ( fractBits &~ fractHOB ) );
- decExp = (int)Math.floor(
- (d2-1.5D)*0.289529654D + 0.176091259 + (double)binExp * 0.301029995663981 );
- int B2, B5; // powers of 2 and powers of 5, respectively, in B
- int S2, S5; // powers of 2 and powers of 5, respectively, in S
- int M2, M5; // powers of 2 and powers of 5, respectively, in M
- int Bbits; // binary digits needed to represent B, approx.
- int tenSbits; // binary digits needed to represent 10*S, approx.
- FDBigInt Sval, Bval, Mval;
- B5 = Math.max( 0, -decExp );
- B2 = B5 + nTinyBits + binExp;
- S5 = Math.max( 0, decExp );
- S2 = S5 + nTinyBits;
- M5 = B5;
- M2 = B2 - nSignificantBits;
- /*
- * the long integer fractBits contains the (nFractBits) interesting
- * bits from the mantissa of d ( hidden 1 added if necessary) followed
- * by (expShift+1-nFractBits) zeros. In the interest of compactness,
- * I will shift out those zeros before turning fractBits into a
- * FDBigInt. The resulting whole number will be
- * d * 2^(nFractBits-1-binExp).
- */
- fractBits >>>= (expShift+1-nFractBits);
- B2 -= nFractBits-1;
- int common2factor = Math.min( B2, S2 );
- B2 -= common2factor;
- S2 -= common2factor;
- M2 -= common2factor;
- /*
- * HACK!! For exact powers of two, the next smallest number
- * is only half as far away as we think (because the meaning of
- * ULP changes at power-of-two bounds) for this reason, we
- * hack M2. Hope this works.
- */
- if ( nFractBits == 1 )
- M2 -= 1;
- if ( M2 < 0 ){
- // oops.
- // since we cannot scale M down far enough,
- // we must scale the other values up.
- B2 -= M2;
- S2 -= M2;
- M2 = 0;
- }
- /*
- * Construct, Scale, iterate.
- * Some day, we'll write a stopping test that takes
- * account of the asymmetry of the spacing of floating-point
- * numbers below perfect powers of 2
- * 26 Sept 96 is not that day.
- * So we use a symmetric test.
- */
- char digits[] = this.digits = new char[18];
- int ndigit = 0;
- boolean low, high;
- long lowDigitDifference;
- int q;
- /*
- * Detect the special cases where all the numbers we are about
- * to compute will fit in int or long integers.
- * In these cases, we will avoid doing FDBigInt arithmetic.
- * We use the same algorithms, except that we "normalize"
- * our FDBigInts before iterating. This is to make division easier,
- * as it makes our fist guess (quotient of high-order words)
- * more accurate!
- *
- * Some day, we'll write a stopping test that takes
- * account of the asymmetry of the spacing of floating-point
- * numbers below perfect powers of 2
- * 26 Sept 96 is not that day.
- * So we use a symmetric test.
- */
- Bbits = nFractBits + B2 + (( B5 < n5bits.length )? n5bits[B5] : ( B5*3 ));
- tenSbits = S2+1 + (( (S5+1) < n5bits.length )? n5bits[(S5+1)] : ( (S5+1)*3 ));
- if ( Bbits < 64 && tenSbits < 64){
- if ( Bbits < 32 && tenSbits < 32){
- // wa-hoo! They're all ints!
- int b = ((int)fractBits * small5pow[B5] ) << B2;
- int s = small5pow[S5] << S2;
- int m = small5pow[M5] << M2;
- int tens = s * 10;
- /*
- * Unroll the first iteration. If our decExp estimate
- * was too high, our first quotient will be zero. In this
- * case, we discard it and decrement decExp.
- */
- ndigit = 0;
- q = b / s;
- b = 10 * ( b % s );
- m *= 10;
- low = (b < m );
- high = (b+m > tens );
- assert q < 10 : q; // excessively large digit
- if ( (q == 0) && ! high ){
- // oops. Usually ignore leading zero.
- decExp--;
- } else {
- digits[ndigit++] = (char)('0' + q);
- }
- /*
- * HACK! Java spec sez that we always have at least
- * one digit after the . in either F- or E-form output.
- * Thus we will need more than one digit if we're using
- * E-form
- */
- if ( decExp < -3 || decExp >= 8 ){
- high = low = false;
- }
- while( ! low && ! high ){
- q = b / s;
- b = 10 * ( b % s );
- m *= 10;
- assert q < 10 : q; // excessively large digit
- if ( m > 0L ){
- low = (b < m );
- high = (b+m > tens );
- } else {
- // hack -- m might overflow!
- // in this case, it is certainly > b,
- // which won't
- // and b+m > tens, too, since that has overflowed
- // either!
- low = true;
- high = true;
- }
- digits[ndigit++] = (char)('0' + q);
- }
- lowDigitDifference = (b<<1) - tens;
- } else {
- // still good! they're all longs!
- long b = (fractBits * long5pow[B5] ) << B2;
- long s = long5pow[S5] << S2;
- long m = long5pow[M5] << M2;
- long tens = s * 10L;
- /*
- * Unroll the first iteration. If our decExp estimate
- * was too high, our first quotient will be zero. In this
- * case, we discard it and decrement decExp.
- */
- ndigit = 0;
- q = (int) ( b / s );
- b = 10L * ( b % s );
- m *= 10L;
- low = (b < m );
- high = (b+m > tens );
- assert q < 10 : q; // excessively large digit
- if ( (q == 0) && ! high ){
- // oops. Usually ignore leading zero.
- decExp--;
- } else {
- digits[ndigit++] = (char)('0' + q);
- }
- /*
- * HACK! Java spec sez that we always have at least
- * one digit after the . in either F- or E-form output.
- * Thus we will need more than one digit if we're using
- * E-form
- */
- if ( decExp < -3 || decExp >= 8 ){
- high = low = false;
- }
- while( ! low && ! high ){
- q = (int) ( b / s );
- b = 10 * ( b % s );
- m *= 10;
- assert q < 10 : q; // excessively large digit
- if ( m > 0L ){
- low = (b < m );
- high = (b+m > tens );
- } else {
- // hack -- m might overflow!
- // in this case, it is certainly > b,
- // which won't
- // and b+m > tens, too, since that has overflowed
- // either!
- low = true;
- high = true;
- }
- digits[ndigit++] = (char)('0' + q);
- }
- lowDigitDifference = (b<<1) - tens;
- }
- } else {
- FDBigInt tenSval;
- int shiftBias;
- /*
- * We really must do FDBigInt arithmetic.
- * Fist, construct our FDBigInt initial values.
- */
- Bval = multPow52( new FDBigInt( fractBits ), B5, B2 );
- Sval = constructPow52( S5, S2 );
- Mval = constructPow52( M5, M2 );
- // normalize so that division works better
- Bval.lshiftMe( shiftBias = Sval.normalizeMe() );
- Mval.lshiftMe( shiftBias );
- tenSval = Sval.mult( 10 );
- /*
- * Unroll the first iteration. If our decExp estimate
- * was too high, our first quotient will be zero. In this
- * case, we discard it and decrement decExp.
- */
- ndigit = 0;
- q = Bval.quoRemIteration( Sval );
- Mval = Mval.mult( 10 );
- low = (Bval.cmp( Mval ) < 0);
- high = (Bval.add( Mval ).cmp( tenSval ) > 0 );
- assert q < 10 : q; // excessively large digit
- if ( (q == 0) && ! high ){
- // oops. Usually ignore leading zero.
- decExp--;
- } else {
- digits[ndigit++] = (char)('0' + q);
- }
- /*
- * HACK! Java spec sez that we always have at least
- * one digit after the . in either F- or E-form output.
- * Thus we will need more than one digit if we're using
- * E-form
- */
- if ( decExp < -3 || decExp >= 8 ){
- high = low = false;
- }
- while( ! low && ! high ){
- q = Bval.quoRemIteration( Sval );
- Mval = Mval.mult( 10 );
- assert q < 10 : q; // excessively large digit
- low = (Bval.cmp( Mval ) < 0);
- high = (Bval.add( Mval ).cmp( tenSval ) > 0 );
- digits[ndigit++] = (char)('0' + q);
- }
- if ( high && low ){
- Bval.lshiftMe(1);
- lowDigitDifference = Bval.cmp(tenSval);
- } else
- lowDigitDifference = 0L; // this here only for flow analysis!
- }
- this.decExponent = decExp+1;
- this.digits = digits;
- this.nDigits = ndigit;
- /*
- * Last digit gets rounded based on stopping condition.
- */
- if ( high ){
- if ( low ){
- if ( lowDigitDifference == 0L ){
- // it's a tie!
- // choose based on which digits we like.
- if ( (digits[nDigits-1]&1) != 0 ) roundup();
- } else if ( lowDigitDifference > 0 ){
- roundup();
- }
- } else {
- roundup();
- }
- }
- }
- public String
- toString(){
- // most brain-dead version
- StringBuffer result = new StringBuffer( nDigits+8 );
- if ( isNegative ){ result.append( '-' ); }
- if ( isExceptional ){
- result.append( digits, 0, nDigits );
- } else {
- result.append( "0.");
- result.append( digits, 0, nDigits );
- result.append('e');
- result.append( decExponent );
- }
- return new String(result);
- }
- public String toJavaFormatString() {
- char result[] = (char[])(perThreadBuffer.get());
- int i = getChars(result);
- return new String(result, 0, i);
- }
- private int getChars(char[] result) {
- assert nDigits <= 19 : nDigits; // generous bound on size of nDigits
- int i = 0;
- if (isNegative) { result[0] = '-'; i = 1; }
- if (isExceptional) {
- System.arraycopy(digits, 0, result, i, nDigits);
- i += nDigits;
- } else {
- if (decExponent > 0 && decExponent < 8) {
- // print digits.digits.
- int charLength = Math.min(nDigits, decExponent);
- System.arraycopy(digits, 0, result, i, charLength);
- i += charLength;
- if (charLength < decExponent) {
- charLength = decExponent-charLength;
- System.arraycopy(zero, 0, result, i, charLength);
- i += charLength;
- result[i++] = '.';
- result[i++] = '0';
- } else {
- result[i++] = '.';
- if (charLength < nDigits) {
- int t = nDigits - charLength;
- System.arraycopy(digits, charLength, result, i, t);
- i += t;
- } else {
- result[i++] = '0';
- }
- }
- } else if (decExponent <=0 && decExponent > -3) {
- result[i++] = '0';
- result[i++] = '.';
- if (decExponent != 0) {
- System.arraycopy(zero, 0, result, i, -decExponent);
- i -= decExponent;
- }
- System.arraycopy(digits, 0, result, i, nDigits);
- i += nDigits;
- } else {
- result[i++] = digits[0];
- result[i++] = '.';
- if (nDigits > 1) {
- System.arraycopy(digits, 1, result, i, nDigits-1);
- i += nDigits-1;
- } else {
- result[i++] = '0';
- }
- result[i++] = 'E';
- int e;
- if (decExponent <= 0) {
- result[i++] = '-';
- e = -decExponent+1;
- } else {
- e = decExponent-1;
- }
- // decExponent has 1, 2, or 3, digits
- if (e <= 9) {
- result[i++] = (char)(e+'0');
- } else if (e <= 99) {
- result[i++] = (char)(e/10 +'0');
- result[i++] = (char)(e%10 + '0');
- } else {
- result[i++] = (char)(e/100+'0');
- e %= 100;
- result[i++] = (char)(e/10+'0');
- result[i++] = (char)(e%10 + '0');
- }
- }
- }
- return i;
- }
- // Per-thread buffer for string/stringbuffer conversion
- private static ThreadLocal perThreadBuffer = new ThreadLocal() {
- protected synchronized Object initialValue() {
- return new char[26];
- }
- };
- public void appendTo(Appendable buf) {
- char result[] = (char[])(perThreadBuffer.get());
- int i = getChars(result);
- if (buf instanceof StringBuilder)
- ((StringBuilder) buf).append(result, 0, i);
- else if (buf instanceof StringBuffer)
- ((StringBuffer) buf).append(result, 0, i);
- else
- assert false;
- }
- public static FloatingDecimal
- readJavaFormatString( String in ) throws NumberFormatException {
- 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 FloatingDecimal(Double.NaN) // NaN has no sign
- : new FloatingDecimal(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 FloatingDecimal( 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.
- …
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