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jdk-24/jdk/test/sun/misc/FloatingDecimal/OldFloatingDecimalForTest.java
Brian Burkhalter a1fe63768f 7032154: Performance tuning of sun.misc.FloatingDecimal/FormattedFloatingDecimal 
Performance improvements for double/float -> String and decimal/hex String -> double/float conversions.

Co-authored-by: Sergey Kuksenko <sergey.kuksenko@oracle.com>
Co-authored-by: Dmitry Nadezhin <dmitry.nadezhin@oracle.com>
Co-authored-by: Olivier Lagneau <olivier.lagneau@oracle.com>
Reviewed-by: martin, iris
2013-06-05 21:01:02 -07:00

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/*
* 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 sun.misc.DoubleConsts;
import sun.misc.FloatConsts;
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 > DoubleConsts.MAX_EXPONENT) { // Infinite result
// overflow to properly signed infinity
return new OldFloatingDecimalForTest(sign * Double.POSITIVE_INFINITY);
} else { // Finite return value
if (exponent <= DoubleConsts.MAX_EXPONENT && // (Usually) normal result
exponent >= DoubleConsts.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 < DoubleConsts.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)(DoubleConsts.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 <= FloatConsts.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;
}
}
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