Brian,
I removed unused methods and fields from FormattedFloatingDecimal.
JDK build passes.
The result of "hg diff" is attached.
-Dima
On Fri, Feb 22, 2013 at 9:49 PM, Brian Burkhalter
<brian.burkhal...@oracle.com> wrote:
> Dima,
>
> If the methods are definitely unused that would be correct. I suppose if a
> clean build of the JDK does not complain then it is acceptable and correct.
>
> Thanks,
>
> Brian
>
> On Feb 22, 2013, at 9:41 AM, Dmitry Nadezhin wrote:
>
> So I think that the required change in FormattedFloatingDecimal is to
> delete methods
> doubleValue(), floatValue() and other unused methods and fields. Am I right
> ?
>
>
diff -r bb97c93e4fd7 src/share/classes/sun/misc/FormattedFloatingDecimal.java
--- a/src/share/classes/sun/misc/FormattedFloatingDecimal.java Thu Feb 21
11:13:23 2013 -0800
+++ b/src/share/classes/sun/misc/FormattedFloatingDecimal.java Sat Feb 23
00:00:12 2013 +0400
@@ -25,10 +25,6 @@
package sun.misc;
-import sun.misc.DoubleConsts;
-import sun.misc.FloatConsts;
-import java.util.regex.*;
-
public class FormattedFloatingDecimal{
boolean isExceptional;
boolean isNegative;
@@ -38,9 +34,6 @@
int nDigits;
int bigIntExp;
int bigIntNBits;
- boolean mustSetRoundDir = false;
- boolean fromHex = false;
- int roundDir = 0; // set by doubleValue
int precision; // number of digits to the right of decimal
public enum Form { SCIENTIFIC, COMPATIBLE, DECIMAL_FLOAT, GENERAL };
@@ -72,10 +65,6 @@
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;
@@ -87,12 +76,6 @@
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,
@@ -241,56 +224,6 @@
}
/*
- * 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
@@ -1146,541 +1079,6 @@
}
};
- /*
- * Take a FormattedFloatingDecimal, 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 FDBigInt arithmetic.
- * Formulate the EXACT big-number result as
- * bigD0 * 10^exp
- */
- FDBigInt bigD0 = new FDBigInt( lValue, digits, kDigits, nDigits );
- exp = decExponent - nDigits;
-
- correctionLoop:
- while(true){
- /* AS A SIDE EFFECT, THIS METHOD WILL SET THE INSTANCE
VARIABLES
- * bigIntExp and bigIntNBits
- */
- FDBigInt bigB = doubleToBigInt( dValue );
-
- /*
- * Scale bigD, bigB appropriately for
- * big-integer operations.
- * Naively, we multipy 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 );
- FDBigInt bigD = multPow52( new FDBigInt( 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.
- //
- FDBigInt 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) ){
- // 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 fequency
- break correctionLoop;
- }
- FDBigInt 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 FormattedFloatingDecimal, 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 converstion
- * 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
- * algorythm 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,
