Java Code Examples for sun.misc.DoubleConsts#SIGN_BIT_MASK
The following examples show how to use
sun.misc.DoubleConsts#SIGN_BIT_MASK .
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Example 1
| Source File: BigInteger.java From jdk1.8-source-analysis with Apache License 2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 2
| Source File: BigInteger.java From dragonwell8_jdk with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 3
| Source File: BigInteger.java From TencentKona-8 with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 4
| Source File: BigInteger.java From jdk8u60 with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 5
| Source File: BigInteger.java From JDKSourceCode1.8 with MIT License | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 6
| Source File: BigInteger.java From openjdk-jdk8u with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 7
| Source File: BigInteger.java From openjdk-jdk8u-backup with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 8
| Source File: BigInteger.java From jdk8u-jdk with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 9
| Source File: BigInteger.java From Java8CN with Apache License 2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 10
| Source File: BigInteger.java From hottub with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 11
| Source File: BigInteger.java From openjdk-8-source with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 12
| Source File: BigInteger.java From openjdk-8 with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 13
| Source File: BigInteger.java From jdk8u_jdk with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 14
| Source File: BigInteger.java From jdk8u-jdk with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 15
| Source File: BigInteger.java From jdk8u-dev-jdk with GNU General Public License v2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
Example 16
| Source File: BigInteger.java From j2objc with Apache License 2.0 | 4 votes |
|
/**
* Converts this BigInteger to a {@code double}. This
* conversion is similar to the
* <i>narrowing primitive conversion</i> from {@code double} to
* {@code float} as defined in section 5.1.3 of
* <cite>The Java™ Language Specification</cite>:
* if this BigInteger has too great a magnitude
* to represent as a {@code double}, it will be converted to
* {@link Double#NEGATIVE_INFINITY} or {@link
* Double#POSITIVE_INFINITY} as appropriate. Note that even when
* the return value is finite, this conversion can lose
* information about the precision of the BigInteger value.
*
* @return this BigInteger converted to a {@code double}.
*/
public double doubleValue() {
if (signum == 0) {
return 0.0;
}
int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1;
// exponent == floor(log2(abs(this))Double)
if (exponent < Long.SIZE - 1) {
return longValue();
} else if (exponent > Double.MAX_EXPONENT) {
return signum > 0 ? Double.POSITIVE_INFINITY : Double.NEGATIVE_INFINITY;
}
/*
* We need the top SIGNIFICAND_WIDTH bits, including the "implicit"
* one bit. To make rounding easier, we pick out the top
* SIGNIFICAND_WIDTH + 1 bits, so we have one to help us round up or
* down. twiceSignifFloor will contain the top SIGNIFICAND_WIDTH + 1
* bits, and signifFloor the top SIGNIFICAND_WIDTH.
*
* It helps to consider the real number signif = abs(this) *
* 2^(SIGNIFICAND_WIDTH - 1 - exponent).
*/
int shift = exponent - DoubleConsts.SIGNIFICAND_WIDTH;
long twiceSignifFloor;
// twiceSignifFloor will be == abs().shiftRight(shift).longValue()
// We do the shift into a long directly to improve performance.
int nBits = shift & 0x1f;
int nBits2 = 32 - nBits;
int highBits;
int lowBits;
if (nBits == 0) {
highBits = mag[0];
lowBits = mag[1];
} else {
highBits = mag[0] >>> nBits;
lowBits = (mag[0] << nBits2) | (mag[1] >>> nBits);
if (highBits == 0) {
highBits = lowBits;
lowBits = (mag[1] << nBits2) | (mag[2] >>> nBits);
}
}
twiceSignifFloor = ((highBits & LONG_MASK) << 32)
| (lowBits & LONG_MASK);
long signifFloor = twiceSignifFloor >> 1;
signifFloor &= DoubleConsts.SIGNIF_BIT_MASK; // remove the implied bit
/*
* We round up if either the fractional part of signif is strictly
* greater than 0.5 (which is true if the 0.5 bit is set and any lower
* bit is set), or if the fractional part of signif is >= 0.5 and
* signifFloor is odd (which is true if both the 0.5 bit and the 1 bit
* are set). This is equivalent to the desired HALF_EVEN rounding.
*/
boolean increment = (twiceSignifFloor & 1) != 0
&& ((signifFloor & 1) != 0 || abs().getLowestSetBit() < shift);
long signifRounded = increment ? signifFloor + 1 : signifFloor;
long bits = (long) ((exponent + DoubleConsts.EXP_BIAS))
<< (DoubleConsts.SIGNIFICAND_WIDTH - 1);
bits += signifRounded;
/*
* If signifRounded == 2^53, we'd need to set all of the significand
* bits to zero and add 1 to the exponent. This is exactly the behavior
* we get from just adding signifRounded to bits directly. If the
* exponent is Double.MAX_EXPONENT, we round up (correctly) to
* Double.POSITIVE_INFINITY.
*/
bits |= signum & DoubleConsts.SIGN_BIT_MASK;
return Double.longBitsToDouble(bits);
}
