--- old/src/share/classes/java/math/BigInteger.java 2013-06-21 11:52:07.000000000 -0700 +++ new/src/share/classes/java/math/BigInteger.java 2013-06-21 11:52:06.000000000 -0700 @@ -35,6 +35,8 @@ import java.io.ObjectStreamField; import java.util.Arrays; import java.util.Random; +import sun.misc.DoubleConsts; +import sun.misc.FloatConsts; /** * Immutable arbitrary-precision integers. All operations behave as if @@ -3452,8 +3454,72 @@ * @return this BigInteger converted to a {@code float}. */ public float floatValue() { - // Somewhat inefficient, but guaranteed to work. - return Float.parseFloat(this.toString()); + if (signum == 0) { + return 0.0f; + } + + int exponent = ((mag.length - 1) << 5) + bitLengthForInt(mag[0]) - 1; + + // exponent == floor(log2(abs(this))) + if (exponent < Long.SIZE - 1) { + return longValue(); + } else if (exponent > Float.MAX_EXPONENT) { + return signum > 0 ? Float.POSITIVE_INFINITY : Float.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 - FloatConsts.SIGNIFICAND_WIDTH; + + int twiceSignifFloor; + // twiceSignifFloor will be == abs().shiftRight(shift).intValue() + // We do the shift into an int directly to improve performance. + + int nBits = shift & 0x1f; + int nBits2 = 32 - nBits; + + if (nBits == 0) { + twiceSignifFloor = mag[0]; + } else { + twiceSignifFloor = mag[0] >>> nBits; + if (twiceSignifFloor == 0) { + twiceSignifFloor = (mag[0] << nBits2) | (mag[1] >>> nBits); + } + } + + int signifFloor = twiceSignifFloor >> 1; + signifFloor &= FloatConsts.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); + int signifRounded = increment ? signifFloor + 1 : signifFloor; + int bits = ((exponent + FloatConsts.EXP_BIAS)) + << (FloatConsts.SIGNIFICAND_WIDTH - 1); + bits += signifRounded; + /* + * If signifRounded == 2^24, 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 Float.MAX_EXPONENT, we round up (correctly) to + * Float.POSITIVE_INFINITY. + */ + bits |= signum & FloatConsts.SIGN_BIT_MASK; + return Float.intBitsToFloat(bits); } /** @@ -3472,8 +3538,80 @@ * @return this BigInteger converted to a {@code double}. */ public double doubleValue() { - // Somewhat inefficient, but guaranteed to work. - return Double.parseDouble(this.toString()); + 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); } /**