/**
* Returns the value of this fraction. If this value is beyond the range of a double,
* Double.INFINITY or Double.NEGATIVE_INFINITY will be returned.
*/
@Override
public double doubleValue() {
// note: must use precision+2 so that new BigFraction(d).doubleValue()
// == d,
// for all possible double values.
return this.toBigDecimal(MathContext.DECIMAL64.getPrecision() + 2).doubleValue();
}
/**
* Returns a BigDecimal representation of this fraction. If possible, the returned value will be
* exactly equal to the fraction. If not, the BigDecimal will have a scale large enough to hold
* the same number of significant figures as both numerator and denominator, or the equivalent of
* a double-precision number, whichever is more.
*/
public BigDecimal toBigDecimal() {
// Implementation note: A fraction can be represented exactly in base-10
// iff its
// denominator is of the form 2^a * 5^b, where a and b are nonnegative
// integers.
// (In other words, if there are no prime factors of the denominator
// except for
// 2 and 5, or if the denominator is 1). So to determine if this
// denominator is
// of this form, continually divide by 2 to get the number of 2's, and
// then
// continually divide by 5 to get the number of 5's. Afterward, if the
// denominator
// is 1 then there are no other prime factors.
// Note: number of 2's is given by the number of trailing 0 bits in the
// number
final int twos = this.denominator.getLowestSetBit();
BigInteger tmpDen = this.denominator.shiftRight(twos); // x / 2^n === x >> n
int fives = 0;
BigInteger[] divMod = null;
// while(tmpDen % 5 == 0) { tmpDen /= 5; fives++; }
while (BigInteger.ZERO.equals((divMod = tmpDen.divideAndRemainder(BIGINT_FIVE))[1])) {
tmpDen = divMod[0];
fives++;
}
if (BigInteger.ONE.equals(tmpDen)) {
// This fraction will terminate in base 10, so it can be represented
// exactly as
// a BigDecimal. We would now like to make the fraction of the form
// unscaled / 10^scale. We know that 2^x * 5^x = 10^x, and our
// denominator is
// in the form 2^twos * 5^fives. So use max(twos, fives) as the
// scale, and
// multiply the numerator and deminator by the appropriate number of
// 2's or 5's
// such that the denominator is of the form 2^scale * 5^scale. (Of
// course, we
// only have to actually multiply the numerator, since all we need
// for the
// BigDecimal constructor is the scale.)
BigInteger unscaled = this.numerator;
final int scale = Math.max(twos, fives);
if (twos < fives) {
unscaled = unscaled.shiftLeft(fives - twos); // x
} else if (fives < twos) {
unscaled = unscaled.multiply(BIGINT_FIVE.pow(twos - fives));
}
return new BigDecimal(unscaled, scale);
}
// else: this number will repeat infinitely in base-10. So try to figure
// out
// a good number of significant digits. Start with the number of digits
// required
// to represent the numerator and denominator in base-10, which is given
// by
// bitLength / log[2](10). (bitLenth is the number of digits in base-2).
final double LG10 = 3.321928094887362; // Precomputed ln(10)/ln(2),
// a.k.a. log[2](10)
int precision = Math.max(this.numerator.bitLength(), this.denominator.bitLength());
precision = (int) Math.ceil(precision / LG10);
// If the precision is less than that of a double, use double-precision
// so
// that the result will be at least as accurate as a cast to a double.
// For
// example, with the fraction 1/3, precision will be 1, giving a result
// of
// 0.3. This is quite a bit different from what a user would expect.
if (precision < MathContext.DECIMAL64.getPrecision() + 2) {
precision = MathContext.DECIMAL64.getPrecision() + 2;
}
return this.toBigDecimal(precision);
}