How to alter double by its smallest increment

问题

Is something broken or I fail to understand what is happening?

static String getRealBinary(double val) {
    long tmp = Double.doubleToLongBits(val);
    StringBuilder sb = new StringBuilder();

    for (long n = 64; --n > 0; tmp >>= 1)
        if ((tmp & 1) == 0)
            sb.insert(0, ('0'));
        else
            sb.insert(0, ('1'));

    sb.insert(0, '[').insert(2, "] [").insert(16, "] [").append(']');
    return sb.toString();
}

public static void main(String[] argv) {
    for (int j = 3; --j >= 0;) {
        double d = j;
        for (int i = 3; --i >= 0;) {
            d += Double.MIN_VALUE;
            System.out.println(d +getRealBinary(d));
        }
    }
}

With output:

2.0[1] [00000000000] [000000000000000000000000000000000000000000000000000]
2.0[1] [00000000000] [000000000000000000000000000000000000000000000000000]
2.0[1] [00000000000] [000000000000000000000000000000000000000000000000000]
1.0[0] [11111111110] [000000000000000000000000000000000000000000000000000]
1.0[0] [11111111110] [000000000000000000000000000000000000000000000000000]
1.0[0] [11111111110] [000000000000000000000000000000000000000000000000000]
4.9E-324[0] [00000000000] [000000000000000000000000000000000000000000000000001]
1.0E-323[0] [00000000000] [000000000000000000000000000000000000000000000000010]
1.5E-323[0] [00000000000] [000000000000000000000000000000000000000000000000011]

回答1:

The general idea is first convert the double to its long representation (using doubleToLongBits as you have done in getRealBinary), increment that long by 1, and finally convert the new long back to the double it represents via longBitsToDouble.

EDIT: Java (since 1.5) provides Math.ulp(double), which I'm guessing you can use to compute the next higher value directly thus: x + Math.ulp(x).

回答2:

Floating point numbers are not spread out uniformly over the number line like integer types are. They are more densely packed near 0 and very far apart as you approach infinity. Therefore there is no constant that you can add to a floating point number to get to the next floating point number.

回答3:

Your code is not well-formed. You try to add the minimum double value and expect the result to be different from the original value. The problem is that double.MinValue is so small that the result is rounded and doesn't get affected.

Suggested reading: http://en.wikipedia.org/wiki/Machine_epsilon

On the Wikipedia article there is the Java code too. Epsilon is by definition the smallest number such as (X + eps * X != X), and eps*X is called "relative-epsilon"

回答4:

Since Java 1.8 there is java.lang.Math.nextUp(double) doing exactly what you want. There is also opposite java.lang.Math.nextDown(double).

回答5:

In case you want to use the BigDecimal class, there is the BigDecimal.ulp() method as well.

来源：https://stackoverflow.com/questions/3900440/how-to-alter-double-by-its-smallest-increment

标签

java

double

ieee-754