CODEWITHSUNDEEP
javadoubleieee-754
Margus
Margus

Reputation: 20038

How to alter double by its smallest increment

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]

Upvotes: 9

Views: 2977

Answers (5)

Maciej Mikosik
Maciej Mikosik

Reputation: 445

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).

Upvotes: 4

Matthias
Matthias

Reputation: 1354

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

Upvotes: 0

Marcelo Cantos
Marcelo Cantos

Reputation: 185842

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).

Upvotes: 9

usr-local-ΕΨΗΕΛΩΝ
usr-local-ΕΨΗΕΛΩΝ

Reputation: 26874

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"

Upvotes: 4

Mark Byers
Mark Byers

Reputation: 838086

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.

Upvotes: 7

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