How can I calculate required bits for an unsigned value?

Question

For a given unsigned int value, I tried following method for the number of bits for storing given value.

    // Returns the number of required bits for (storing) specified unsigned value.
    static int size(final int value) {
        if (value < 0) {
            throw new IllegalArgumentException("value(" + value + ") is negative");
        }
        return (int) Math.ceil(Math.log10(value) / Math.log10(2));
    }

And here comes what I got.

size for 1310394095: 31              (RANDOM)
size for 1: 0                        (1; Not Good; should be 1)
size_NotNegative_Zero() is @Disabled (0; ERROR; Expecting actual: -2147483648 ...)
size for 2147483647: 31              (Integer.MAX_VALUE)

The 31 for Integer.MAX_VALUE seems OK.

How can I fix this?

Does this help? https://stackoverflow.com/a/40746823/5811874 — Hanchen Jiang, Jan 02 '22 at 05:46

score 0 · Answer 1 · answered Jan 02 '22 at 06:12

Although I flagged my own question as a duplicate, I'm posting what I found.

The problem is that the input is not just positive but also including zero and that's why the way of simply dividing log_whateverv with log_whatever2 won't work cuz log_whatever0 is not defined.

Here comes what I concluded with https://stackoverflow.com/a/680040/330457.

    static int size(final int value) {
        if (value < 0) {
            throw new IllegalArgumentException("value(" + value + ") is negative");
        }
        if (value == 0) {
            return 1;
        }
        return Integer.SIZE - Integer.numberOfLeadingZeros(value);
    }

size for 909663241: 30  (RANDOM)
size for 1: 1           (ONE)
size for 0: 1           (ZERO)
size for 2147483647: 31 (Integer.MAX_VALUE)

How can I calculate required bits for an unsigned value?

1 Answers1