Finding absolute value of a number without using Math.abs()

Question

Is there any way to find the absolute value of a number without using the Math.abs() method in java.

Answer 1

Like this:

if (number < 0) {
    number *= -1;
}

Answer 2

Yes:

abs_number = (number < 0) ? -number : number;

For integers, this works fine (except for Integer.MIN_VALUE, whose absolute value cannot be represented as an int).

For floating-point numbers, things are more subtle. For example, this method -- and all other methods posted thus far -- won't handle the negative zero correctly.

To avoid having to deal with such subtleties yourself, my advice would be to stick to Math.abs().

Answer 3

If you look inside Math.abs you can probably find the best answer:

Eg, for floats:

    /*
     * Returns the absolute value of a {@code float} value.
     * If the argument is not negative, the argument is returned.
     * If the argument is negative, the negation of the argument is returned.
     * Special cases:
     * <ul><li>If the argument is positive zero or negative zero, the
     * result is positive zero.
     * <li>If the argument is infinite, the result is positive infinity.
     * <li>If the argument is NaN, the result is NaN.</ul>
     * In other words, the result is the same as the value of the expression:
     * <p>{@code Float.intBitsToFloat(0x7fffffff & Float.floatToIntBits(a))}
     *
     * @param   a   the argument whose absolute value is to be determined
     * @return  the absolute value of the argument.
     */
    public static float abs(float a) {
        return (a <= 0.0F) ? 0.0F - a : a;
    }

Answer 4

Although this shouldn't be a bottle neck as branching issues on modern processors isn't normally a problem, but in the case of integers you could go for a branch-less solution as outlined here: http://graphics.stanford.edu/~seander/bithacks.html#IntegerAbs.

(x + (x >> 31)) ^ (x >> 31);

This does fail in the obvious case of Integer.MIN_VALUE however, so this is a use at your own risk solution.