Question

Working on a method that would take in an integer (num) and an integer (n) that would take the absolute value of an n-bit number. I believe my logic is correct, have done it on paper and it worked out but the code seems like it is off. All help is greatly appreciated!

/**
 * Take the absolute value of an n-bit number.
 * 
 * Examples:
 *     abs(0x00001234, 16); // => 0x00001234
 *     abs(0x00001234, 13); // => 0x00000DCC
 * 
 * Note:  values passed in will only range from 1 to 31 for n.
 * 
 * @param num An n-bit 2's complement number.
 * @param n The bit length of the number.
 * @return The n-bit absolute value of num.
 */
public static int abs(int num, int n)
{
    int shifter = num << (n+1);
    int newInt = num & ~shifter;
    return newInt;

}

Answer 1

I don’t think there’s a single bitmask that will work for both positive and negative cases.

First test if the number is negative by checking that the nth bit is 1; if not, return the original, else return the two’s complement.

Something like this looks like it works:

public static int abs(int num, int n)
{
    int shifter = -1 << (n - 1);
    if ((num & shifter) == 0) 
        return num;
    shifter = shifter << 1;
    return (~num + 1) & ~shifter;
}

For example, suppose you pass in 0x1FFF as a 16 bit number, so it’s positive.
-1 << 15 would be 0xFFFF8000 (0’s for the lowest 15 bits, 1’s for the rest), 0xFFFF8000 & 0x00001FFF is 0, and your return the original.

If on the other hand 0x1FFF is treated as only 13 bits, then it’s negative. num & shifter will be 1 because both have the 13th bit set. Now do the two’s complement by flipping bits and adding ones. Because you’ll be flipping all 32 bits, you need to use a bitmask to zero out all the remaining ones. The original shifter works if you push it one more bit to the left and invert it.