CODEWITHSUNDEEP
javabit
DonMarco
DonMarco

Reputation: 415

Get the position of a bit and if it is set

I have a function which checks, whether a bit in an int is set or not. But I think there will be a much faster implementation, since this one is linear and can't be the most efficient one, although I know the int should be between 1 and 1024.

 public static int getBitPos(final int n) {
        if (Integer.bitCount(n) != 1)
            return Constants.UNDEFINED;
        else {
            for (int i = 0; i < Integer.MAX_VALUE; ++i) {
                if (testBit(n, i))
                    return i;
            }
        }
        return Constants.UNDEFINED;
    }

Where testBit is the following standard function:

public static boolean testBit(final int n, final int pos) {
    int mask = 1 << pos;
    return (n & mask) == mask;
}

But there mast be a faster way, isn't there? If I have the value 17 and I want to know if the 4th bit (n = 8) is set? There should be a faster way to check whether the bit for n=8 is set...

Hope you can help me...

EDIT 1: Thanks for the support. The comments and answers brought me to my mistake. I was setting the values wrongly, which made it more complicated than needed. I was never good at bit shifting. I set the value like this, if I wanted the second bit to be set:

value = 2;

If I wanted the 4th bit to be set too, I added the value according to the 4th bit:

value += 8;

So value was 10, and the 2nd and 4th bit were set. So I saved the numbers in my class, instead of the bit-positions (8 as value, instead of 4 for the 4th bit, ...). After changing this, I could get rid of my unnecessary function, which was way over the top! Thanks for all help!

Upvotes: 0

Views: 1265

Answers (1)

jlordo
jlordo

Reputation: 37813

Your code always returns the lowest bit that is 1, if there is only one. You can achieve the same by doing this:

int foo = whatever;
int lowestSetBit = Integer.numberOfTrailingZeros(foo) + 1;

Your code would be

public static int getBitPos(final int n) {
    if (Integer.bitCount(n) == 1)
        return Integer.numberOfTrailingZeros(n) + 1;
    return Constants.UNDEFINED;
}

Upvotes: 5

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