Java – store the int value of the bitmask – extract the bit of 1 value

I'm calculating the int equivalent for the location set and storing it in memory From there, I want to determine all 1 value bits in the original bitmask Example:

33 – > [1,6] 97 – > [1,6,7]

The idea of implementing in Java?

Solution

On BitSet

Using Java util. BitSet to store a set of bits

Here is how to convert from int to BitSet according to which bit settings in int:

static BitSet fromInt(int num) {
    BitSet bs = new BitSet();
    for (int k = 0; k < Integer.SIZE; k++) {
        if (((num >> k) & 1) == 1) {
            bs.set(k);
        }
    }
    return bs;
}

So now you can do the following:

System.out.println(fromInt(33)); // prints "{0,5}"
System.out.println(fromInt(97)); // prints "{0,5,6}"

For completeness, here is the reverse transformation:

static int toInt(BitSet bs) {
    int num = 0;
    for (int k = -1; (k = bs.nextSetBit(k + 1)) != -1; ) {
        num |= (1 << k);
    }
    return num;
}

Therefore, combining the two together, we always get the original number:

System.out.println(toInt(fromInt(33))); // prints "33"
System.out.println(toInt(fromInt(97))); // prints "97"

0-based index

Note that this uses a 0 - based index, the more common bit index (and most other indexes in Java) This is also more correct Below, ^ denotes exponentiation:

33 = 2^0 + 2^5 = 1 + 32          97 = 2^0 + 2^5 + 2^6 = 1 + 32 + 64
33 -> {0,5}                     97 -> {0,6}

However, if you insist on using a 1-based index, you can use BS set(k 1); (1 < (< (k-1)) in the above fragment However, I will strongly oppose this proposal Related questions > What does the ^ operator do in Java? – It's not actually exponentiation