CODEWITHSUNDEEP
javabit-manipulationbit-fields
tinkuge
tinkuge

Reputation: 97

Efficient way to manipulate bits in byte array representing a bitfield

So, I have a byte array that represents a bitfield. A bit 1 at any index of a byte array implies that I have the corresponding piece and vice versa for a 0. Now, I need to change the bit value of 0 to 1, whenever I have the corresponding piece.

My question is, is it better if I convert the byte array to an int array and then change the corresponding value of the array index or is it simpler to do it in a byte array?

If it's the former, how can I convert the byte array to an integer array? If it's the latter, how do I change the value of the corresponding byte array?

Upvotes: 3

Views: 1618

Answers (1)

phflack
phflack

Reputation: 2727

To check if bit n is true

boolean get(int n, byte[] bitField)
{
    return (bitField[n >> 3] & 1 << (n & 0x7)) != 0; //or use n / 8 and n % 8
}

To set bit n

void set(int n, byte[] bitField, boolean value)
{
    if(value)
        bitField[n >> 3] |= 1 << (n & 0x7);
    else
        bitField[n >> 3] &= ~(1 << (n & 0x7));
}

If you use a BitSet, it's a bit simpler

To instantiate

BitSet bitField = new BitSet(); //can specify size

To check if bit n is true

bitField.get(n);

To set bit n

bitField.set(n, value); //can also use set(int) and clear(int) instead

Upvotes: 6

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