Change bits value in Byte

Question

I have some data in field type Byte ( I save eight inputs in Byte, every bit is one input ). How to change just one input in that field ( Byte) but not to lose information about

Answer 1

To set a bit use :

public final static byte setBit(byte _byte,int bitPosition,boolean bitValue)
{
    if (bitValue)
        return (byte) (_byte | (1 << bitPosition));
    return (byte) (_byte & ~(1 << bitPosition));
}

To get a bit value use :

public final static Boolean getBit(byte _byte, int bitPosition)
{
    return (_byte & (1 << bitPosition)) != 0;
}

Answer 2

Note that the "Byte" wrapper class is immutable, and you will need to work with "byte".

Answer 3

Declare b as the primitive type byte:

byte b = ...;

Then you can use the compound assignment operators that combine binary operations and assignment (this doesn't work on Byte):

b |= (1 << bitIndex); // set a bit to 1
b &= ~(1 << bitIndex); // set a bit to 0

Without the assignment operator you would need a cast, because the result of the | and & operations is an int:

b = (byte) (b | (1 << bitIndex));
b = (byte) (b & ~(1 << bitIndex));

The cast is implicit in the compound assignment operators, see the Java Language Specification.

Answer 4

You really owe it to yourself to look into masking functions for and, or, and xor -- they allow you to simultaneously verify, validate, or change... one, some, or all of the bits in a byte structure in a single statement.

I'm not a java programmer by trade, but it's derived from C and a quick search online seemed to reveal support for those bitwise operations.

See this Wikipedia article for more information about this technique.

Answer 5

To set the seventh bit to 1:

b = (byte) (b | (1 << 6));

To set the sixth bit to zero:

b = (byte) (b & ~(1 << 5));

(The bit positions are effectively 0-based, so that's why the "seventh bit" maps to 1 << 6 instead of 1 << 7.)