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 others ( example change seventh bit to one, or change sixth bit to zero )?

Better use the primitive type byte. The wrapper class Byte should only be used where needed, e.g. when you want to put bytes into collections. — starblue
– starblue, Commented Jan 30, 2011 at 18:13

dkarp · Accepted Answer · 2011-01-30 18:29:12Z

30

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.)

edited Jan 30, 2011 at 18:29

answered Jan 30, 2011 at 17:33

dkarp

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1 Comment

dkarp Over a year ago

(This takes advantage of auto-boxing and -unboxing to convert from Byte to byte and back. If you're running in an older JVM, you'll have to convert them explicitly.)

starblue · Accepted Answer · 2011-01-30 18:28:54Z

15

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.

edited Jan 30, 2011 at 18:28

answered Jan 30, 2011 at 18:21

starblue

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3 Comments

Pacerier Over a year ago

Performance-wise, is it true to say that the compound assignment one and the non-compound one is identical ?

starblue Over a year ago

Yes, most likely they compile to the same machine code. But I haven't checked, why don't you do that?

Pacerier Over a year ago

well I'm not even sure what should I Google to learn how to do that.

user3162712 · Accepted Answer · 2016-09-10 07:26:11Z

2

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;
}

answered Sep 10, 2016 at 7:26

user3162712

Comments

jhouse · Accepted Answer · 2011-01-30 17:36:38Z

0

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

answered Jan 30, 2011 at 17:36

jhouse

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Comments

Hardryv · Accepted Answer · 2012-03-28 23:50:39Z

0

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.

answered Mar 28, 2012 at 23:50

Hardryv

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