bits array to bytes - java

Question

how can I store a sequence of bits in an array of bytes[] ? in more detail: I have an array

byte[] bytes;

how will I store individual values (0s and 1s) in a single byte of the array? I am looking for something like:

bit[] bits = new bit[8];
bits[0] = 1;
bits[3] = 1;
bytes[3] = bits[];

I hope that makes sense. thanks

Could you put the 0s and 1s into a String and use Byte.parseByte with a radix of 2? — Dawood ibn Kareem
– Dawood ibn Kareem, Commented Feb 28, 2014 at 2:37
I need to make alterations to the byte sequence every now and then so it wont be a good idea to do it like that as i would have to rebuild the sequence over and over instead of just adding/swapping the new value to the required index posission — Rakim
– Rakim, Commented Feb 28, 2014 at 2:40
Yes, but OP hasn't told us what his/her input is. Maybe he/she is reading a file full of 0s and 1s, so already has String. In which case, there's no real cost in memory (at least in Java 6; Java 7 manages memory differently for strings). — Dawood ibn Kareem
– Dawood ibn Kareem, Commented Feb 28, 2014 at 2:43
to make it more clear I am using an array of bytes because I am storing binary values of strings in the rest of the indexes. it is just one or two indexes that I need to store bits in them. — Rakim
– Rakim, Commented Feb 28, 2014 at 2:49

Hot Licks · Accepted Answer · 2014-02-28 03:20:03Z

1

void setBit(int[] array, int index, boolean value) {
    int x = index / 32;
    int y = index % 32;
    int mask = 1 << y;
    if (value) {
        array[x] |= mask;
    }
    else {
        array[x] &= (0xFFFFFFFF ^ mask);
    }
}

boolean getBit(int[] array, int index) {
    int x = index / 32;
    int y = index % 32;
    int mask = 1 << y;
    return (array[x] & mask) != 0;
}

answered Feb 28, 2014 at 3:20

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Comments

PlasmaPower · Accepted Answer · 2014-02-28 02:46:45Z

0

A boolean can be used to represent a bit, but it takes up more memory. You could use bitwise operators to save on memory, here is an example:

int n = 0;
n |= 1 << 1; // Set bit 1 to true, n is 2
n |= 1 << 3; // Set bit 3 to true, n is 10
bool b0 = n & (1 << 3) > 0; // True
n &= 1 << 3; // Set bit 3 to false, n is 2
bool b1 = n & (1 << 1) > 0; // True
bool b2 = n & (1 << 2) > 0; // False
bool b3 = n & (1 << 3) > 0; // False

edited Feb 28, 2014 at 2:46

answered Feb 28, 2014 at 2:38

PlasmaPower

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3 Comments

Dawood ibn Kareem Over a year ago

You could replace each Math.pow with 1 << something

Rakim Over a year ago

It is more a matter of memory in my case than a matter of true & false so a Boolean wont work for me..

PlasmaPower Over a year ago

Bitwise operators will be the most memory efficient way possible.

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