CODEWITHSUNDEEP
java
nunos
nunos

Reputation: 21389

Byte Array and Int conversion in Java

I am having some difficulty with these two functions: byteArrayToInt and intToByteArray.

The problem is that if I use one to get to another and that result to get to the former, the results are different, as you can see from my examples below.

I cannot find the bug in the code. Any ideas are very welcome. Thanks.

public static void main(String[] args)
{
    int a = 123;
    byte[] aBytes = intToByteArray(a);
    int a2 = byteArrayToInt(aBytes);

    System.out.println(a);         // prints '123'
    System.out.println(aBytes);    // prints '[B@459189e1'
    System.out.println(a2);        // prints '2063597568
            System.out.println(intToByteArray(a2));  // prints '[B@459189e1'
}

public static int byteArrayToInt(byte[] b) 
{
    int value = 0;
    for (int i = 0; i < 4; i++) {
        int shift = (4 - 1 - i) * 8;
        value += (b[i] & 0x000000FF) << shift;
    }
    return value;
}

public static byte[] intToByteArray(int a)
{
    byte[] ret = new byte[4];
    ret[0] = (byte) (a & 0xFF);   
    ret[1] = (byte) ((a >> 8) & 0xFF);   
    ret[2] = (byte) ((a >> 16) & 0xFF);   
    ret[3] = (byte) ((a >> 24) & 0xFF);
    return ret;
}

Upvotes: 48

Views: 155950

Answers (10)

Maarten Bodewes
Maarten Bodewes

Reputation: 93948

That's a lot of work for:

public static int byteArrayToLeInt(byte[] b) {
    final ByteBuffer bb = ByteBuffer.wrap(b);
    bb.order(ByteOrder.LITTLE_ENDIAN);
    return bb.getInt();
}

public static byte[] leIntToByteArray(int i) {
    final ByteBuffer bb = ByteBuffer.allocate(Integer.SIZE / Byte.SIZE);
    bb.order(ByteOrder.LITTLE_ENDIAN);
    bb.putInt(i);
    return bb.array();
}

This method uses the Java ByteBuffer and ByteOrder functionality in the java.nio package. This code should be preferred where readability is required. It should also be very easy to remember.

I've shown Little Endian byte order here. To create a Big Endian version you can simply leave out the call to order(ByteOrder).

In code where performance is higher priority than readability (about 10x as fast):

public static int byteArrayToLeInt(byte[] encodedValue) {
    int value = (encodedValue[3] << (Byte.SIZE * 3));
    value |= (encodedValue[2] & 0xFF) << (Byte.SIZE * 2);
    value |= (encodedValue[1] & 0xFF) << (Byte.SIZE * 1);
    value |= (encodedValue[0] & 0xFF);
    return value;
}

public static byte[] leIntToByteArray(int value) {
    byte[] encodedValue = new byte[Integer.SIZE / Byte.SIZE];
    encodedValue[3] = (byte) (value >> Byte.SIZE * 3);
    encodedValue[2] = (byte) (value >> Byte.SIZE * 2);   
    encodedValue[1] = (byte) (value >> Byte.SIZE);   
    encodedValue[0] = (byte) value;
    return encodedValue;
}

Just reverse the byte array index to count from zero to three to create a Big Endian version of this code.

Notes:

  • In Java 8 you can also make use of the Integer.BYTES constant, which is more succinct than Integer.SIZE / Byte.SIZE.

Upvotes: 51

Debosmit Ray
Debosmit Ray

Reputation: 5403

Instead of allocating space, et al, an approach using ByteBuffer from java.nio....

byte[] arr = { 0x01, 0x00, 0x00, 0x00, 0x48, 0x01};

// say we want to consider indices 1, 2, 3, 4 {0x00, 0x00, 0x00, 0x48};
ByteBuffer bf = ByteBuffer.wrap(arr, 1, 4); // big endian by default
int num = bf.getInt();    // 72

Now, to go the other way. 

ByteBuffer newBuf = ByteBuffer.allocate(4);
newBuf.putInt(num);
byte[] bytes = newBuf.array();  // [0, 0, 0, 72] {0x48 = 72}

Upvotes: -1

kissrain
kissrain

Reputation: 105

I found a simple way in com.google.common.primitives which is in the [Maven:com.google.guava:guava:12.0.1]

long newLong = Longs.fromByteArray(oldLongByteArray);
int newInt = Ints.fromByteArray(oldIntByteArray);

Have a nice try :)

Upvotes: 2

Ebrahim
Ebrahim

Reputation: 1

/*sorry this is the correct */

     public byte[] IntArrayToByteArray(int[] iarray , int sizeofintarray)
     {
       final ByteBuffer bb ;
       bb = ByteBuffer.allocate( sizeofintarray * 4);
       for(int k = 0; k < sizeofintarray ; k++)
       bb.putInt(k * 4, iar[k]);
       return bb.array();
     }

Upvotes: 0

Matt Klein
Matt Klein

Reputation: 8424

I like owlstead's original answer, and if you don't like the idea of creating a ByteBuffer on every method call then you can reuse the ByteBuffer by calling it's .clear() and .flip() methods:

ByteBuffer _intShifter = ByteBuffer.allocate(Integer.SIZE / Byte.SIZE)
                                   .order(ByteOrder.LITTLE_ENDIAN);

public byte[] intToByte(int value) {
    _intShifter.clear();
    _intShifter.putInt(value);      
    return _intShifter.array();
}

public int byteToInt(byte[] data)
{
    _intShifter.clear();
    _intShifter.put(data, 0, Integer.SIZE / Byte.SIZE);
    _intShifter.flip();
    return _intShifter.getInt();
}

Upvotes: 2

Jamel Toms
Jamel Toms

Reputation: 4565

You can also use BigInteger for variable length bytes. You can convert it to Long, Integer or Short, whichever suits your needs.

new BigInteger(bytes).intValue();

or to denote polarity:

new BigInteger(1, bytes).intValue();

To get bytes back just:

new BigInteger(bytes).toByteArray()

Upvotes: 24

Yanick Rochon
Yanick Rochon

Reputation: 53521

Your methods should be (something like) 

public static int byteArrayToInt(byte[] b) 
{
    return   b[3] & 0xFF |
            (b[2] & 0xFF) << 8 |
            (b[1] & 0xFF) << 16 |
            (b[0] & 0xFF) << 24;
}

public static byte[] intToByteArray(int a)
{
    return new byte[] {
        (byte) ((a >> 24) & 0xFF),
        (byte) ((a >> 16) & 0xFF),   
        (byte) ((a >> 8) & 0xFF),   
        (byte) (a & 0xFF)
    };
}

These methods were tested with the following code :

Random rand = new Random(System.currentTimeMillis());
byte[] b;
int a, v;
for (int i=0; i<10000000; i++) {
    a = rand.nextInt();
    b = intToByteArray(a);
    v = byteArrayToInt(b);
    if (a != v) {
        System.out.println("ERR! " + a + " != " + Arrays.toString(b) + " != " + v);
    }
}
System.out.println("Done!");

Upvotes: 60

TacB0sS
TacB0sS

Reputation: 10266

I took a long look at many questions like this, and found this post... I didn't like the fact that the conversion code is duplicated for each type, so I've made a generic method to perform the task:

public static byte[] toByteArray(long value, int n)
{
    byte[] ret = new byte[n];
    ret[n-1] = (byte) ((value >> (0*8) & 0xFF);   
    ret[n-2] = (byte) ((value >> (1*8) & 0xFF);   
    ...
    ret[1] = (byte) ((value >> ((n-2)*8) & 0xFF);   
    ret[0] = (byte) ((value >> ((n-1)*8) & 0xFF);   
    return ret;
}

See full post.

Upvotes: 0

Dapeng
Dapeng

Reputation: 1716

here is my implementation 

public static byte[] intToByteArray(int a) {
    return BigInteger.valueOf(a).toByteArray();
}

public static int byteArrayToInt(byte[] b) {
    return new BigInteger(b).intValue();
}

Upvotes: -2

aroth
aroth

Reputation: 54796

You're swapping endianness between your two methods. You have intToByteArray(int a) assigning the low-order bits into ret[0], but then byteArrayToInt(byte[] b) assigns b[0] to the high-order bits of the result. You need to invert one or the other, like:

public static byte[] intToByteArray(int a)
{
    byte[] ret = new byte[4];
    ret[3] = (byte) (a & 0xFF);   
    ret[2] = (byte) ((a >> 8) & 0xFF);   
    ret[1] = (byte) ((a >> 16) & 0xFF);   
    ret[0] = (byte) ((a >> 24) & 0xFF);
    return ret;
}

Upvotes: 44

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