CODEWITHSUNDEEP
javaintbyte
ccit
ccit

Reputation: 309

java integer to byte, and byte to integer

I am aware -that in Java- int is 4 bytes. But I wish to covert an int to n-bytes array, where n can be 1, 2, 3, or 4 bytes. I want to have it as signed byte/bytes, so that if I need to convert them back to int (event if it was 1 byte), I get the same signed int. I am fully aware about the possibility of precision loss when converting from int to 3 or lower bytes.

I managed to convert from int to n-byte, but converting it back for a negative number yield unsigned results.

Edit:

int to bytes (parameter n specify the number of bytes required 1,2,3, or 4 regardless of possible precession loss)

public static byte[] intToBytes(int x, int n) {
    byte[] bytes = new byte[n];
    for (int i = 0; i < n; i++, x >>>= 8)
        bytes[i] = (byte) (x & 0xFF);
    return bytes;
}

bytes to int (regardless of how many bytes 1,2,3, or 4)

public static int bytesToInt(byte[] x) {
    int value = 0;
    for(int i = 0; i < x.length; i++)
        value += ((long) x[i] & 0xffL) << (8 * i);
    return value;
}

There is probably a problem in the bytes to int converter.

Upvotes: 6

Views: 34146

Answers (3)

Greg Kopff
Greg Kopff

Reputation: 16545

BigInteger.toByteArray() will do this for you ...

Returns a byte array containing the two's-complement representation of this BigInteger. The byte array will be in big-endian byte-order: the most significant byte is in the zeroth element. The array will contain the minimum number of bytes required to represent this BigInteger, including at least one sign bit, which is (ceil((this.bitLength() + 1)/8)). (This representation is compatible with the (byte[]) constructor.)

Example code:

final BigInteger bi = BigInteger.valueOf(256);
final byte[] bytes = bi.toByteArray();

System.out.println(Arrays.toString(bytes));

Prints:

[1, 0]

To go from the byte array back to a int, use the BigInteger(byte[]) constructor:

final int i = new BigInteger(bytes).intValue();
System.out.println(i);

which prints the expected:

256

Upvotes: 6

Marko Topolnik
Marko Topolnik

Reputation: 200148

Anyway, this is the code I threw together:

public static void main(String[] args) throws Exception {
  final byte[] bi = encode(-1);
  System.out.println(Arrays.toString(bi));
  System.out.println(decode(bi));
}
private static int decode(byte[] bi) {
  return bi[3] & 0xFF | (bi[2] & 0xFF) << 8 |
         (bi[1] & 0xFF) << 16 | (bi[0] & 0xFF) << 24;
}
private static byte[] encode(int i) {
  return new byte[] { (byte) (i >>> 24), (byte) ((i << 8) >>> 24),
                      (byte) ((i << 16) >>> 24), (byte) ((i << 24) >>> 24)
  };
}

Upvotes: 6

Joop Eggen
Joop Eggen

Reputation: 109547

Something like:

int unsignedByte = ((int)bytes[i]) & 0xFF;

int n = 0;
n |= unsignedByte << 24;

Upvotes: 1

Related Questions