Question

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.

Answer 1

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