Java: Convert HEX byte array to 16 fixed length binary String

Question

I have a byte array obtained by automatically MYBATIS conversion. Into ORACLE DB column (RAW type) I have HEX value (example: 0x0, 0x81, 0xC801). This value is stored with maximum 2 bytes (example 1100100000000001 -> 0xC801, 0000000000000000 -> 0x0)

I have a problem when i read and convert this value.

I need to convert the byte array obtained by automatically conversion, to 16 fixed length binary string.

For example if the value in DB is 0x0:

byte array is [0] (automatically MYBATIS conversion)
binary string that I need must be 0000000000000000

For example if the value in DB is 0xC801:

byte array is [0, -56, 1] (automatically MYBATIS conversion)
binary string that I need must be 1100100000000001

How i can do that?

I tried with this solution but doesn't work with HEX value like 0x0 or 0x81 because 0x0 is mapped to "0" and 0x81 is mapped to "100000001". 0 padding is missing... and i don't know how to add the padding insite this script.

private static final char[] HEX_ARRAY = "0123456789ABCDEF".toCharArray();

public static String bytesToBinary(byte[] bytes) {
    return hexToBin(bytesToHex(bytes));
}

public static String hexToBin(String s) {

    if (StringUtils.isNotEmpty(s)) {
        return new BigInteger(s, 16).toString(2);
    }

    return "";
}

public static String bytesToHex(byte[] bytes) {

    if (bytes != null && bytes.length > 0) {

        char[] hexChars = new char[bytes.length * 2];

        for (int j = 0; j < bytes.length; j++) {
            int v = bytes[j] & 0xFF;
            hexChars[j * 2] = HEX_ARRAY[v >>> 4];
            hexChars[j * 2 + 1] = HEX_ARRAY[v & 0x0F];
        }

        return new String(hexChars);
    }

    return "";
}

Do you always want to convert just two bytes. Are you always returned a 3 byte array? — WJS
– WJS, Commented Oct 23, 2020 at 12:04
@WJS Read the question again. "byte array is [0]" and "byte array is [0, -56, 1]". Question has 2 examples, of length 1 and 3, respectively, so no, it is not always just two bytes. — Andreas
– Andreas, Commented Oct 23, 2020 at 12:08
@Andreas The title says convert to a 16 bit string. That implies 2 bytes to me. — WJS
– WJS, Commented Oct 23, 2020 at 12:09
@WJS Yes, and the question text shows 2 examples of byte arrays with 1 and 3 bytes. Since the first byte is 0, that is still at most a 16 bit value. — Andreas
– Andreas, Commented Oct 23, 2020 at 12:11
Which is what I said. You only want to convert at most two bytes (in spite of the array size). — WJS
– WJS, Commented Oct 23, 2020 at 12:12

WJS · Accepted Answer · 2020-10-23 12:15:40Z

2

Try using BigInteger

byte[] bytes = {0, -56, 1};
        
                              // array, offset, length 
String result = new BigInteger(bytes).setBit(16).toString(2).substring(1);
System.out.println(result);

Prints

1100100000000001

edited Oct 23, 2020 at 12:15

answered Oct 23, 2020 at 11:51

WJS

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4 Comments

Andreas Over a year ago

Close, but with input {0} the output is 0, not the requested 0000000000000000. Can be fixed e.g. as follows: new BigInteger(bytes).setBit(16).toString(2).substring(1)

Andreas Over a year ago

If you use new BigInteger(bytes) instead of new BigInteger(bytes,0,3), then the code can handle any byte array of length 1-4, instead of only handling arrays of length 3.

Andreas Over a year ago

Also, the BigInteger(byte[] val, int off, int len) constructor wasn't even added until Java 9, so by changing to the BigInteger(byte[] val) constructor it'll work for all Java versions (well, Java 1.1+).

Ralph Over a year ago

Thank you very much guys! SOLVED with your solution :)

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