How store hexadecimal strings in a Byte array?

Asked Mar 23 '15 at 11:46

Active Mar 23 '15 at 12:38

Viewed 46 times

I need help to create the header as explain below, I am not confident with byte conversion and big endian convention.

I think my problem comes from the lines byte[] VERSION = "01".getBytes(); I don't know how to 'force' the String to be stored in one byte.

Header creation

public static final int HEADER_SIZE = 8;
byte[] header = new byte[HEADER_SIZE];

// Header struture
/*
*byte 0   : Version
*byte 1   : Inverse Version
*byte 2-3 : Type
*byte 4-7 : Length
*/

// Hex values
byte[] VERSION = "01".getBytes();
byte[] INVERSE_VERSION = "FE".getBytes();
byte[] TYPE = "9000".getBytes();
byte[] LENGTH = "02".getBytes();

ByteBuffer buf = ByteBuffer.wrap(this.header);
    buf.order(ByteOrder.BIG_ENDIAN);
    buf.put(VERSION);
    buf.put(INVERS_VERSION);
    buf.put(TYPE);
    //buf.put(LENGTH); <-- Overflow

//out
System.out.println(new String(buf.array(), "ASCII"));

Actual Output

01FE9000

Output Expected

01FE900020000000

edited Mar 23 '15 at 12:38

asked Mar 23 '15 at 11:46

Jean

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did you missed this code `byte[] LENGTH = "02".getBytes();` – Bhargav Modi Mar 23 '15 at 11:49
1

how do you expect this `01FE900000000002` and not this `01FE900002000000`? – guido Mar 23 '15 at 12:14
Hexadecimal is a representation, not a data type. – Hot Licks Mar 23 '15 at 12:41
`public static void main(String[] args){ long hex1 = 0x01; long hex2 = 0xFF; long hex3 = 0x9000; long hex4 = 0x20000000; long hexResult = hex4 | (hex3<<32) | (hex2< – Jean Mar 23 '15 at 13:53
Since I can't post an answer I posted here what helped me after few experiments... – Jean Mar 23 '15 at 13:54

How store hexadecimal strings in a Byte array?

0 Answers0