Long to Byte Array Invalid

Question

I'm starting to use bytes and hexes to try to store some data easier. This is currently what I'm doing:

byte[] data = new byte[] {0x20, 0x40};
long cosmetics = 0;

for(byte d : data) {
    cosmetics = cosmetics | d;
    System.out.println(d + ": " + cosmetics);
}

String hex = Long.toHexString(cosmetics);
System.out.println("hex: 0x" + hex);
System.out.println("from hex: " + Long.decode("0x"+hex));

byte[] bytes = longToBytes(cosmetics);
String s = "";
for(byte b : bytes)
  s += b+", ";
System.out.println("bytes: " + s);

This all works fine, hex: 0x60 and from hex = 96 , just as it is supposed to be (afaik).

However when I attempt to convert 96 back to the byte array, using longToBytes(cosmetics) :

public static byte[] longToBytes(long x) {
    ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES);
    buffer.putLong(x);
    return buffer.array();
}

It doesn't return the array I initially used, it gives: 0, 0, 0, 0, 0, 0, 0, 96

But what I want it to give me, is the array that I initially used:

byte[] data = new byte[] {0x20, 0x40};

Answer 1

A long has 8 bytes, and you put the byte 0x20|0x40 = 0x60 = 96 as long into the array.

Java per default orders the bytes bigendian , so the least significant byte, 96, comes last.

To do it the other way around:

public static byte[] longToBytes(long x) {
    return ByteBuffer.allocate(Long.BYTES)
            .order(ByteOrder.LITTLE_ENDIAN)
            .putLong(x)
            .array();
}

Should give

96, 0, 0, 0, 0, 0, 0, 0

---

After refined question

One cannot determine that 96 stemmed from 0x20|0x40, but I'll assume you want the individual bit masks.

byte[] longToBytes(long x) {
    ByteArrayOutputStream baos = new ByteArrayOutputStream();
    int mask = 1;
    for (int i = 0; i < 8; ++i) {
        if ((mask & x) != 0) {
            baos.write(0xFF & (int)mask);
        }
        mask <<= 1;
    }
    return baos.toArray();
}

The parameter could/should be a byte or 0-256 restricted int for a sensible result.