CODEWITHSUNDEEP
java
user697111
user697111

Reputation: 2262

Is it possible to have an unsigned ByteBuffer in java?

Subject says it all. I'm working with OpenGL and OpenCL and would make life easier if I could just use an unsigned ByteBuffer to store data.

Upvotes: 34

Views: 23504

Answers (4)

francojohnc
francojohnc

Reputation: 815

how about this Short.toUnsignedInt(buffer.getShort());

Upvotes: 2

John3136
John3136

Reputation: 29266

Java doesn't support unsigned types. The typical solution is to go to the next biggest type (in your case: short), and just mask it so you only use the lower 'n' (in your case 8) bits.

... but that kind of breaks when you try to apply to buffers :-(

Upvotes: 1

kandarp
kandarp

Reputation: 5047

unsigned ByteBuffer example:

import java.nio.ByteBuffer;

public class test {
    public static short getUnsignedByte(ByteBuffer bb) {
        return ((short) (bb.get() & 0xff));
    }

    public static void putUnsignedByte(ByteBuffer bb, int value) {
        bb.put((byte) (value & 0xff));
    }

    public static short getUnsignedByte(ByteBuffer bb, int position) {
        return ((short) (bb.get(position) & (short) 0xff));
    }

    public static void putUnsignedByte(ByteBuffer bb, int position, int value) {
        bb.put(position, (byte) (value & 0xff));
    }

    // ---------------------------------------------------------------

    public static int getUnsignedShort(ByteBuffer bb) {
        return (bb.getShort() & 0xffff);
    }

    public static void putUnsignedShort(ByteBuffer bb, int value) {
        bb.putShort((short) (value & 0xffff));
    }

    public static int getUnsignedShort(ByteBuffer bb, int position) {
        return (bb.getShort(position) & 0xffff);
    }

    public static void putUnsignedShort(ByteBuffer bb, int position, int value) {
        bb.putShort(position, (short) (value & 0xffff));
    }

    // ---------------------------------------------------------------

    public static long getUnsignedInt(ByteBuffer bb) {
        return ((long) bb.getInt() & 0xffffffffL);
    }

    public static void putUnsignedInt(ByteBuffer bb, long value) {
        bb.putInt((int) (value & 0xffffffffL));
    }

    public static long getUnsignedInt(ByteBuffer bb, int position) {
        return ((long) bb.getInt(position) & 0xffffffffL);
    }

    public static void putUnsignedInt(ByteBuffer bb, int position, long value) {
        bb.putInt(position, (int) (value & 0xffffffffL));
    }

    // ---------------------------------------------------

    public static void main(String[] argv) throws Exception {
        ByteBuffer buffer = ByteBuffer.allocate(20);

        buffer.clear();
        test.putUnsignedByte(buffer, 255);
        test.putUnsignedByte(buffer, 128);
        test.putUnsignedShort(buffer, 0xcafe);
        test.putUnsignedInt(buffer, 0xcafebabe);

        for (int i = 0; i < 8; i++) {
            System.out.println("" + i + ": "
                    + Integer.toHexString((int) getUnsignedByte(buffer, i)));
        }

        System.out.println("2: "
                + Integer.toHexString(getUnsignedShort(buffer, 2)));
        System.out.println("4: " + Long.toHexString(getUnsignedInt(buffer, 4)));
    }
}

Upvotes: 75

Andreas Dolk
Andreas Dolk

Reputation: 114817

It's not a problem of ByteBuffer - even if it was unsigned - every byte that you read from it will be signed, just because byte is signed and we can't change that.

Upvotes: -4

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