Java 24-bit number signing

Question

I'm currently doing some work that has me working with some 24-bit integers.

Essentially I need to be able to get both the signed, and unsigned values from these 24-bits.

Currently I'm using the following code to put the three bytes together and return me their value.

private static int bytesToInt(byte[] input) {
    if (input.length == 3) {
        return (input[2] & 0xFF) << 16 | (input[1] & 0xFF) <<8 | (input[0] & 0xFF);
    }
}

The input I'm giving it are the bytes: 0x42 0x87 0xfe and the returned result is: 16680770

This (I believe) is the correct unsigned value, however I also need the signed value of it, which I think is -96446

I'm guessing I'll have to do some bitshifting here to solve it, but I'm not sure how to accomplish this.

I've tried casting the result into an and a long, but neither return the signed value. I've also tried Math.abs(result), but I don't think I'm using that correctly.

Any input would be appreciated.

Thanks

Answer 1

private static int bytesToUInt(byte[] input) {
    if (input.length == 3) {
        return (input[2] & 0xFF) << 16 | (input[1] & 0xFF) <<8 | (input[0] & 0xFF);
    }
}
private static int bytesToSInt(byte[] input) {
    if (input.length == 3) {
        return (input[2]) << 16 | (input[1] & 0xFF) <<8 | (input[0] & 0xFF);
    }
}

Answer 2

One option is just to fill the top 8 bits with 1:

int x = 16680770;
int y = x | 0xff000000;
System.out.println(y); // -96446

Or you can just subtract 1 << 24:

int y = x - (1 << 24);