Bit operations converting to an integer

Question

I have some binary operations that are not working like I expect. I have byte array with the first 2 bytes having these values : 0x5, and 0xE0. I want to combine them into an integer value that should be 0x5E0. I tried doing :

int val = (b[i]) << 8 | b[i+1];

but the value is coming out 0xFFFFFFEE0 and the first byte 0x5 is getting lost

I thought this would be easy? What am I doing wrong?

Answer 1

Try: int val = ((b[i] & 0xff) << 8) | (b[i + 1] & 0xff). Bytes are (unfortunately) signed in Java, so if the high bit is set, it gets sign-extended when converted to an integer.

Answer 2

The problem is that byte data type is signed. Therefore, b[i+1] gets sign-extended before performing the operation, becoming 0xFFFFFFE0. When it gets OR-ed with 0x0500 from b[i]<<8, the 0x0500 gets lost.

You can fix this by AND-ing with 0xFF before performing the operation:

public static int toInt16(byte high, byte low) {
    int res = (high << 8);
    res |= (low & 0xFF);
    return res & 0xFFFF;
}

Demo.