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I have question regarding my code here:

public class Main
{
    public static void main(String[] args) {
        System.out.println("Hello World\n");
        
        int x = 36;
        byte b1 = ((byte) x) & ((byte) 0xff); // it seems it is the part after &, but I have 0xff cast to byte by using (byte)0xff, so not sure where exactly the error is coming from.
 
        
        System.out.println(b1);
        
    }
}

I am not sure exactly which part is causing the error of:

incompatible types: possible lossy conversion from int to byte

This is the error message output from the program:

error pointing to the ampersand in "byte b1 = ((byte) x) & ((byte) 0xff);"

Laurel
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hodondo
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2 Answers2

1

You appear to be confused.

There is no point in your code. taking any number, calculating that & 0xFF, and then storing it in a byte, is always a noop - it does nothing.

You additionally get an error because & inherently always produces at least an int (it'll upcast anything smaller to match), so you're trying to assign an int to a byte.

What are you trying to accomplish?

"I want to have my byte be unsigned"!

No can do. Java doesn't have unsigned bytes. A java byte is signed. Period. It can hold a value between -128 and +127. For calculation purposes, -128 and 255 are identical (they are both the bit sequence 1111 1111 - in hex, 0xFF, and they act identically under all relevant arithmetic, though it does get tricky when converting them to another numeric type int).

"I just want to store 255"!

Then use int. This is where most & 0xFF you'll ever see in java code comes from: When you have a byte value which java inherently treats as signed, but you wish to treat it as unsigned and, therefore (given that in java bytes can't do that), you want to upcast it to an int, containing the unsigned representation. This is how to do that:

int x = y & 0xFF;

Where y is any byte.

You presumably saw this somewhere and are now trying to apply it, but assigning the result of y & 0xFF to a byte doesn't mean anything. You'd assign it to an int variable, or just use it as expression in a further calculation (y & 0xFF is an int - make sure you add the appropriate parentheses, & has perhaps unexpected precedence).

int x = 36;

    byte b1 = ((byte) x) & ((byte) 0xff);

Every imaginable way of this actually working would mean that b1 is... still 36.

rzwitserloot
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To compute x & y where the two operands are bytes, they must first be promoted to int values. There is no & between bytes. The result is therefore of type int

That is, what you wrote is effectively evaluated as if you'd written it as the following, making explicit what the language gives you implicitly:

byte b1 = ((int) (byte) x) & ((int) (byte) 0xff);

Just do the arithmetic and then cast the result to byte.

byte b1 = (byte)(x & 0xff);

Link to Java Language Specification

Edited to add, thanks to @rzwitserloot, that masking a byte value with 0xff is however pointless. If you need the assignment from an integer to a byte, just write the cast:

byte b1 = (byte)x;
dangling else
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  • You're literally telling OP to do a completely pointless act. If the seemingly correct answer leads to self-evident nonsense I'm not sure it's a good answer. Good effort - just, that final little cherry on top: At the very end you have a code snippet that does what OP wants - and... in fact, does nothing. Conclusion: OP doesn't actually want what they think they want. (`& 0xFF`ing a byte, storing the result in a byte, __does nothing__). – rzwitserloot May 20 '22 at 02:58
  • True enough - I was focused on the reason for the compilation problem and totally missed the useless semantics of it. I'll add a note. But the assignment still does something - sets variable b1. – dangling else May 20 '22 at 12:15