Java: Arithmetic operation on byte

Question

What happened to my code? The following code worked for integer type of data, but couldn't work for byte type of data.

public class Exchange {
    public static void main(String[] args) {
    //int a = 23, b = 44;
    byte a = 23, b = 44;
    a = a + b;
    b = a - b;
    a = a - b;
    System.out.println("a=" + a + "b=" + b);
    }
}

I know that the data-type byte can hold data of the range -2^(8-1) to -1+2^(8-1). But I'm using 23 & 44, so it's less than 127.

Here I got error message "incompatible types: possible lossy conversion from int to byte".

And if you were adding a=127 and b=127 would the result fit in one byte? — Elliott Frisch
– Elliott Frisch, Commented Feb 20, 2016 at 4:51
@NawinMandal Use Alt+Enter keywords, it helps you to show you a way. In your case I just copied your code and cast your arithmetic operation before running that application. — ersks
– ersks, Commented Feb 20, 2016 at 5:07
alternative: a += b; b = (byte) (a-b); a -= b; -- compound assignments (like +=) automatically do the casting -- the second calculation can be replaced by b -= a; b *= -1; or b += a-2*b; to make it even more unclear, I mean, make it not using casts -- I would still prefer using a temporary variable, instead of a bunch of operations just to save one stack position — user85421
– user85421, Commented Aug 1 at 18:16

Nir Alfasi · Accepted Answer · 2016-02-20 04:49:54Z

4

If you want to perform an arithmetic operation on byte and assign it back to a byte variable you should explicitly let the compiler know that "you know what you're doing" otherwise you'll get the the error that you're losing information by converting int (the outcome of the arithmetic operation) to byte (on the left side).

To fix this, cast the outcome of the arithmetic operation back to byte:

    byte a = 23, b = 44;
    a = (byte) (a + b);
    b = (byte) (a - b);
    a = (byte) (a - b);

answered Feb 20, 2016 at 4:49

Nir Alfasi

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Comments

ersks · Accepted Answer · 2016-02-20 04:48:49Z

2

Simply, Cast the result of your arithmetic operation to byte as follows:

public class Exchange {
    public static void main(String[] args) {
    //int a = 23, b = 44;
    byte a = 23, b = 44;
    a = (byte) a + b;
    b = (byte) a - b;
    a = (byte) a - b;
    System.out.println("a=" + a + "b=" + b);
    }
}

Tip:- Use Alt+Enter for hints

answered Feb 20, 2016 at 4:48

ersks

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3 Comments

user508633 Over a year ago

stackoverflow.com/questions/14125843/… explains exactly why the widening on arithmetic occurs.

Nawin Mandal Over a year ago

@user508633, that link is not working for me! It's maybe helpful, but I found convoluted.

Nir Alfasi Over a year ago

@NawinMandal doing "alt-enter" without understanding the meaning of what you're doing is a bad habit for a developer. Pro tip: take the time to go over the link in the first comment, make sure you understand what widening means otherwise, you're doomed to run into nasty bugs in the future...

Raghu K Nair · Accepted Answer · 2016-02-20 04:50:11Z

2

Corrected code is here

 public static void main(String[] args) {
    //int a = 23, b = 44;
    byte a = 23, b = 44;
    a = (byte) (a + b);
    b = (byte) (a - b);
    a = (byte) (a - b);
    System.out.println("a=" + a + "b=" + b);
}

answered Feb 20, 2016 at 4:50

Raghu K Nair

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Comments

Vamshi Krishna · Accepted Answer · 2025-07-31 14:42:56Z

0

Because, in java all the arithmetic expressions are by default treated as "int" data type, that means whatever the expression you write in RHS will be promoted to "int" type, unless it contains even more wider type data, so implicit conversion from int to byte is not possible, you need to cast explicitly or use matching data type.

byte a = 23, b = 44;
a = (byte) (a + b);
b = (byte) (a - b);
a = (byte) (a - b);

answered Jul 31 at 14:42

Vamshi Krishna

11 bronze badge

2 Comments

user85421 Jul 31 at 15:54

I would remove the all from "all the arithmetic expressions" (for example when any long, float or double variable is involved, it is not treated as int {not mentioning String, since that would not be arithmetic} )

Vamshi Krishna Aug 1 at 16:59

Yes I agree, I also mentioned "unless it contains even more wider type data"

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