Why b=b+1 when b is a byte won't compile but b+=1 compiles

Question

This is my code:

class Example{
    public static void main(String args[]){
        byte b=10;
        //b=b+1; //Illegal
        b+=1;   //Legal
        System.out.println(b);
    }
}

I want to know why I'm getting a compilation error if I use b=b+1, but on the other hand b+=1 compiles properly while they seem to do the same thing.

Answer 1

This is an interesting question. See JLS 15.26.2. Compound Assignment Operators:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

So when you are writing b+=1;, you are actually casting the result into a byte, which is the similar expressing as (byte)(b+1) and compiler will know what you are talking about. In contrast, when you use b=b+1 you are adding two different types and therefore you'll get an Incompatible Types Exception.

Answer 2

the Error you get is because of the operations with different data types and that can cause an overflow.

when you do this:

byte b = 127;
b=b+1; 

you generate an overflow, so the solution would be casting the result

b=(byte) (b+1); 

Answer 3

Because can't convert int to byte

You can try:

b=(byte) (b+1);