Why there is no error when a final int is assigned to a byte [duplicate]

Question

Why do I get an error when

       int i=123;
       byte b=i;

But not in this case

      final int i=123;
      byte b=i;

Answer 1

When you initialize a final variable with a constant expression, it will become a compile-time constant. Essentially, when the code is compiled, it will just hardcode the value everywhere your variable is added. You can see this in the byte code:

 0  bipush 123
 2  istore_1 [i]
 3  bipush 123
 5  istore_2 [b]

As you can see, it pushes the value 123 directly into the byte (same as byte b = 123), and that is a valid value for a byte. It would not work with a value that is outside the allowed range for bytes.

If the variable is not final (or not initialized with a constant expression), then the compiler will see it as a normal variable, and normal rules for assigning are applied. Meaning that to assign an int to a byte it needs to be casted:

int i = 123;
byte b = (byte) i;

Which produces this bytecode:

0  bipush 123
2  istore_1 [i]
3  iload_1 [i]
4  i2b
5  istore_2 [b]