Primitive cast and assignments in Java

Question

I understand why the following is wrong:

byte a = 3; 
byte b = 8; 
byte c = a + b;  // compile error

It won\'t compile. Expressions alway

Answer 1

The only thing I can guess is that the compiler equates this expression to the following:

Yes it does. As long as the right side expression is made of constants (which fit into the required primitive type -- see @Jason's answer for what the JLS says about this exactly), you can do that. This will not compile because 128 is out of range:

byte a = 128;

Note that if you transform your first code snippet like this:

final byte a = 3; 
final byte b = 8; 
byte c = a + b;

it compiles! As your two bytes are final and their expressions are constants, this time, the compiler can determine that the result will fit into a byte when it is first initialized.

This, however, will not compile:

final byte a = 127; // Byte.MAX_VALUE
final byte b = 1;
byte c = a + b // Nope...

The compiler will error out with a "possible loss of precision".