Why isn't a final variable always a constant expression?

Question

In the below code:

final int a;
a=2;
byte b=a;   // error: possible loss of precision

Why do I get this error? Isn't a final variable compile time constant expression and hence implicitly narrowed to byte during the assignment?

In other words isn't the above code equivalent to:

final int a=2;
byte b=a;

Answer 1

The compiler isn't that smart.

We can tell that the value will always be 2. But what if we had something like this?

class ABC{
    final int a;

    public ABC(){
       if(Math.random() < .5){
          a = 2;
       }
       else{
          a = 12345;
       }

       byte b = a;
    }
}

The compiler is not smart enough to tell these two cases apart, so it gives you an error instead.