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

问题

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;

回答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.

回答2:

From the JLS

A blank final is a final variable whose declaration lacks an initializer.

A constant variable is a final variable of primitive type or type String that is initialized with a constant expression (§15.28).

Your variable

final int a;

is a blank final variable. It lacks an initializer. The second paragraph doesn't apply to it because it is not initialized at declaration. It therefore isn't a constant expression.

This applies to fields as well.