Why is the final variable always a constant expression?

Question

In the code below:

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

Why am I getting this error? Is a final compile time variable a constant constant and, therefore, implicitly tapers to a byte at the time of assignment?

In other words, the code not shown above is equivalent:

 final int a=2; byte b=a; 

Answer 1

The compiler is not so smart.

We can say 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 talk about all these cases, so it gives you an error.

Answer 2

From JLS

Empty final is the final variable, in the declaration of which there is no initializer.

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

Your variable

 final int a; 

is an empty final variable. He lacks an initializer. The second paragraph does not apply to it, since it is not initialized upon declaration. Therefore, this is not a constant expression.

This also applies to fields.

Answer 3

Since the final variables can be delayed initialized and the compiler cannot determine for b that it matters in the case branch.