switch question

[Dinesh Tahiliani]

A case constant must evaluate to the same type as the switch expression can
use, with one additional�and big�constraint: the case constant must be a
compile time constant! Since the case argument has to be resolved at compile
time, that means you can use only a constant or final variable that is assigned a literal value. It is not enough to be final, it must be a compile time constant. For example:

final int a = 1;
final int b;
b = 2;
int x = 0;
switch (x) {
case a: // ok
case b: // compiler error

Can anyone please explain me the sentence and why it is giving compile error for case b. Please ranchers...

[camilo lopes]

because when you declared one variable as a final, is necessary that you define one value. Because with the modifier final the variable not have value default and in case switch �s necessary that each case have one value constant.

[fred rosenberger]

The compiler is stupid, and cannot figure out that the two lines go together... especially since they could theoretically be hundreds of lines apart.

[Prakash Subramanian]

In the below code you can see why a compile time constant is being insisted upon. Even though it is defined as final, it varies at runtime with each instantiation. Hope this example clarifies your question.

class SwitchTester {
  final int P = 10; //compile time constant
  final int K;
  static int counter = 0;
 
  public SwitchTester() {
    K = ++counter;
  }
 
  void testSwitch(int x) {
    switch(x) {
      case P:
        //this is fine
        break;
      case K:
        //the above will produce a compile time error
        break;
     }
  }
}

[Jesper de Jong]

Originally posted by Dinesh Tahiliani:
Can anyone please explain me the sentence and why it is giving compile error for case b.

Obviously because it's not a compile-time constant.