OR operator in switch-case?

Let's take a simple switch-case that looks like:

@Override
public void onClick(View v) {
    switch (v.getId()) {
        case R.id.someValue :
        case R.id.someOtherValue:
            // do stuff
            break;
    }
}

I wonder why it is not allowed to use the || operator? Like

switch (v.getId()) {
    case R.id.someValue || R.id.someOtherValue:
        // do stuff
        break;
}

The switch-case construct is pretty similar to an if-else statement, you can use the OR operator in an if however. What are the backgrounds for a switch-case to not accept this operator?

Solution 1:

dude do like this

    case R.id.someValue :
    case R.id.someOtherValue :
       //do stuff

This is same as using OR operator between two values Because of this case operator isn't there in switch case

Solution 2:

What are the backgrounds for a switch-case to not accept this operator?

Because case requires constant expression as its value. And since an || expression is not a compile time constant, it is not allowed.

From JLS Section 14.11:

Switch label should have following syntax:

SwitchLabel:
case ConstantExpression :
case EnumConstantName :
default :

Under the hood:

The reason behind allowing just constant expression with cases can be understood from the JVM Spec Section 3.10 - Compiling Switches:

Compilation of switch statements uses the tableswitch and lookupswitch instructions. The tableswitch instruction is used when the cases of the switch can be efficiently represented as indices into a table of target offsets. The default target of the switch is used if the value of the expression of the switch falls outside the range of valid indices.

So, for the cases label to be used by tableswitch as a index into the table of target offsets, the value of the case should be known at compile time. That is only possible if the case value is a constant expression. And || expression will be evaluated at runtime, and the value will only be available at that time.

From the same JVM section, the following switch-case:

switch (i) {
    case 0:  return  0;
    case 1:  return  1;
    case 2:  return  2;
    default: return -1;
}

is compiled to:

0   iload_1             // Push local variable 1 (argument i)
1   tableswitch 0 to 2: // Valid indices are 0 through 2  (NOTICE This instruction?)
      0: 28             // If i is 0, continue at 28
      1: 30             // If i is 1, continue at 30
      2: 32             // If i is 2, continue at 32
      default:34        // Otherwise, continue at 34
28  iconst_0            // i was 0; push int constant 0...
29  ireturn             // ...and return it
30  iconst_1            // i was 1; push int constant 1...
31  ireturn             // ...and return it
32  iconst_2            // i was 2; push int constant 2...
33  ireturn             // ...and return it
34  iconst_m1           // otherwise push int constant -1...
35  ireturn             // ...and return it

So, if the case value is not a constant expressions, compiler won't be able to index it into the table of instruction pointers, using tableswitch instruction.

Solution 3:

You cannot use || operators in between 2 case. But you can use multiple case values without using a break between them. The program will then jump to the respective case and then it will look for code to execute until it finds a "break". As a result these cases will share the same code.

switch(value) 
{ 
    case 0: 
    case 1: 
        // do stuff for if case 0 || case 1 
        break; 
    // other cases 
    default: 
        break; 
}