Why can I place a case inside another case in a switch statement?

Question

Here is an example of a switch statement. I don't get why it works this way:

int main(){

    int number1 = 100, number2 = 200;

    switch(number1){
        case 100:{
            cout << " outer switch number is: " << number1 << endl;
        case 200:{ // any value other than 100
                cout << "inner switch number is: " << number2 << endl;
            }
            break;            
        }
        break;    
    }

    return 0;
}

• The program above prints: 100 and then prints the next statement of case 200. Also if any value other than 200 is used in the second case, it still gets executed. I know there is no break after case 100. But why don't I get a compile-time error instead?

• To be clearer, why will any other value inside the inner case also succeed? E.g.,

  case 70000:
  

Answer 1

But why don't I get a compile-time error instead?

Because you didn't do anything illegal. A case statement is just a labeled statement. The expression in the switch is evaluated to figure which label to jump to (note the "jump" and "label") and then that statement starts executing. It's just a goto in disguise, to be honest. It's a bit more constrained, but still a jump.

The only constraint, since this is C++, is that you may not skip the initialization of an object by jumping ahead of it.

This feature was famously used in Duff's Device (albeit in C).

To be more clear, why any other value inside the inner case will also succeed?

Because after the jump is performed, those labels don't matter. They just mark specific statements. After the jump, execution proceeds normally. And normal execution doesn't avoid specific statements just because they are labeled.