Switch Statement Default Case Fall Through

Question

Here is a small confusion so kindly pardon my ignorance. Here is a code snippet.

public class SwitchTest {
    public static void main(String[] args) {

        int x = 2;

        switch (x) {
            case 1:
                System.out.println("1");
                break;
            default:
                System.out.println("helllo");
            case 2:
                System.out.println("Benjamin");
                break;

        }

    }
}

Here, if value of x is 2, only Benjamin is printed. That's perfectly fine. Now lets suppose, i change value of x to 3, not matching any case, than its a fall through from default case. Ain't compiler needs to match every case for 3, by that time CASE 2 will be passed, than why it goes back to default and prints hello Benjamin. Can someone explain please?

Thanks,

Answer 1

You need to add a break; statement to break out of the switch block.

switch (x) {
    case 1:
        System.out.println("1");
        break;
    default:
        System.out.println("helllo");
        break; // <-- add this here
    case 2:
        System.out.println("Benjamin");
        break;

    }

Generally speaking, it is also better coding practice to have your default: case be the last case in the switch block.

In this case, the switch is following the pattern:

x==1? No, check next case
default? Not done yet, check other cases
x==2? No, check next case
//No more cases, so go back to default
default? Yes, do default logic
// no break statement in default, so it falls through to case 2: logic without checking the case
output case 2 logic
break

Notice how the block will jump over the default case, and save it until a later time unless we have exhausted all other possible cases.