Question

Why does this,

public class Bar 
{
    public static void main(String[] args) 
    {
        int i = 1;

        switch(i)
        {
            case 0:
            case 1:
            case 2:
                System.out.println("Case 2 being executed");
                break;
            default:
                break;
        }
    }
}

output this,

Case 2 being executed

?

How is it even possible to enter the case block for an input value of 2 when the input value is explicitly 1? Note that I’m aware I can avoid this behavior by adding a break statement in the case block for 1.

Answer 1

How is it even possible to enter the case block for an input value of
2 when the input value is explicitly 1?

This behaviour is called fall-through which is quite common mistake with beginners working with switch-case. Actually, case 1: does execute first. But, since there is no break statement in case 1, your switch-case goes onto executing the next cases, until it finds a break statement. So, it will even execute the code for case 2: and hence the output. And then it breaks after executing case 2, as it encounters a break.

So, change your swich-case to: –

switch(i)
{
        case 0: break;
        case 1: break;
        case 2:
            System.out.println("Case 2 being executed");
            break;
        default:
            break;
}

to see the intended behaviour.