Java switch statement - using a variable, in a case, declared in another case

Question

I'm not new to Java but this is a thing that i have never seen before.
Supposing have a switch case, declaring a variable in case 0 makes the variable visible in the other cases that is a strange thing and i think is not in the java style.

public static void main(String[] args) {
    String _case = "case";

    switch (_case) {
        case "1":
            String foo = "foo";
            break;
        case "2":
            String bar = "bar";
            break;
        case "case":
            foo = "foo";
            System.out.println(foo);
            // System.out.println(bar); //Variable bar might not have been initialized
            break;
    }
}

If case 1 is not validated why and how the String foo variable becomes declared?

Execute the code online

I found the answer, I was surprisedl I could've sworn I was correct — Matthew Kerian, Jul 05 '19 at 19:49
yes it was very strange for me too seeing something like this :D — Roberto Manfreda, Jul 05 '19 at 20:00
Another question == another post... 1 question per post please. — Nic3500, Jul 06 '19 at 00:56

Ferdz · Accepted Answer · 2019-07-05T19:57:40.753

0

Because you set foo = "foo" inside of your last case, it guarantees the variable was instantiated at that point. There's no moment in which bar is instantiated in the last case, giving you this error. This is due to the fact that the scope of a variable in a case is equal to that of the entire switch statement. More info on other questions about this

edited Jul 05 '19 at 19:57

answered Jul 05 '19 at 19:38

Ferdz

1,182
1
13
31

Yes i know, but i declared the foo in the case one! Why it's visible in the last case? This is my question – Roberto Manfreda Jul 05 '19 at 19:46
2

@RobertoManfreda The scope of a variable in a case is equal to the *entire* switch statement – Ferdz Jul 05 '19 at 19:51

Java switch statement - using a variable, in a case, declared in another case

1 Answers1