Why "switch" in java does check more cases than required?

Question

I already searched in stackoverflow for reasons concerning the question, why switch does not break after having found a matching case.

For example, I have written some code which simulates a dice. For every throw, I count up for statistical reasons (one, ... = int).

switch (actualThrow) {
case (1): one++; 
case (2): two++; 
case (3): three++; 
case (4): four++;
case (5): five++; 
case (6): six++; 
}

I DO know, that case(3)-case(6) will be checked, if my throw was a 3 (Why do we need break after case statements?) and it will be prevented by inserting break;, but I want to understand, why obviously the following cases seem to be checked, but execute, although the condition is not fulfilled.

The various cases in a switch *fall through* until the switch ends or they reach a break. The condition check only determines the *starting* case. — thegrinner, Jun 07 '13 at 20:33
So you understand that you need `break`s but want to know why the subsequent cases are executed? I'm sorry, but I don't understand the question. — jerry, Jun 07 '13 at 20:34
i understand, that subsequents are executed, so i need `break`s. i do not understand, WHY the checking fails. — Gary Klasen, Jun 07 '13 at 20:38
Because there is no checking after the jump to the matching case. The cases are merely entry points. — Andy Thomas, Jun 07 '13 at 20:40
Voting to reopen. None of the answers in the duplicate cite the JLS that explains why this is the case. — Brian Roach, Jun 07 '13 at 20:40
It's not checking anymore and it's not failing. It looks for the first label that matches your `actualThrow` and continues until it finds the end of statement (for the switch statement) or a `break;`. That's the behavior given to a switch statement. — BLaZuRE, Jun 07 '13 at 20:40
Because (1): there can be a "default" that needs to be executed; (2): there can be more than one case per block (ie, case 1: case 2: case 3: lowroll++;); and, (default): it is what it is. — michael, Jun 07 '13 at 20:42
@michael_n I disagree with part (1) of your comment. I don't think the fallthrough is simply because there *might* be a `default:` label. Also, a `default:` statement can be placed at the top, it's simply convention we put it at the bottom. — BLaZuRE, Jun 07 '13 at 20:49
@BLaZuRE I agree with you; I frankly don't completely understand what the questioner is really looking for, other than some way to better understand why it is the way it is; ie, more practical examples rather than a spec clarification. — michael, Jun 07 '13 at 20:54
@BLaZuRE based on the accepted answer, sounds like my case #2 is really what was needed to clarify the situation. — michael, Jun 07 '13 at 20:59

score 5 · Accepted Answer · answered Jun 07 '13 at 20:33

Imagine this scenario:

switch(trafficLight.state) {
    case GREEN:
        car.drive();
        break;
    case YELLOW:
        car.drive();
        break;
    case RED:
        car.stop();
        break;
}

Seems kinda wasteful since GREEN and YELLOW both do the same thing. Java lets us consolidate so that multiple conditions take the same action:

switch(trafficLight.state) {
    case GREEN:
    case YELLOW:
        // Now both GREEN and YELLOW will use this code
        car.drive();
        break;
    case RED:
        car.stop();
        break;
}

Less code, same effect. That's all. That's the simple case mind you, there are other ways to use this that are less obvious and more prone to abuse.

Shouldn't it be `car.floorIt()` for `case YELLOW` ? – splungebob Jun 07 '13 at 20:39 — splungebob, Jun 07 '13 at 20:39

score 0 · Answer 2 · answered Jun 07 '13 at 20:34

When parsing through a switch-case statement, each case is simply a label. If anything matches a case then it will just go through until it either finds a break or the end of the statement (a }).

It is similar to the Java opcode (assembly code essentially). You use labels to jump to. Besides that, they don't really serve a purpose in the code.

Why "switch" in java does check more cases than required?

2 Answers2