This question already has answers here:
Switch statement fall-through...should it be allowed? [closed]
(12 answers)
Closed 9 years ago.
After evaluating a case in a switch statement in Java (and I am sure other languages) the following case's are also evaluated unless a control statement like break, or return is used.
I understand this is probably an implementation detail, but what is/are the reasons for having this functionality happen?
Thanks!
Because it is useful to "fallthrough" from one case to another. If you don't need this (as is often the case), a simple break will prevent this. On the other hand, if case didn't fallthrough by default, there wouldn't be any easy way to do that.
It saves me a lot of duplicated code when the hardware changes.
void measureCPD (void) {
char setting;
switch (DrawerType) {
case 1:
setting = SV1 | SV7;
break;
case 0:
case 2:
setting = SV1;
break;
case 5:
PORTA |= SVx;
case 3:
case 4:
setting = SV1 | SV7;
break;
default: // Illegal drawer type
setting = SV1;
}
SetValves (setting);
}
It's because the case labels are goto destination labels.
There are times where you might want to have multiple case statement execute the same code.
So you would fall through and do the break at the end.
I tend to think of it in analogy to assembly programming, the case's are labels where you jump into, and it runs through the ones below it (program flows from up to down), so you will need a jump (break;) to get out.
Related
I saw this code in the OCA Java 8 exam study guide which confused me.
int dayOfWeek = 5;
switch(dayOfWeek) {
case 0:
System.out.println("Sunday");
default:
System.out.println("Weekday");
case 6:
System.out.println("Saturday");
break;
}
The book says it prints Weekday and Saturday.
Why is this?
I thought the case had to be found to get selected since it is 5 is is not selected so nothing should happen.
As no case matches value of dayOfWeek, the default case is executed: printing "Weekday"
As the default case has no break, the next case that follows it is also executed, until it reaches end of switch or a break: printing "Saturday"
Simple: because you "fall" through cases. If you want to not fall through, you have to use break.
So the default is really that you execute all cases following the first match - until you hit the first break statement.
You can mention the cases for which you want to explicitly do something.
Otherwise, for all the remaining cases they will be handled in default.
Also, check the related Must the "default" case come last in a switch? for more explanation.
And as others have also pointed here, if you don't put break then it will go through all the remaining cases too without fail till it encounters a break.
I have a primitive question but I can't actually recall the answer.
What if I use Switch Cases instead of IF\ELSE IF. Will all cases get evaluated or break will break out of the whole switch cases and return the first satisfied condition only.
In other words, what if I want to check if someone has a car, bike and plane. And in my particular case someone has all of them would switch return all three as true or it will return the first case only and ignore the rest because of the break?
Sorry for inconvenience.
Thank you
From the Official Java Tutorials :
Each break statement terminates the enclosing switch statement.
In your particular case, if someone has a car, bike or a plane, you should construct more complex if\else statement.
But if you still prefer to use switch, you can do the following:
switch (possession)
{
case CAR:
case BIKE:
case PLANE:
doSomething();
break;
}
It will break out at the first matching. the best thing to do in your case is using logical operators in if statements.
if (hasCar && hasBike && hasPlane)
{
...
}
The break terminates the switch-statement.
Besides, switch evaluates a single variable, your case sounds a bit more complex to me.
certainly break will break the switch-case flow after encountering first break statementif no break is found then it will start the execution of all statement starting from first matching case,you can implement the logic inside case.and one thing more switch case is little bit faster than if-else
please see Why switch is faster than if
With switch the first case it finds a match runs and then all the following cases regardless of matching or not, provided you don't use break. By using break,only the actual match case runs and it is almost always only one. Therefore I do not regard in your problem using switch as a good approach, since you can handle it better with if-else.
Perhaps you would need to rewrite a new style of Data Structure to confirm this. The easyest way that I think is with a Boolean Array (bool[]);
Continuing with your example if you has the
Owner Class
, with
bool bike, bool car, bool motocycle... properties, witch you can write a
public bool[] ownArray()
function, that can be evaluated.
I hope this helps.
myEnum = 3;
...
switch (myEnum) {
case 1: System.out.println("This will not print"); break;
case 2: System.out.println("This will not print"); break;
case 3: System.out.println("This will print!"); //no break!
case 4: System.out.println("So will this, because there wasn't a break"); break;
case 5: System.out.println("But this won't, there was a break");
}
Switch-case in java could evaluate one case at a time. The moment it gets into one of the case, it would keep on executing the next statements (no matter, if those statement were part of other case ); until it hit a break;
In nutshell, A switch statement gives you the option to test for a range of values for your variables. They can be used instead of long, complex if … else if statements. The structure of the switch statement is this:
switch ( variable_to_test ) {
case value1: code_here1;
break;
case value2: code_here2;
case value3: code_here3;
break;
default:
values_not_caught_above;
}
In the above example if value2 is satifies, it will execute code_here2 and code_here3 as well (because there is no break statement)
Sadly, you can't test for a range of values after case, just the one value. So you can't do this:
case (user <= 18):
This question already has answers here:
Why do we need break after case statements?
(17 answers)
Closed 9 years ago.
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.
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.
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.
in JavaScript and also I've seen it in PHP : You can use a logical expression inside you cases :
For Example :
switch(true){
case (d<10):
document.write("Less than 10");
break;
case (d==10):
document.write("Equal to 10");
break;
case (d>10):
document.write("Greater than 10");
break;
default:
document.write("Some dumb error. Probably not a number.");
}
I was wondering if we can do this in Java ... say I needed a to compare a very large range.
case 1 :
case 2 :
...
case 5000:
case 5001:
default:
It would be tedious to do. I could wrap my cases inside if statements but i was wondering it could be done like in JavaScript or PHP.
It is nice to have, but you cannot. Quoting JLS:
The body of a switch statement is known as a switch block.
Any statement immediately contained by the switch block may be labeled with one or more switch labels, which are case or default labels.
These labels are said to be associated with the switch statement, as are the values of the constant expressions (§15.28) or enum constants (§8.9.1) in the case labels.
These labels have to be constant expressions or enum constants.
No You cannot do that using switch in java. You have to use if-else to accomplish this.
See this link for more details.
I'm using int[] arrays as a reference. I wondering if my use of case statements is sound or if it will cause errors down the line.
This is my code:
int switcheroo = intarray[0];
int foo = intarray[1];
boolean size = false;
boolean biggersize = false;
switch (switcheroo) {
case 0:
switch (foo) {
case 1:
doSomething(switcheroo); //change switcheroo somehow.
break;
case 2:
doSomethingElse(switcheroo); //change switcheroo differently.
break;
}
case 1:
size = true;
break;
case 2:
biggersize = true;
break;
default:
break;
}
Unless it's a coincidence, this is working to ripple the changes from the nested case statement into the other cases, as I want.
My questions are:
Will this nesting causes trouble further down the line?
Is the lack of a break; after a case bad practice?
Thanks.
Edit: The methods which change switcheroo in the middle of the switch statements were put there for responses to that. I will not be doing this is my program.
Nesting won't cause trouble down the line exactly, but it can be confusing to read. Adding comments and/or other documentation will really help future coders (and yourself in a week!) understand this by looking at it.
The lack of a break isn't a bad practice by itself, but it is something that -most- case statements have, so I would add a comment at the end like // no break, allow fall-through.
So both cases boil down to good documentation.
These points are perpendicular to the fact that I don't think this code does what you think it will do.
The case clause is not reevalauted every time you come across one - they're just points to jump to for the switch. So in your example, you will always end up in case 1 if you start at case 0 - you'll never end up at case 2 from case 0.
If I were to restructure this, here's what I would do. Instead of using int, I would use enum:
enum Foo { GOOD_FOO, BAD_FOO }
enum Switcharoo { BAR, BAZ, BAQ, ESCAPE }
enum Size { NONE, REGULAR, BIGGER }
Foo foo = ... // assigned somewhere
Switcharoo roo = ... // assigned somewhere
Size size = NONE;
// use a while loop to reevalulate roo with each pass
while(roo != Switcharoo.ESCAPE) {
switch(roo){
case BAR:
switch(foo) {
case GOOD_FOO: foo = doSomething(foo); break;
case BAD_FOO: foo = doSomethingElse(foo); break;
}
break;
case BAZ:
roo = Switcharoo.ESCAPE;
size = Size.REGULAR;
break;
case BAQ:
roo = Switcharoo.ESCAPE;
size = Size.BIGGER;
break;
}
}
RE: Is the lack of a break; after a case bad practice?
If you don't include a break after each case statement the flow will continue trough the next statements, so the code for more than one option will be executed.
From the Java Tutorial:
Another point of interest is the break statement. Each break statement
terminates the enclosing switch statement. Control flow continues with
the first statement following the switch block. The break statements
are necessary because without them, statements in switch blocks fall
through: All statements after the matching case label are executed in
sequence, regardless of the expression of subsequent case labels,
until a break statement is encountered.
Check http://docs.oracle.com/javase/tutorial/java/nutsandbolts/switch.html to learn more details about switch statements.
In case foo is neither 1 nor 2, is switcheroo supposed to continue to case 1? If yes, your code is correct. If not, you need to add a break before case 1
switch operator is a kind of conditional if operator. And a bunch of if operators may report that you need to involve object polymorphism in your code instead of making sequental if-switch operators.
Consider re-arranging your data to classes/objects and process them using polymorphism approach. It will make your code more reliable, more manageable and just more beautiful.
Im not sure if i understood your first question, can you give me some more information (concrete maybe)?
Concerning the second question, there's nothing wrong about not having a break statement after a case, as long as you know that the code will continue to execute till it finds a break statement.
well..It seems the only thing you lack here is readability. Do this type of nesting only if you are forces to do.
Is the lack of a break; after a case bad practice?
not bad rather it is a good practice to avoid unsual behaviour. Your code will continue executing to your last case in absensce of this Break.
Add break in your first outer case too.
I'd recommend avoiding nesting and fall-through in combination. At least personally, I tend to expect that every case ends with a break, so I'd use fallthrough only in cases where it's very easy to spot, and leads to clearer code than an if..else, such as:
switch (foo) {
case 0:
case 1:
doSomething();
break;
case 2:
doSomethingElse();
break;
default:
break;
}
That is, with either empty or very simple (one statement) case bodies. In your case, refactoring the body of the first case into a method would probably work:
void changeSwitcheroo(int foo, int switcheroo) {
switch (foo) {
case 1:
doSomething(switcheroo); //change switcheroo somehow.
break;
case 2:
doSomethingElse(switcheroo); //change switcheroo differently.
break;
}
}
// ...
int switcheroo = intarray[0];
int foo = intarray[1];
switch (switcheroo) {
case 0:
changeSwitcheroo(foo, switcheroo);
case 1:
size = true;
break;
case 2:
biggersize = true;
break;
default:
break;
}
(This is mostly arguing readability and style. The meaning of the switch will still be "if switcheroo is 1, jump into the middle of case 0.)
If there are only 2 alternatives, prefer an if/else. Fewer lines of code and no break issues.
Will this nesting causes trouble further down the line?
A problem is that such switch nesting is hard to read: I had to look at the code twice before I noticed that you were using a nested switch. This can be a maintenance issue, but that isn't a reason not to use a nested switch, just use this sparingly, adjust your indentation to clarify what's going on, and comment on the usage.
Is the lack of a break; after a case bad practice?
Falling through by omitting the break can also lead to maintenance/debugging issues, but I don't consider it bad practice. It is, after all, intrinsic in the design of the switch statement. A fall through comment is always welcome so that the next guy - or you, revisiting your code months later - knows that the fall-through is intentional.