I would like to do something like this:
int i = 0;
switch(difficulty) {
case 1: i++; break;
case 2: i--; break;
default: case 1;
}
Is something like this possible? I want to prevent duplicate code. I know that in this particular example there is little reason to do this, since the duplicated code would be small. The only thing I can come up with is something like this [using the fall through abilities of switch cases]:
switch(difficulty) {
case 2: i--; break;
default:
case 1: i++; break;
}
I would rather not do it like this, since it would make more sense to increase case numbers and have the default at the bottom.
But I wonder that if I would do this, would it mess up with the goto statements under the hood? In particular, will it not take longer to decide which goto statement to use since the numbers or out of order? Does the order matter in the switch statement? Imagine all cases have the same odds of being called, would it matter if you had them in random order instead of linear order?
[Edit: for my side question about efficiency I found this: Does the order matter of switch-statements, the short answer is no: Does switch case order affect speed?
How does Java's switch work under the hood?
This should suit your needs:
switch(difficulty) {
case 2: i--; break;
default: i++; break;
}
This is probably what you need/want, and is valid code
int i = 0;
switch (difficulty) {
default:
case 1: i++; break;
case 2: i--; break;
}
As far as I know, the order of the cases does not alter in any way the switch statement, either your parameter is fit for a case or it isn't, so it doesn't matter, in your case, which case you have first.
You can see the documentation as reference if you wish.
https://docs.oracle.com/javase/tutorial/java/nutsandbolts/switch.html
(moved from comment, since it seemed to be helpful)
Add a private method and call it from both case 1 and default, that way you won't have duplicated code.
No, in switch statement order doesn't matter at all.
Also the compiler always maintain a hash table for jumping to the switch statement. So ordering or increase of casing is not a matter at all. It's always a O(1) operation.
However, this will reduce the switching for your case:
switch(difficulty) {
case 2: i--; break;
default: i++; break;
}
Related
I have two sets of conditions. One has two possible values and the other has more (in this example I wrote 3 cases but there could be up to 8). Which code runs faster and is less error prompt (more accurate)?
code a)
if (letter.equals(a)) {
switch (number) {
case 1:
.........
case 2:
.........
case 3:
.........
}
} else if (letter.equals(b)) {
switch (number) {
case 1:
.........
case 2:
.........
case 3:
.........
}
}
code b)
switch (number) {
case 1:
if (letter.equals(a)) {
.........
} else if (letter.equals(b)) {
.........
}
case 2:
if (letter.equals(a)) {
.........
} else if (letter.equals(b)) {
.........
}
case 3:
if (letter.equals(a)) {
.........
} else if (letter.equals(b)) {
.........
}
}
Please tell me if you think there is a better option other than these two. (I could also create a parameter that gets both letter and number and create 6 cases using that.)
Thank you in advance!
Which code runs faster and is less error prompt (more accurate)?
Answer: In this specific case performance is not the issue. Because no matter how you are going to use this actual execution numbers are same. But you can improve the code readability and making it less errorprone.
Instead of worrying about performance, start with SOLID principle. Why don't you just break this big method into some smaller method which has a concrete responsibility. It will make code more beautiful and less error prone. For example:
Methods:
void processA(int number){
switch (number) {
case 1:
.........
case 2:
.........
case 3:
.........
}
}
void processB(int number){
switch (number) {
case 1:
.........
case 2:
.........
case 3:
.........
}
}
///
now from the main method you could simply call:
if (letter.equals(a)) {
// call the method which will process A
processA(number);
} else if (letter.equals(b)) {
// call the method which will process A
processB(number);
}
This is an example of a micro-optimisation that probably has no effect on your code, so 'which is faster' doesn't really matter here.
If you want to find out which of a particular combination of cases in your codebase then you probably want to use a benchmarking tool like JMH to figure out which is more appropriate for you. But unless this code is on the critical path it won't make a difference.
Code needs to be:
Correct
Readable
Performant
It's much better to focus on 1 & 2 than the last.
In this particular case, a switch-over-numbers is probably faster for the JVM under the covers, while switch-over-expressions will naturally involve more jumping around in the branches. I would therefore naively expect switch-over-int-followed-by-if will marginally edge out the other way around, but without proof this would be a hunch and not something to rely on. It's also quite likely that whatever benchmark is written there will be some kind of flaws, such that the JVM is automatically promoting the optimal test case combination to the first test, in which case it won't tell you what you think it is doing.
Ultimately, write it for readability/maintainability rather than performance, until you can prove this is on the hot loop, and then measure it explicitly.
There is another option:
switch (String.format("%1s%1d", letter, number) {
case "a1":
...
case "b1":
...
case "a2"
...
case "b2":
...
case "a3":
...
case "b3":
...
}
Better? Maybe, maybe not. But it is an alternative...
For class I'm working on an interpreter, currently working on the scanner. Being a class that will be called many many times, I would like for it to be optimized for speed. In the scanner, to categorize an operator you have to compare the current token to the 6 or so operators. What method is best for speed, but for also for readability.
Many cases in an if statement
Loop through a char array of every operator and compare
Switch statement
These are the only cases I could think of. Which is best, or if you have a better approach please share. I implemented #2 because it takes up the least amount of lines of code.
Any sensible hand-written scanner is based on a switch statement. Note that if you return special characters directly to the parser as themselves, you can economize on case actions:
switch (ch) // the next incoming character
{
case '+':
case '-':
case '*':
case '/':
case '%':
// etc.
return ch;
case 'A':
case 'B':
// ...
case 'Z':
case 'a':
case 'b':
// ...
case 'z':
// start of an identifier: accumulate it, with a do/while loop,
// save it somewhere, return IDENTIFIER
return IDENTIFIER;
case '0':
case '1':
// ...
case '9':
// start of a numeric literal: ...
return NUMERIC_LITERAL;
// etc.
}
For just a few items, the difference is small. If you have many items you should definitely use a switch.
If a switch contains more than five items, it's implemented using a lookup table or a hash list. This means that all items get the same access time, compared to a list of if where the last item takes much more time to reach as it has to evaluate every previous condition first.
And also as #Qix commented switch it more readable.
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):
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.
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.