CODEWITHSUNDEEP
javascriptloopsswitch-statement
Ka Mok
Ka Mok

Reputation: 1988

Should you use break in switch case loops?

Using FizzBuzz as an example, this switch case inside the for loop uses break after each case. 

for (num = 1; num <= 100000; num ++) {
    switch (true) {
        case (num % 15 == 0):
            console.log("FizzBuzz");
            break;
        case (num % 3 == 0):
            console.log("Fizz");
            break;
        case (num % 5 == 0):
            console.log("Buzz");
            break;
        default:
            console.log(num);
            break;
    }
}

In Mozilla's docs, they say that...

The optional break statement associated with each case label ensures that the program breaks out of switch once the matched statement is executed and continues execution at the statement following switch. If break is omitted, the program continues execution at the next statement in the switch statement.

I'm assuming what they mean is that when the script checks that num % 15 == 0, if there's no break and eventhough it matches true, it goes to num % 3 ==0 and so on. So then I did two tests with 100,000 numbers. One with the breaks and one without them.

I found about 2 seconds difference between break and no break, with break shorter and faster. But that's over many iterations.

My question is, does it really matter for most developers building small scale apps to use breaks in their loops? Is it a matter of habit and preference? Thoughts?

Upvotes: 1

Views: 4585

Answers (4)

Chris Poe
Chris Poe

Reputation: 808

To elaborate on Callum Morrisson's answer:

One case where you can remove break gracefully is when you are returning a value in a case.

For example,

switch(expression) {
    case x:
        return *something*
    case y:
        return *something different*
}

This does not cause fallthrough due to return breaking you out of the statement. You can think of return as doubling for a break in this example.

If you were to run a linter on a switch statement with both a return and a break within a case, you would receive an unreachable code error since the return is breaking you out of the statement before the break has a chance to execute.

// eslint
switch(expression) {
    case x:
        return *something*
        break; // error unreachable code
    case y:
        return *something different*
        break; // error unreachable code
}

Upvotes: 0

Mike Robinson
Mike Robinson

Reputation: 8945

If you do not include a break statement, you have undoubtedly introduced a bug(!) into your program!

Here's why:

If you do not include a break statement, the logic will "simply fall-through" into the code for the additional conditions . . . but it will not(!) evaluate those conditions! It will simply crash right through, like the proverbial bull in the proverbial china-shop.

In other words, without the break statements, the case for (num % 15 == 0) would console.log() four(!!) (not "one") messages. Similarly, (num % 3 == 0) would log three, and so on.

"Heh... Don't go there."

===== EDIT: It may be helpful to consider that the original ("C...") concept of switch was that of "a multi-way goto" occurring at the top of the construct, which branched to one of the case-points within the body. The break statement generated a goto that sent you to the end of the switch. If you did not include it, the "helpful" assumption was that you intended to do what otherwise would happen:   that one case would simply "fall through" to the next one. (Which, absent break, is precisely what it does.) "Caveat Coder!"

Upvotes: 1

user3282374
user3282374

Reputation:

When a case condition is met in a switch statement it will execute all of the code blocks within that switch statement until it hits an exit command (think break or return). So the break is not only changing performance speed it's also changes functionality.

An easy way to think about switch statements is as alternative's to chained if-else statements.

your switch code can just as easily be written 

if (num % 15 == 0) {
   console.log("FizzBuzz");
} else if (num % 3 == 0) {
    console.log("Fizz");
} else if (num % 5 == 0) {
    console.log("Buzz");
} else {
    console.log(num);
}

if you remove the break; statements you'll change how the code actually functions. It will perform more like 

if (num % 15 == 0) {
   console.log("FizzBuzz");
}

if (num % 3 == 0 || num % 15 == 0) {
    console.log("Fizz");
}

if (num % 5 == 0 || num % 3 == 0 || num % 15 == 0) {
    console.log("Buzz");
} 

console.log(num);

The main difference being the lack of an else condition.

Upvotes: 12

Eli Sadoff
Eli Sadoff

Reputation: 7308

You get a different result with and without break. Without a break it passes through so on num = 15 for example it will print out "FizzBuzz", "Fizz", "Buzz", and "15". Whereas with break it will just print out "FizzBuzz". Having said that you don't need a break after the default case, if it is at the end of the switch.

Upvotes: 2

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