if else or switch case

Question

I need to check a small piece of logic and would highly appreciate if someone can give me some valuable input.

I have two ways of checking my logic and want to know which is more efficient.

1st way:

if(url.equalsIgnoreCase("1")){
    url = "aaa";
}
else if(url.equalsIgnoreCase("2")){
    url = "bbb";
}
else if(url.equalsIgnoreCase("3")){
    url = "ccc";
}
else if(url.equalsIgnoreCase("4")){
    url = "ddd";
}
else if(url.equalsIgnoreCase("5")){
    url = "eee";
}
else if(url.equalsIgnoreCase("6")){
    url = "fff";
}

2nd Way:

int temp = Integer.parseInt(url);
switch (temp) {
case 1:
    url = "aaa";
    break;
case 2:
    url = "bbb";
    break;
case 3:
    url = "ccc";
    break;
case 4:
    url = "ddd";
    break;
case 5:
    url = "eee";
    break;
case 6:
    url = "fff";
    break;
}

Please let me know which is more efficient. Is it bad to use Integer.parseInt(string)?

@micro.pravi .. its partially duplicate of your suggested link. I mainly wanted to know if adding Interger.parseInt() will make it more efficient or not..so only I posted a fresh question.. — Sushil, Sep 06 '13 at 10:02

score 30 · Accepted Answer · edited Sep 06 '13 at 08:43

30

If your values really are 1-6, the clearest and most efficient way is using an array :

String[] URLS = {...};
url = URLS[Integer.parseInt(url) - 1];

edited Sep 06 '13 at 08:43

Philipp Sander

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answered Sep 06 '13 at 08:23

njzk2

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10

+1 for thinking out of the box :) – Suresh Atta Sep 06 '13 at 08:24
2

i have to admit: smooth... but this is not really answering his questions (still +1) – Philipp Sander Sep 06 '13 at 08:34
1

Most of the answers are correct, but this is really out of box thinking :) – Sushil Sep 06 '13 at 10:03
While some might like this solution and praise it for its originality (that comes mostly from not answering the question), it might not be the most efficient one when it comes to compiler optimizations. – Hermes Mar 01 '17 at 11:00

Philipp Sander · Answer 2 · 2013-09-06T08:25:47.690

14

Please let me know which is more efficient.

a switch statement is more efficient is your case

Is it bad to use Integer.parseInt(string)?

No. it's fine. but when you're using java7 you can use String-constants values in your switch cases but not on Android.

Asside from the efficiency: switch looks cleaner in most cases.

edited Sep 06 '13 at 08:25

answered Sep 06 '13 at 08:19

Philipp Sander

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(but java 7 is not yet on android) – njzk2 Sep 06 '13 at 08:22
@njzk2 Tagged java :) – Suresh Atta Sep 06 '13 at 08:23
@sᴜʀᴇsʜᴀᴛᴛᴀ : but tagged android too :) – njzk2 Sep 06 '13 at 08:24
i know that it's not on android. i just wanted to mention that it works on java7 – Philipp Sander Sep 06 '13 at 08:26
1

just wanted to make sure that's clear for people reading your answer. (would certainly be nice to finally have string switches in android, though) – njzk2 Sep 06 '13 at 08:35

score 4 · Answer 3 · edited May 23 '17 at 12:10

4

In terms of efficiency check this: Case vs If Else If: Which is more efficient?, but Way 2 looks to be a more clean and readable code.

edited May 23 '17 at 12:10

Community

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answered Sep 06 '13 at 08:19

Nargis

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score 3 · Answer 4 · answered Sep 06 '13 at 08:18

3

As a general rule, the switch statement produces more efficient bytecode. With Java 7, switch statements with String was introduced, so you don't need to cast it.

answered Sep 06 '13 at 08:18

christopher

26,815
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score 3 · Answer 5 · answered Sep 06 '13 at 08:22

3

In this case, switch is more efficient.

In fact, If you are using Java7, you may directly use string case rather than using Integer.parseInt().

answered Sep 06 '13 at 08:22

Pramod Ravikant

1,039
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score 2 · Answer 6 · answered Sep 06 '13 at 08:23

It is not bad using parseInt but it will throw an exception if the string is not an integer.

otherwise, and I think you can see this for yourself, the switch/case is much more readible.

if your if construct would have a final else catching all other cases (including non numeric strings)

then you can do, assuming your ints are always positive

int temp = -1;
try {
    temp = Integer.parseInt(str);
} catch (NumberFormatException ex) {
    // ignore exception and use -1 as original value for default case
}

score 2 · Answer 7 · answered Sep 06 '13 at 08:27

Readability and debugability (is that even a word?) are rather subjective but some people (including me) find the switch statement to be more clear. However, in many cases a compiler has a better chance of generating faster code using a switch compared to the if else construct.

score 2 · Answer 8 · answered Sep 06 '13 at 08:28

The switch statement is faster for the following two reasons.

The switch statement is generally faster than if-else-if construct because the control is directly transferred to the respective case. While in the cases of if-else-if, the all the checks are going to be performed to reach the first matching condition. For example, if you assign temp = 6 the switch statement will execute the respective block directly. The if-else-if construct will iterate through all the conditions.
Calling the equalsIgnoreCase() is more costly than performing the equality check that happens at the background of the case matching.

score 1 · Answer 9 · answered Sep 06 '13 at 08:20

1

for such a long list is a switch statement definitely the better choice. The longer the junction is if the better cut off a switch in comparison

answered Sep 06 '13 at 08:20

JavaDM

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score 1 · Answer 10 · edited May 23 '17 at 10:31

1

Efficiency depends on the matching condition in both cases.Please conform your answer here.

edited May 23 '17 at 10:31

Community

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answered Sep 06 '13 at 08:26

NCA

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score 1 · Answer 11 · answered Sep 06 '13 at 08:27

I think an alternative approach would be to use a pre initialized map. It should have the (string)numbers as the key and the url results as value. Then you can simply do

url = map.get(Integer.parseInt(url));

This is probably even a much cleaner version then the long switch statement.

score 1 · Answer 12 · answered Sep 06 '13 at 08:28

The switch has more features than the if else, because you can fall through cases.

See for details: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/switch.html

class SwitchDemo2 {
    public static void main(String[] args) {

        int month = 2;
        int year = 2000;
        int numDays = 0;

        switch (month) {
            case 1: case 3: case 5:
            case 7: case 8: case 10:
            case 12:
                numDays = 31;
                break;
            case 4: case 6:
            case 9: case 11:
                numDays = 30;
                break;
            case 2:
                if (((year % 4 == 0) && 
                     !(year % 100 == 0))
                     || (year % 400 == 0))
                    numDays = 29;
                else
                    numDays = 28;
                break;
            default:
                System.out.println("Invalid month.");
                break;
        }
        System.out.println("Number of Days = "
                           + numDays);
    }
}

will output

Number of Days = 29

So it also depends on your implementation requirements which one you use.

if else or switch case

12 Answers12