Is it better to access several times a value at array[i] or to stock this value in a variable (v=array[i]) and access this variable v instead?

Question

I'd like to know if it costs more to access several times a value at array[i] or to stock the value as a new variable (v=array[i]) and access the variable v several times instead ?

for example (in java), would it be better to write :

int [] array = {1,2,3,4,5};

for(int i=0;i<5;i++){
    if (array[i]<0){
        System.out.println("negativ");
    }else if(array[i]>0){
        System.out.println("positiv");
        if (array[i]==42){
            System.out.println("great answer");
        }
    }else{
        System.out.println("zero");
    }
}

or

int [] array = {1,2,3,4,5};
int v;

for(int i=0;i<5;i++){
    v = array[i];
    if (v<0){
        System.out.println("negativ");
    }else if(v>0){
        System.out.println("positiv");
        if (v==42){
            System.out.println("great answer");
        }
    }else{
        System.out.println("zero");
    }
}

Thank you for your help.

It isn't better or worse. Prefer whichever you find most readable. — Elliott Frisch, Jun 12 '17 at 23:41
Thank you for your answer. So it doesn't affect the cost to use one way or an other ? — Mireille, Jun 12 '17 at 23:45

score 1 · Answer 1 · answered Jun 12 '17 at 23:54

You end up taking two more bytes with the second form:

  34: istore_2
  35: iload_2

Variable access ends up being an iaload vs an iload_2 - not a huge difference. If you had enough variables (more than 4) that you had to use iload (index) rather than iload_ then you would see even larger code for the second form.

To be explicit: you don't save anything by introducing the variable, and in fact add instructions (where the variable gets set).

Zakir · Answer 2 · 2017-06-12T23:56:08.700

0

Both the styles have their pros and cons
For array[i] access it might be little cumbersome to read but is a safer bet. In your example you are only reading from the array, think of a case where you have to write as well. You might end up updating just v and that might prove to be a quite costly bug !!

Another point to note is your example is about primitive array hence the storage/access overhead I guess is pretty negligible

edited Jun 12 '17 at 23:56

answered Jun 12 '17 at 23:48

Zakir

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Thank you very much for your answers, this has been very helpful :-) – Mireille Jun 13 '17 at 04:54
Glad that it helped you - would you mind accepting/up voting it - might help others reaching here with the same question – Zakir Jun 13 '17 at 16:58
I tried to up vote it but I don't have enough of reputation for my votes to be counted yet (sorry). I'll come back once I'll get some :-) – Mireille Jun 13 '17 at 21:15

K.Nicholas · Answer 3 · 2017-06-13T00:14:26.710

0

Neither. The java language has the enhanced for loops optimization for this:

int [] array = {1,2,3,4,5};

for(int v: array){
    if (v<0){
        System.out.println("negativ");
    }else if(v>0){
        System.out.println("positiv");
        if (v==42){
            System.out.println("great answer");
        }
    }else{
        System.out.println("zero");
    }
}

edited Jun 13 '17 at 00:14

answered Jun 12 '17 at 23:58

K.Nicholas

10,956
4
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`for(int v: arrays)` should be `for(int v: array)` – Fureeish Jun 13 '17 at 00:09
the loop was not the question.. it was variable assignment and usage – AnthonyJClink Jun 13 '17 at 00:11
1

The variable is set in the loop. – K.Nicholas Jun 13 '17 at 00:17

user207421 · Answer 4 · 2017-06-13T00:27:31.280

0

The JVM will optimize both to the same thing so it makes no difference. Prefer the most readable.

edited Jun 13 '17 at 00:27

answered Jun 13 '17 at 00:19

user207421

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1

@Tom Oops, missed that :-| – user207421 Jun 13 '17 at 00:27

Is it better to access several times a value at array[i] or to stock this value in a variable (v=array[i]) and access this variable v instead?

4 Answers4