Java variable declaration efficiency

Question

As I understand, in case of an array, JAVA checks the index against the size of the Array. So instead of using array[i] multiple times in a loop, it is better to declare a variable which stores the value of array[i], and use that variable multiple times.

My question is, if I have a class like this:

public class MyClass(){

    public MyClass(int value){
      this.value = value;
    }

    int value;
}

If I create an instance of this class somewhere else: (MyClass myobject = new MyClass(7)), and I have to use the objects value multiple times, is it okay to use myobject.value often or would it be better to declare a variable which stores that value and use that multiple times, or would it be the same?

This is exactly the kind of micro-optimization that you should not waste a second thinking about until you have a profiler telling you it's in a hot spot of your application. — Michael Borgwardt
– Michael Borgwardt, Commented Sep 2, 2012 at 21:42

Keppil · Accepted Answer · 2012-09-02 21:19:37Z

4

In your case, it wouldn't make any difference, since referencing myobject.value is as fast and effective as referencing a new int variable.

Also, the JVM is usually able to optimize these kinds of things, and you shouldn't spend time worrying about it unless you have a highly performance critical piece of code. Just concentrate on writing clear, readable code.

answered Sep 2, 2012 at 21:19

Keppil

46.3k9 gold badges101 silver badges121 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

SJuan76 · Accepted Answer · 2012-09-02 21:21:11Z

2

The short answer is yes (in fact, in the array case, it does not only have to check the index limit but to calculate the actual memory position of the reference you are looking for -as in i=7, get the base position of the array and add 7 words-).

The long answer is that, unless you are really using that value a lot (and I mean a lot) and you are really constrained due to speed, it is not worth the added complexity of the code. Add to that that the local variable means that your JVM uses more memory, may hit a cache fault, and so on.

In general, you should worry more about the efficiency of your algorithm (the O(n)) and less about these tiny things.

answered Sep 2, 2012 at 21:21

SJuan76

24.9k7 gold badges52 silver badges95 bronze badges

2 Comments

nes1983 Over a year ago

Nonsense. There is no speed difference.

rgamber Over a year ago

@nes1983 So the JVM takes care of these things completely. And there will be no speed difference even for the array[i] case.

nes1983 · Accepted Answer · 2012-09-03 06:28:42Z

The Java compiler is no bozo. He will do that optimization for you. There is 0 speed difference between all the options you give, usually.

I say 'usually' because whether or not accessing the original object or your local copy isn't always the same. If your array is globally visible, and another thread is accessing it, the two forms will yield different results, and the compiler cannot optimize one into the other. It is possible that something confuses the compiler into thinking there may be a problem, even though there isn't. Then it won't apply a legal optimization.

However, if you aren't doing funny stuff, the compiler will see what you're doing and optimize variable access for you. Really, that's what a compiler does. That's what it's for.

You need to optimize at least one level above that. This one isn't for you.

Collectives™ on Stack Overflow

Java variable declaration efficiency

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related