Fast way for making the sum of an integer and each element of an array

Question

I would like to optimize the below code that is very simple, Do you have any idea?

public void addValueToTab(int _valueToAdd){
    for(int k=0; k < myArray.length; k++){
        myArray[k]+= _valueToAdd;
    }
}

myArray is an int[]

the length of myArray can vary between 50 to 1000.

Thanks a lot for your help,

Loop unrolling. Counting down rather than up (testing against 0 is faster than testing against other values). Outside of that, you're running up against the limits of the language. Important question: Have you profiled your code and determined that you're spending enough time in these few lines to justify focusing on them rather than on higher-level algorithms? — keshlam
– keshlam, Commented Jan 15, 2014 at 3:26
@keshlam i've tested. JIT vectorizes only the simpliest loop, it is really the fastest option — leventov
– leventov, Commented Jan 15, 2014 at 8:47
Interesting. I keep forgetting that vectorization's now a factor. That re-emphasizes that micro-optimization is usually the wrong place to start, and that any source-code-level optimization needs to be guided and checked with proper profiling tools. — keshlam
– keshlam, Commented Jan 15, 2014 at 13:45

maaartinus · Accepted Answer · 2014-01-15 03:47:18Z

5

I'd bet that with a server JVM and hot code, there's nothing you could do to speed it up (since the JVM does all the optimizations already). Especially, myArray.length gets cached in a local variable and the loop gets unrolled several times.

In case you care about performance with a weaker JIT compiler, you could do the two above things manually, but why should you care?

A side note

Interestingly, there's a similar loop where the JVM is pretty loosy:

for (int i = 0; i < value.length; i++) {
    h = 31 * h + val[i];
}

It's not an unimportant piece of code, it's the body of String.hashCode. By unrolling it manually you can get a speedup of factor 2+.

answered Jan 15, 2014 at 3:47

maaartinus

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Rominus Over a year ago

Thank you for your help on this one.

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