Fastest way to set all values of an array?

Question

I have a char [], and I want to set the value of every index to the same char value.
There is the obvious way to do it (iteration):

  char f = '+';   char [] c = new char [50];   for(int i = 0; i < c.length; i++){       c[i] = f;   } 

But I was wondering if there's a way that I can utilize System.arraycopy or something equivalent that would bypass the need to iterate. Is there a way to do that?

EDIT : From Arrays.java

public static void fill(char[] a, int fromIndex, int toIndex, char val) {         rangeCheck(a.length, fromIndex, toIndex);         for (int i = fromIndex; i < toIndex; i++)             a[i] = val;     } 

This is exactly the same process, which shows that there might not be a better way to do this.
+1 to everyone who suggested fill anyway - you're all correct and thank you.

Answer 1

Try Arrays.fill(c, f) : Arrays javadoc

Answer 2

As another option and for posterity I was looking into this recently and found a solution that allows a much shorter loop by handing some of the work off to the System class, which (if the JVM you're using is smart enough) can be turned into a memset operation:-

/*  * initialize a smaller piece of the array and use the System.arraycopy   * call to fill in the rest of the array in an expanding binary fashion  */ public static void bytefill(byte[] array, byte value) {   int len = array.length;    if (len > 0){     array[0] = value;   }    //Value of i will be [1, 2, 4, 8, 16, 32, ..., len]   for (int i = 1; i < len; i += i) {     System.arraycopy(array, 0, array, i, ((len - i) < i) ? (len - i) : i);   } } 

This solution was taken from the IBM research paper "Java server performance: A case study of building efficient, scalable Jvms" by R. Dimpsey, R. Arora, K. Kuiper.

Simplified explanation

As the comment suggests, this sets index 0 of the destination array to your value then uses the System class to copy one object i.e. the object at index 0 to index 1 then those two objects (index 0 and 1) into 2 and 3, then those four objects (0,1,2 and 3) into 4,5,6 and 7 and so on...

Efficiency (at the point of writing)

In a quick run through, grabbing the System.nanoTime() before and after and calculating a duration I came up with:-

• This method : 332,617 - 390,262 ('highest - lowest' from 10 tests)
• Float[] n = new Float[array.length]; //Fill with null : 666,650
• Setting via loop : 3,743,488 - 9,767,744 ('highest - lowest' from 10 tests)
• Arrays.fill : 12,539,336

The JVM and JIT compilation

It should be noted that as the JVM and JIT evolves, this approach may well become obsolete as library and runtime optimisations could reach or even exceed these numbers simply using fill(). At the time of writing, this was the fastest option I had found. It has been mentioned this might not be the case now but I have not checked. This is the beauty and the curse of Java.