设置数组所有值的最快方法？

我有一个 char []，并且我想将每个索引的值设置为相同的 char值。
有一种显而易见的方法可以做到这一点(迭代) :

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

但我想知道是否有一种方法，我可以利用 System.arraycopy或等价的东西，将绕过需要迭代。有办法吗？

编辑: 来自 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;
}

这是完全相同的过程，这表明可能没有更好的方法来做到这一点。
每个提出 fill建议的人 + 1-你们都是正确的，谢谢。

212160 次浏览
小开
最佳答案

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

小开

Use Arrays.fill

  char f = '+';
char [] c = new char [50];
Arrays.fill(c, f)
小开

Arrays.fill might suit your needs

小开

Arrays.fill(myArray, 'c');

Arrays.fill

Although it is quite possible that this is doing the loop in the background and is therefore not any more efficient than what you have (other than the lines of code savings). If you really care about efficiency, try the following in comparison to the above:

int size = 50;
char[] array = new char[size];
for (int i=0; i<size; i++){
array[i] = 'c';
}

Notice that the above doesn't call array.size() for each iteration.

小开

See Arrays.fill method:

char f = '+';
char [] c = new char [50];
Arrays.fill(c, f);
小开

If you have another array of char, char[] b and you want to replace c with b, you can use c=b.clone();.

小开 
   /**
* Assigns the specified char value to each element of the specified array
* of chars.
*
* @param a the array to be filled
* @param val the value to be stored in all elements of the array
*/
public static void fill(char[] a, char val) {
for (int i = 0, len = a.length; i < len; i++)
a[i] = val;
}

That's the way Arrays.fill does it.

(I suppose you could drop into JNI and use memset.)

小开

You could use arraycopy but it depends on whether you can predefine the source array, - do you need a different character fill each time, or are you filling arrays repeatedly with the same char?

Clearly the length of the fill matters - either you need a source that is bigger than all possible destinations, or you need a loop to repeatedly arraycopy a chunk of data until the destination is full.

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


char[] d = new char[50];
System.arraycopy(c, 0, d, 0, d.length);
小开

Java Programmer's FAQ Part B Sect 6 suggests:

public static void bytefill(byte[] array, byte value) {
int len = array.length;
if (len > 0)
array[0] = value;
for (int i = 1; i < len; i += i)
System.arraycopy( array, 0, array, i,
((len - i) < i) ? (len - i) : i);
}

This essentially makes log2(array.length) calls to System.arraycopy which hopefully utilizes an optimized memcpy implementation.

However, is this technique still required on modern Java JITs such as the Oracle/Android JIT?

小开

System.arraycopy is my answer. Please let me know is there any better ways. Thx

private static long[] r1 = new long[64];
private static long[][] r2 = new long[64][64];


/**Proved:
* {@link Arrays#fill(long[], long[])} makes r2 has 64 references to r1 - not the answer;
* {@link Arrays#fill(long[], long)} sometimes slower than deep 2 looping.<br/>
*/
private static void testFillPerformance() {
SimpleDateFormat sdf = new SimpleDateFormat("HH:mm:ss");
System.out.println(sdf.format(new Date()));
Arrays.fill(r1, 0l);


long stamp0 = System.nanoTime();
//      Arrays.fill(r2, 0l); -- exception
long stamp1 = System.nanoTime();
//      System.out.println(String.format("Arrays.fill takes %s nano-seconds.", stamp1 - stamp0));


stamp0 = System.nanoTime();
for (int i = 0; i < 64; i++) {
for (int j = 0; j < 64; j++)
r2[i][j] = 0l;
}
stamp1 = System.nanoTime();
System.out.println(String.format("Arrays' 2-looping takes %s nano-seconds.", stamp1 - stamp0));


stamp0 = System.nanoTime();
for (int i = 0; i < 64; i++) {
System.arraycopy(r1, 0, r2[i], 0, 64);
}
stamp1 = System.nanoTime();
System.out.println(String.format("System.arraycopy looping takes %s nano-seconds.", stamp1 - stamp0));


stamp0 = System.nanoTime();
Arrays.fill(r2, r1);
stamp1 = System.nanoTime();
System.out.println(String.format("One round Arrays.fill takes %s nano-seconds.", stamp1 - stamp0));


stamp0 = System.nanoTime();
for (int i = 0; i < 64; i++)
Arrays.fill(r2[i], 0l);
stamp1 = System.nanoTime();
System.out.println(String.format("Two rounds Arrays.fill takes %s nano-seconds.", stamp1 - stamp0));
}

12:33:18
Arrays' 2-looping takes 133536 nano-seconds.
System.arraycopy looping takes 22070 nano-seconds.
One round Arrays.fill takes 9777 nano-seconds.
Two rounds Arrays.fill takes 93028 nano-seconds.

12:33:38
Arrays' 2-looping takes 133816 nano-seconds.
System.arraycopy looping takes 22070 nano-seconds.
One round Arrays.fill takes 17042 nano-seconds.
Two rounds Arrays.fill takes 95263 nano-seconds.

12:33:51
Arrays' 2-looping takes 199187 nano-seconds.
System.arraycopy looping takes 44140 nano-seconds.
One round Arrays.fill takes 19555 nano-seconds.
Two rounds Arrays.fill takes 449219 nano-seconds.

12:34:16
Arrays' 2-looping takes 199467 nano-seconds.
System.arraycopy looping takes 42464 nano-seconds.
One round Arrays.fill takes 17600 nano-seconds.
Two rounds Arrays.fill takes 170971 nano-seconds.

12:34:26
Arrays' 2-looping takes 198907 nano-seconds.
System.arraycopy looping takes 24584 nano-seconds.
One round Arrays.fill takes 10616 nano-seconds.
Two rounds Arrays.fill takes 94426 nano-seconds.

小开

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.

小开

Arrays.fill is the best option for general purpose use. If you need to fill large arrays though as of latest idk 1.8 u102, there is a faster way that leverages System.arraycopy. You can take a look at this alternate Arrays.fill implementation:

According to the JMH benchmarks you can get almost 2x performance boost for large arrays (1000 +)

In any case, these implementations should be used only where needed. JDKs Arrays.fill should be the preferred choice.

小开

As of Java-8, there are four variants of the setAll method which sets all elements of the specified array, using a provided generator function to compute each element.

Of those four overloads only three of them accept an array of primitives declared as such:

Examples of how to use the aforementioned methods:

// given an index, set the element at the specified index with the provided value
double [] doubles = new double[50];
Arrays.setAll(doubles, index -> 30D);


// given an index, set the element at the specified index with the provided value
int [] ints = new int[50];
Arrays.setAll(ints, index -> 60);


// given an index, set the element at the specified index with the provided value
long [] longs = new long[50];
Arrays.setAll(longs, index -> 90L);

The function provided to the setAll method receives the element index and returns a value for that index.

you may be wondering how about characters array?

This is where the fourth overload of the setAll method comes into play. As there is no overload that consumes an array of character primitives, the only option we have is to change the declaration of our character array to a type Character[].

If changing the type of the array to Character is not appropriate then you can fall back to the Arrays.fill method.

Example of using the setAll method with Character[]:

// given an index, set the element at the specified index with the provided value
Character[] character = new Character[50];
Arrays.setAll(characters, index -> '+');

Although, it's simpler to use the Arrays.fill method rather than the setAll method to set a specific value.

The setAll method has the advantage that you can either set all the elements of the array to have the same value or generate an array of even numbers, odd numbers or any other formula:

e.g.

int[] evenNumbers = new int[10];
Arrays.setAll(evenNumbers, i -> i * 2);

There's also several overloads of the parallelSetAll method which is executed in parallel, although it's important to note that the function passed to the parallelSetAll method must be side-effect free.

Conclusion

If your goal is simply to set a specific value for each element of the array then using the Arrays.fill overloads would be the most appropriate option. However, if you want to be more flexible or generate elements on demand then using the Arrays.setAll or Arrays.parallelSetAll (when appropriate) would be the option to go for.

小开

I have a minor improvement on Ross Drew's answer.

For a small array, a simple loop is faster than the System.arraycopy approach, because of the overhead associated with setting up System.arraycopy. Therefore, it's better to fill the first few bytes of the array using a simple loop, and only move to System.arraycopy when the filled array has a certain size.

The optimal size of the initial loop will be JVM specific and system specific of course. 

private static final int SMALL = 16;


public static void arrayFill(byte[] array, byte value) {
int len = array.length;
int lenB = len < SMALL ? len : SMALL;


for (int i = 0; i < lenB; i++) {
array[i] = value;
}


for (int i = SMALL; i < len; i += i) {
System.arraycopy(array, 0, array, i, len < i + i ? len - i : i);
}
}

提交答案