ementation details follow: static final int THRESHOLD = 1000; void sortSequentially(int lo, int hi) { Arrays.sort(array, lo, hi); } void merge(int lo, int mid, int hi) { long[] buf = Arrays.copyOfRange(array, lo, mid); for (int i = 0, j = lo, k = mid; i < buf.length; j++) array[j] = (k == hi || buf[i] < array[k]) ? buf[i++] : array[k++]; } }} You could then sort {@code anArray} by creating {@code newSortTask(anArray)} and invoking it in a ForkJoinPool. As a moreconcrete simple example, the following task increments each element of an array:
{@code}class IncrementTask extends RecursiveAction final long[] array; final int lo, hi; IncrementTask(long[] array, int lo, int hi) { this.array = array; this.lo = lo; this.hi = hi; } protected void compute() { if (hi - lo < THRESHOLD) { for (int i = lo; i < hi; ++i) array[i]++; } else { int mid = (lo + hi) >>> 1; invokeAll(new IncrementTask(array, lo, mid), new IncrementTask(array, mid, hi)); } } }} The following example illustrates some refinements and idioms that may lead to better performance: RecursiveActions need not be fully recursive, so long as they maintain the basic divide-and-conquer approach. Here is a class that sums the squares of each element of a double array, by subdividing out only the right-hand-sides of repeated divisions by two, and keeping track of them with a chain of {@code next} references. It uses a dynamicthreshold based on method {@code getSurplusQueuedTaskCount}, but counterbalances potential excess partitioning by directly performing leaf actions on unstolen tasks rather than further subdividing.
{@code}double sumOfSquares(ForkJoinPool pool, double[] array) int n = array.length; Applyer a = new Applyer(array, 0, n, null); pool.invoke(a); return a.result; } class Applyer extends RecursiveAction { final double[] array; final int lo, hi; double result; Applyer next; // keeps track of right-hand-side tasks Applyer(double[] array, int lo, int hi, Applyer next) { this.array = array; this.lo = lo; this.hi = hi; this.next = next; } double atLeaf(int l, int h) { double sum = 0; for (int i = l; i < h; ++i) // perform leftmost base step sum += array[i] * array[i]; return sum; } protected void compute() { int l = lo; int h = hi; Applyer right = null; while (h - l > 1 && getSurplusQueuedTaskCount() <= 3) { int mid = (l + h) >>> 1; right = new Applyer(array, mid, h, right); right.fork(); h = mid; } double sum = atLeaf(l, h); while (right != null) { if (right.tryUnfork()) // directly calculate if not stolen sum += right.atLeaf(right.lo, right.hi); else { right.join(); sum += right.result; } right = right.next; } result = sum; } }}
@since 1.7
@author Doug Lea