据我了解ForkJoinPool，该池创建固定数量的线程（默认值：核心数）并且永远不会创建更多线程（除非应用程序通过 using 指示需要这些线程managedBlock）。
然而，使用ForkJoinPool.getPoolSize()我发现在一个创建 30,000 个任务 ( RecursiveAction) 的程序中，ForkJoinPool执行这些任务平均使用 700 个线程（每次创建任务时都会计算线程数）。这些任务不做 I/O，而是纯粹的计算；唯一的任务间同步是调用ForkJoinTask.join()和访问AtomicBooleans，即没有线程阻塞操作。
由于join()按照我的理解不会阻塞调用线程，因此池中的任何线程都没有理由阻塞，因此（我假设）没有理由创建任何进一步的线程（这显然仍在发生） .
那么，为什么要ForkJoinPool创建这么多线程呢？哪些因素决定了创建的线程数？
我曾希望可以在不发布代码的情况下回答这个问题，但这里是应要求提供的。这段代码是从一个四倍大小的程序中摘录出来的，精简到了基本部分；它不会按原样编译。如果需要，我当然也可以发布完整的程序。
该程序使用深度优先搜索在迷宫中搜索从给定起点到给定终点的路径。保证存在解决方案。主要逻辑在: A的compute()方法中，它从某个给定点开始，并继续从当前点可到达的所有相邻点。而不是创造一个新的SolverTaskRecursiveActionSolverTask在每个分支点（这将创建太多任务），它将除一个之外的所有邻居推送到回溯堆栈以供稍后处理，并且仅继续仅一个未推送到堆栈的邻居。一旦它以这种方式到达死胡同，最近推送到回溯堆栈的点就会被弹出，并且从那里继续搜索（相应地削减从 taks 的起点构建的路径）。一旦任务发现其回溯堆栈大于某个阈值，就会创建一个新任务；从那时起，该任务在继续从其回溯堆栈中弹出直到用完为止，但在到达分支点时不会将任何其他点推入其堆栈，而是为每个这样的点创建一个新任务。因此，可以使用堆栈限制阈值来调整任务的大小。
我上面引用的数字（“30,000 个任务，平均 700 个线程”）来自于搜索 5000x5000 个单元的迷宫。所以，这里是基本代码：
class SolverTask extends RecursiveTask<ArrayDeque<Point>> {
// Once the backtrack stack has reached this size, the current task
// will never add another cell to it, but create a new task for each
// newly discovered branch:
private static final int MAX_BACKTRACK_CELLS = 100*1000;
/**
* @return Tries to compute a path through the maze from local start to end
* and returns that (or null if no such path found)
*/
@Override
public ArrayDeque<Point> compute() {
// Is this task still accepting new branches for processing on its own,
// or will it create new tasks to handle those?
boolean stillAcceptingNewBranches = true;
Point current = localStart;
ArrayDeque<Point> pathFromLocalStart = new ArrayDeque<Point>(); // Path from localStart to (including) current
ArrayDeque<PointAndDirection> backtrackStack = new ArrayDeque<PointAndDirection>();
// Used as a stack: Branches not yet taken; solver will backtrack to these branching points later
Direction[] allDirections = Direction.values();
while (!current.equals(end)) {
pathFromLocalStart.addLast(current);
// Collect current's unvisited neighbors in random order:
ArrayDeque<PointAndDirection> neighborsToVisit = new ArrayDeque<PointAndDirection>(allDirections.length);
for (Direction directionToNeighbor: allDirections) {
Point neighbor = current.getNeighbor(directionToNeighbor);
// contains() and hasPassage() are read-only methods and thus need no synchronization
if (maze.contains(neighbor) && maze.hasPassage(current, neighbor) && maze.visit(neighbor))
neighborsToVisit.add(new PointAndDirection(neighbor, directionToNeighbor.opposite));
}
// Process unvisited neighbors
if (neighborsToVisit.size() == 1) {
// Current node is no branch: Continue with that neighbor
current = neighborsToVisit.getFirst().getPoint();
continue;
}
if (neighborsToVisit.size() >= 2) {
// Current node is a branch
if (stillAcceptingNewBranches) {
current = neighborsToVisit.removeLast().getPoint();
// Push all neighbors except one on the backtrack stack for later processing
for(PointAndDirection neighborAndDirection: neighborsToVisit)
backtrackStack.push(neighborAndDirection);
if (backtrackStack.size() > MAX_BACKTRACK_CELLS)
stillAcceptingNewBranches = false;
// Continue with the one neighbor that was not pushed onto the backtrack stack
continue;
} else {
// Current node is a branch point, but this task does not accept new branches any more:
// Create new task for each neighbor to visit and wait for the end of those tasks
SolverTask[] subTasks = new SolverTask[neighborsToVisit.size()];
int t = 0;
for(PointAndDirection neighborAndDirection: neighborsToVisit) {
SolverTask task = new SolverTask(neighborAndDirection.getPoint(), end, maze);
task.fork();
subTasks[t++] = task;
}
for (SolverTask task: subTasks) {
ArrayDeque<Point> subTaskResult = null;
try {
subTaskResult = task.join();
} catch (CancellationException e) {
// Nothing to do here: Another task has found the solution and cancelled all other tasks
}
catch (Exception e) {
e.printStackTrace();
}
if (subTaskResult != null) { // subtask found solution
pathFromLocalStart.addAll(subTaskResult);
// No need to wait for the other subtasks once a solution has been found
return pathFromLocalStart;
}
} // for subTasks
} // else (not accepting any more branches)
} // if (current node is a branch)
// Current node is dead end or all its neighbors lead to dead ends:
// Continue with a node from the backtracking stack, if any is left:
if (backtrackStack.isEmpty()) {
return null; // No more backtracking avaible: No solution exists => end of this task
}
// Backtrack: Continue with cell saved at latest branching point:
PointAndDirection pd = backtrackStack.pop();
current = pd.getPoint();
Point branchingPoint = current.getNeighbor(pd.getDirectionToBranchingPoint());
// DEBUG System.out.println("Backtracking to " + branchingPoint);
// Remove the dead end from the top of pathSoFar, i.e. all cells after branchingPoint:
while (!pathFromLocalStart.peekLast().equals(branchingPoint)) {
// DEBUG System.out.println(" Going back before " + pathSoFar.peekLast());
pathFromLocalStart.removeLast();
}
// continue while loop with newly popped current
} // while (current ...
if (!current.equals(end)) {
// this task was interrupted by another one that already found the solution
// and should end now therefore:
return null;
} else {
// Found the solution path:
pathFromLocalStart.addLast(current);
return pathFromLocalStart;
}
} // compute()
} // class SolverTask
@SuppressWarnings("serial")
public class ParallelMaze {
// for each cell in the maze: Has the solver visited it yet?
private final AtomicBoolean[][] visited;
/**
* Atomically marks this point as visited unless visited before
* @return whether the point was visited for the first time, i.e. whether it could be marked
*/
boolean visit(Point p) {
return visited[p.getX()][p.getY()].compareAndSet(false, true);
}
public static void main(String[] args) {
ForkJoinPool pool = new ForkJoinPool();
ParallelMaze maze = new ParallelMaze(width, height, new Point(width-1, 0), new Point(0, height-1));
// Start initial task
long startTime = System.currentTimeMillis();
// since SolverTask.compute() expects its starting point already visited,
// must do that explicitly for the global starting point:
maze.visit(maze.start);
maze.solution = pool.invoke(new SolverTask(maze.start, maze.end, maze));
// One solution is enough: Stop all tasks that are still running
pool.shutdownNow();
pool.awaitTermination(Integer.MAX_VALUE, TimeUnit.DAYS);
long endTime = System.currentTimeMillis();
System.out.println("Computed solution of length " + maze.solution.size() + " to maze of size " +
width + "x" + height + " in " + ((float)(endTime - startTime))/1000 + "s.");
}