【发布时间】：2012-05-29 10:44:22
【问题描述】：
据我了解ForkJoinPool，该池创建固定数量的线程（默认值：内核数）并且永远不会创建更多线程（除非应用程序通过使用managedBlock 表明需要这些线程）。
但是，使用ForkJoinPool.getPoolSize() 我发现在一个创建 30,000 个任务 (RecursiveAction) 的程序中，执行这些任务的 ForkJoinPool 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() 并访问AtomicBooleans，即没有线程阻塞操作。
由于join() 没有按照我的理解阻塞调用线程，因此池中的任何线程都没有理由阻塞，因此（我假设）不应该再创建任何线程（这显然正在发生）。
那么，为什么ForkJoinPool 会创建这么多线程？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。
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SolverTask 的compute() 方法中：RecursiveAction 从某个给定点开始，并继续从当前点可到达的所有相邻点。它不是在每个分支点创建一个新的SolverTask（这会创建太多的任务），而是将除一个之外的所有邻居推送到回溯堆栈中以供稍后处理，并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。
我在上面引用的数字（“平均 30,000 个任务，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.");
}
【问题讨论】：
标签： java parallel-processing threadpool fork-join