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java并发编程学习之线程池AbstractExecutorService二

AbstractExecutorService 抽象类,实现了 ExecutorService 的接口。

newTaskFor

将任务封装成 FutureTask

protected <T> RunnableFuture<T> newTaskFor(Runnable runnable, T value) {return new FutureTask<T>(runnable, value);
}
protected <T> RunnableFuture<T> newTaskFor(Callable<T> callable) {return new FutureTask<T>(callable);
}

submit

提交任务

public Future<?> submit(Runnable task) {if (task == null) throw new NullPointerException();
    RunnableFuture<Void> ftask = newTaskFor(task, null);
    execute(ftask);
    return ftask;
}
public <T> Future<T> submit(Runnable task, T result) {if (task == null) throw new NullPointerException();
    RunnableFuture<T> ftask = newTaskFor(task, result);
    execute(ftask);
    return ftask;
}
public <T> Future<T> submit(Callable<T> task) {if (task == null) throw new NullPointerException();
    RunnableFuture<T> ftask = newTaskFor(task);
    execute(ftask);
    return ftask;
}

invokeAny

主要方法在 doInvokeAny

//tasks 任务
//timed 是否超时
//nanos 超时时间
private <T> T doInvokeAny(Collection<? extends Callable<T>> tasks,
                              boolean timed, long nanos)
    throws InterruptedException, ExecutionException, TimeoutException {if (tasks == null)
        throw new NullPointerException();
    int ntasks = tasks.size();// 任务书
    if (ntasks == 0)
        throw new IllegalArgumentException();
    ArrayList<Future<T>> futures = new ArrayList<Future<T>>(ntasks);
    // 用于存放结果的,先完成的放前面。所以第一个任务没完成的时候,会继续提交后续任务
    ExecutorCompletionService<T> ecs =
        new ExecutorCompletionService<T>(this);

    try {     
        // 异常信息
        ExecutionException ee = null;
        // 过期时间
        final long deadline = timed ? System.nanoTime() + nanos : 0L;
        Iterator<? extends Callable<T>> it = tasks.iterator();// 获取第一个任务
        提交任务
        futures.add(ecs.submit(it.next()));
        --ntasks;// 因为提交了一个,任务数 -1
        int active = 1;// 正在执行的任务

        for (;;) {Future<T> f = ecs.poll();
            if (f == null) {// 第一个没完成
                if (ntasks > 0) {// 还有没提交的任务
                    --ntasks;// 任务数 -1
                    futures.add(ecs.submit(it.next()));// 提交任务
                    ++active;// 正在执行的任务 +1
                }
                else if (active == 0)// 当前没任务了,但是都失败了,异常被捕获了
                    break;
                else if (timed) {f = ecs.poll(nanos, TimeUnit.NANOSECONDS);// 等待
                    if (f == null)// 返回空,超时抛出异常,结束
                        throw new TimeoutException();
                    nanos = deadline - System.nanoTime();// 剩余时间}
                else
                    f = ecs.take();// 阻塞等待获取}
            if (f != null) {// 说明已经执行完
                --active;// 任务数 -1
                try {return f.get();// 返回执行结果
                } catch (ExecutionException eex) {ee = eex;} catch (RuntimeException rex) {ee = new ExecutionException(rex);
                }
            }
        }

        if (ee == null)
            ee = new ExecutionException();
        throw ee;

    } finally {
        // 取消其他任务,毕竟第一个结果已经返回了
        for (int i = 0, size = futures.size(); i < size; i++)
            futures.get(i).cancel(true);
    }
}

public <T> T invokeAny(Collection<? extends Callable<T>> tasks)
    throws InterruptedException, ExecutionException {
    try {return doInvokeAny(tasks, false, 0);
    } catch (TimeoutException cannotHappen) {
        assert false;
        return null;
    }
}

public <T> T invokeAny(Collection<? extends Callable<T>> tasks,
                       long timeout, TimeUnit unit)
    throws InterruptedException, ExecutionException, TimeoutException {return doInvokeAny(tasks, true, unit.toNanos(timeout));
}

invokeAll

返回所有任务的结果

public <T> List<Future<T>> invokeAll(Collection<? extends Callable<T>> tasks)
    throws InterruptedException {if (tasks == null)
        throw new NullPointerException();
    ArrayList<Future<T>> futures = new ArrayList<Future<T>>(tasks.size());//
    boolean done = false;
    try {for (Callable<T> t : tasks) {// 封装任务,并提交
            RunnableFuture<T> f = newTaskFor(t);
            futures.add(f);
            execute(f);
        }
        for (int i = 0, size = futures.size(); i < size; i++) {Future<T> f = futures.get(i);
            if (!f.isDone()) {
                try {f.get();// 阻塞,等待结果
                } catch (CancellationException ignore) {} catch (ExecutionException ignore) {}}
        }
        done = true;
        return futures;
    } finally {if (!done)// 有异常,取消
            for (int i = 0, size = futures.size(); i < size; i++)
                futures.get(i).cancel(true);
    }
}

public <T> List<Future<T>> invokeAll(Collection<? extends Callable<T>> tasks,
                                     long timeout, TimeUnit unit)
    throws InterruptedException {if (tasks == null)
        throw new NullPointerException();
    long nanos = unit.toNanos(timeout);
    ArrayList<Future<T>> futures = new ArrayList<Future<T>>(tasks.size());
    boolean done = false;
    try {for (Callable<T> t : tasks)
            futures.add(newTaskFor(t));

        final long deadline = System.nanoTime() + nanos;
        final int size = futures.size();

        // Interleave time checks and calls to execute in case
        // executor doesn't have any/much parallelism.
        
        for (int i = 0; i < size; i++) {execute((Runnable)futures.get(i));
            nanos = deadline - System.nanoTime();
            if (nanos <= 0L)
                return futures;// 每个提交都要判断,超时了返回 Future
        }

        for (int i = 0; i < size; i++) {Future<T> f = futures.get(i);
            if (!f.isDone()) {if (nanos <= 0L)
                    return futures;
                try {f.get(nanos, TimeUnit.NANOSECONDS);
                } catch (CancellationException ignore) {} catch (ExecutionException ignore) {} catch (TimeoutException toe) {return futures;}
                nanos = deadline - System.nanoTime();}
        }
        done = true;
        return futures;
    } finally {if (!done)
            for (int i = 0, size = futures.size(); i < size; i++)
                futures.get(i).cancel(true);
    }
}
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