线程池提交任务方法
作者：互联网
excute方法: 源码
submit方法通过提交参数构造FutrueTask,然后执行excute(FutrueTask)方法,返回一个future对象
public Future<?> submit(Runnable task) {
if (task == null) throw new NullPointerException();
RunnableFuture<Void> ftask = newTaskFor(task, null);//new FutureTask<T>(runnable, null)
execute(ftask);
return ftask;
}
submit(Runnable task)
public <T> Future<T> submit(Runnable task, T result) {
if (task == null) throw new NullPointerException();
RunnableFuture<T> ftask = newTaskFor(task, result);//new FutureTask<T>(runnable, value)
execute(ftask);
return ftask;
}
submit(Runnable task, T result)
public <T> Future<T> submit(Callable<T> task) {
if (task == null) throw new NullPointerException();
RunnableFuture<T> ftask = newTaskFor(task);
execute(ftask);
return ftask;
}
submit(Callable task)
任务批量提交
超时时间：执行invokeAll或者invokeAny的时间
会等待所有任务完成，如果异常终止(包括超时停止)，取消所有任务
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) {
//构造FutureTask，添加进futures集合里
RunnableFuture<T> f = newTaskFor(t);
futures.add(f);
execute(f);//执行任务
}
for (int i = 0, size = futures.size(); i < size; i++) {
Future<T> f = futures.get(i);
if (!f.isDone()) {//任务已经完成
try {
f.get();//这里只是起到阻塞的作用，如果线程中断，则跳到finally
} catch (CancellationException ignore) {
} catch (ExecutionException ignore) {//不处理任务执行抛出
}
}
}
//如果没走到这里，就说明抛出了异常，可能是线程中断异常或其他异常。
done = true;
return futures;
} finally {
//未正常结束，取消所有任务，通过future get时抛出CancellationException(
if (!done)
for (int i = 0, size = futures.size(); i < size; i++)
futures.get(i).cancel(true);
}
}
invokeAll(Collection<? extends Callable> tasks)
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)//超时返回
return futures;
}
for (int i = 0; i < size; i++) {
Future<T> f = futures.get(i);
//如果任务未完成，则进去阻塞到任务完成
if (!f.isDone()) {
//超时返回
if (nanos <= 0L)
return futures;
try {
f.get(nanos, TimeUnit.NANOSECONDS);
} catch (CancellationException ignore) {
} catch (ExecutionException ignore) {
} catch (TimeoutException toe) {
return futures;//get超时返回
}
nanos = deadline - System.nanoTime();
}
}
done = true;//未走到这里，标识抛出异常了，在finally里取消所有任务
return futures;
} finally {
//进而取消所有任务，取消的任务不能再执行了
if (!done)
for (int i = 0, size = futures.size(); i < size; i++)
futures.get(i).cancel(true);
}
}
invokeAll(Collection<? extends Callable> tasks, long timeout, TimeUnit unit)
只要有任意一个任务完成，就会返回其结果值，如果超时，则抛出 TimeoutException异常，最终都会取消所有任务
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;
}
}
invokeAny(Collection<? extends Callable> tasks)
public <T> T invokeAny(Collection<? extends Callable<T>> tasks,
long timeout, TimeUnit unit)
throws InterruptedException, ExecutionException, TimeoutException {
return doInvokeAny(tasks, true, unit.toNanos(timeout));
}
invokeAny(Collection<? extends Callable> tasks, long timeout, TimeUnit unit)
真正执行invokeAny的方法
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();
if (ntasks == 0)
throw new IllegalArgumentException();
ArrayList<Future<T>> futures = new ArrayList<Future<T>>(ntasks);
//可以优先获取到已经完成的任务
ExecutorCompletionService<T> ecs =
new ExecutorCompletionService<T>(this);
// For efficiency, especially in executors with limited
// parallelism, check to see if previously submitted tasks are
// done before submitting more of them. This interleaving
// plus the exception mechanics account for messiness of main
// loop.
try {
// Record exceptions so that if we fail to obtain any
// result, we can throw the last exception we got.
//任务执行时抛出的异常
ExecutionException ee = null;
final long deadline = timed ? System.nanoTime() + nanos : 0L;
Iterator<? extends Callable<T>> it = tasks.iterator();
// Start one task for sure; the rest incrementally
futures.add(ecs.submit(it.next()));
--ntasks;
int active = 1;//正在执行的任务数量
for (;;) {
//获取已完成的任务， 这里是通过ExecutorCompletionService的方法来获取
Future<T> f = ecs.poll();
//没有已经完成的任务
if (f == null) {
if (ntasks > 0) {//当前未提交的任务
--ntasks;//未提交任务-1
futures.add(ecs.submit(it.next()));//提交任务
++active;
}
else if (active == 0) //如果没有任务正在执行
break;
else if (timed) {//如果设置了超时时间
//阻塞获取
f = ecs.poll(nanos, TimeUnit.NANOSECONDS);
//获取不到，就抛出超时异常
if (f == null)
throw new TimeoutException();
nanos = deadline - System.nanoTime();
}
else
f = ecs.take();//阻塞获取
}
if (f != null) {//有任务完成了
--active;//任务正在执行数量-1
try {
return f.get();//直接返回结果值
} catch (ExecutionException eex) {
ee = eex;
} catch (RuntimeException rex) {
ee = new ExecutionException(rex);
}
}
}
if (ee == null)
ee = new ExecutionException();
throw ee;
} finally {
//最终取消所有任务
for (int i = 0, size = futures.size(); i < size; i++)
futures.get(i).cancel(true);
}
}
doInvokeAny
标签：tasks,futures,任务,task,线程,提交,new,null,size 来源： https://www.cnblogs.com/shuiyingyuan/p/15262979.html