在所有子线程完成执行之后如何运行主线程 [英] how to run the main thread after all child threads have completed there exceution

问题描述

我有一个要求，其中28个线程必须完成某些功能.我已经在匿名内部类(如:

I have a requirement in which 28 threads have to complete some functionality. I have created these threads as in anonymous inner classes like :

Thread t=new Thread(new Runnable(){public void run() 
                        {//code
                        }}
                                );

                        t.start();

现在，我希望所有这些线程完成工作后再开始执行.

Now I want that the further execution should start after all these threads have finished there work.

注意:我对join()方法感到困惑，因为它使我的线程按顺序运行.

Note : I am confused about join() method as it makes my threads run sequentially.

所以有人可以建议我一旦这些线程完成工作后如何运行主线程.

So can anyone suggest me how can I make main thread run once these threads are done with work.

推荐答案

注意:我对join()方法感到困惑，因为它会使我的线程按顺序运行.

Note : I am confused about join() method as it makes my threads run sequentially.

如果您有这样的代码，它将这样做:

It will do that if you have code like this:

for (Runnable runnable : runnables) {
    Thread t = new Thread(runnable);
    t.start();
    t.join();
}

但是，您可以启动要并行运行的所有线程，然后然后在所有线程上调用join.例如:

However, you can start all the threads you want to run in parallel, then call join on them all. For example:

List<Thread> threads = new ArrayList<>();
for (Runnable runnable : runnables) {
    Thread t = new Thread(runnable);
    t.start();
    threads.add(t);
}
// Now everything's running - join all the threads
for (Thread thread : threads) {
     thread.join();
}

// Now you can do whatever you need to after all the
// threads have finished.

当然，还有许多其他方法-直接启动线程可能不如使用更高级别的抽象在您的代码中那么合适；这取决于您要实现的目标.上面的代码应该可以正常工作-假设所有Runnable都可以并行运行，而不会通过同步彼此阻塞.

There are many other approaches, of course - starting threads directly may well not be as suitable in your code as using a higher level abstraction; it depends on what you're trying to achieve. The above should work fine though - assuming all the Runnables are able to run in parallel without blocking each other through synchronization.

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