等待所有线程结束后，主线程才退出

[liyoubaidu]

console程序：

main()
{
for(int i=0;i<100;i++)
{
用线程池开启一个新线程;
}

等待上面开启的所有线程执行完后，才继续往下面执行。//这一句要怎么实现。
}

[人到中年琐事多]

Thread.join()

[starj1]

轮询吧，看每个线程的ThreadState。用Thread.Join()是不现实的。

[photoplan]

查查 WaitHandle

[ProjectDD]

//MSDN :
//注意看 ThreadPool如何实例化的，以及 WaitAll();静态方法的调用 这就是你要的等待所有线程完成的阻塞。
using System;
using System.Threading;

public sealed class App
{
// Define an array with two AutoResetEvent WaitHandles.
static WaitHandle[] waitHandles = new WaitHandle[]
{
new AutoResetEvent(false),
new AutoResetEvent(false)
};

// Define a random number generator for testing.
static Random r = new Random();

static void Main()
{
// Queue up two tasks on two different threads;
// wait until all tasks are completed.
DateTime dt = DateTime.Now;
Console.WriteLine("Main thread is waiting for BOTH tasks to complete.");
ThreadPool.QueueUserWorkItem(new WaitCallback(DoTask), waitHandles[0]);
ThreadPool.QueueUserWorkItem(new WaitCallback(DoTask), waitHandles[1]);
WaitHandle.WaitAll(waitHandles);
// The time shown below should match the longest task.
Console.WriteLine("Both tasks are completed (time waited={0})",
(DateTime.Now - dt).TotalMilliseconds);

// Queue up two tasks on two different threads;
// wait until any tasks are completed.
dt = DateTime.Now;
Console.WriteLine();
Console.WriteLine("The main thread is waiting for either task to complete.");
ThreadPool.QueueUserWorkItem(new WaitCallback(DoTask), waitHandles[0]);
ThreadPool.QueueUserWorkItem(new WaitCallback(DoTask), waitHandles[1]);
int index = WaitHandle.WaitAny(waitHandles);
// The time shown below should match the shortest task.
Console.WriteLine("Task {0} finished first (time waited={1}).",
index + 1, (DateTime.Now - dt).TotalMilliseconds);
}

static void DoTask(Object state)
{
AutoResetEvent are = (AutoResetEvent) state;
int time = 1000 * r.Next(2, 10);
Console.WriteLine("Performing a task for {0} milliseconds.", time);
Thread.Sleep(time);
are.Set();
}
}

// This code produces output similar to the following:
//
// Main thread is waiting for BOTH tasks to complete.
// Performing a task for 7000 milliseconds.
// Performing a task for 4000 milliseconds.
// Both tasks are completed (time waited=7064.8052)
//
// The main thread is waiting for either task to complete.
// Performing a task for 2000 milliseconds.
// Performing a task for 2000 milliseconds.
// Task 1 finished first (time waited=2000.6528).

[liyoubaidu]

up

[barbara2008]

设置一个事件

[dc212009]

用类似信号量的东西ManualResetEvent，
来控制吧。

[liyoubaidu]

还有没有其他的解决方案呢。

[88csdn]

呵呵，最好就是信号量了，要不就写个while(...){sleep(1000);}但是这样不推荐哦