当所有的线程完成 [英] When All Threads Are Complete

问题描述

这是在使用多线程我第一次真正尝试，我想知道什么时候我所有的工作组正在运行完我如何告诉：

 的for（int i = 0; I< taskGroups.Count（）;我++）{
的ThreadStart T = {委托RunThread（taskGroups [I]）; }; 
新的Thread（T）。开始（）; 
} 
如果（allThreadsComplete）{//？ 
 
} 
  

任何帮助将非常感激。

附录：

 的ThreadStart [] =线程新的ThreadStart [taskGroups.Count（）]; 
的for（int i = 0; I< taskGroups.Count（）;我++）{
线由[i] =新的ThreadStart [] 
线由[i] =委托{RunThread（taskGroups [一世]）; }; 
新的Thread（T）。开始（）; 
} 
布尔threadsComplete = FALSE; 
，而（！threadsComplete）{
的for（int i = 0; I< taskGroups.Count（）;我++）{
如果（线程[I] .STATE ==完成）
 threadsComplete = TRUE; 
} 
} 
  

解决方案

您。需要存储所有的线程，然后调用的Thread.join（）

事情是这样的：

 列表<螺纹>线程=新的List<螺纹>（）; 
的for（int i = 0; I< taskGroups.Count（）;我++）{
 INT TEMP = I; //这个修复该问题与我所共享
线程线程=新主题（（）=> RunThread（taskGroups [临时]））; 
 threads.Add（螺纹）; 
 thread.Start（）; 
} 
 
的foreach（在线程的线程VAR）{
的Thread.join（）; 
} 
  

This is my first real attempt at using multithreading, I want to know how I can tell when all of my tasks groups are done running:

for (int i = 0; i < taskGroups.Count(); i++) {
    ThreadStart t = delegate { RunThread(taskGroups[i]); };
    new Thread(t).Start();
}
if(allThreadsComplete){ //???

}

Any help would be much appreciated

Addendum:

ThreadStart[] threads = new ThreadStart[taskGroups.Count()];
for (int i = 0; i < taskGroups.Count(); i++) {
    threads[i] = new ThreadStart[]
    threads[i] = delegate { RunThread(taskGroups[i]); };
    new Thread(t).Start();
}
bool threadsComplete = false;
while(!threadsComplete){
    for(int i=0;i<taskGroups.Count();i++){
        if(threads[i].State == complete)
        threadsComplete = true;
    }
}

解决方案

You need to store all your threads, and then call Thread.Join().

Something like this:

List<Thread> threads = new List<Thread>();
for (int i = 0; i < taskGroups.Count(); i++) {
   int temp = i; //This fixes the issue with i being shared
   Thread thread = new Thread(() => RunThread(taskGroups[temp]));
   threads.Add(thread);
   thread.Start();
}

foreach (var thread in threads) {
    thread.Join();
}

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