Question

我正在使用tpl并行执行数据。问题是，有时它会毫无理由地挂起来提供这样的输出：

The thread '<No Name>' (0x4aa4) has exited with code 0 (0x0).
The thread '<No Name>' (0x2bf4) has exited with code 0 (0x0).
The thread '<No Name>' (0x417c) has exited with code 0 (0x0).
The thread '<No Name>' (0x432c) has exited with code 0 (0x0).
The thread '<No Name>' (0x3ad0) has exited with code 0 (0x0).
The thread '<No Name>' (0x4440) has exited with code 0 (0x0).
The thread '<No Name>' (0x24e8) has exited with code 0 (0x0).
The thread '<No Name>' (0x3354) has exited with code 0 (0x0).
The thread '<No Name>' (0x4a30) has exited with code 0 (0x0).

程序仍然会执行，当我在visual studio中暂停它时，它会暂停我有ForEach的并行程序：

Parallel.ForEach
(
   hold1.remainingHolds.ToArray(), 
   subSequence =>
   {
      //only if subsequence has 1 common substring
      if (checkIfCommon(remainingSequence, subSequence.ToString()) == 1)
      {
         goDownLinkType2(subSequence.ToString().Split(','), 0, 
           (SuffixNode)hold1.childHolds[subSequence.ToString().Split(',')[0]]);
       }
    }
  );

我不认为它是一个死锁，因为不存在可以有一个线程等待不同资源并且互相锁定的情况

它应该进入这个方法并且recursivley去一个后缀树，直到没有线程将任何对象添加到arraylist

 Boolean flag = false;
 public void goDownLinkType2
      (string[] splittedString, 
       int index, SuffixNode currentNode)
       {
          Boolean writingFlag = false;
          if (index == 2)
          {
             while(writingFlag == false)
             {
                while(flag == true)
                {
        //blocked
                 }
                 if (flag == false)
                 {
                    flag = true;
                 if(!secondChoiceResults.Contains
                     (currentNode.representingStance.SequenceOfHolds))
                 {
                    Console.WriteLine("new addition");
                    secondChoiceResults.Add
                       (currentNode.representingStance.SequenceOfHolds,
                        currentNode.representingStance);
                  }
                  flag = false;
                  writingFlag = true;
               }
            }

         }
         else
         {
            int nextIndex = index + 1;
            goDownLinkType2
              (splittedString, 
               nextIndex,
               (SuffixNode)currentNode.childHolds[splittedString[nextIndex]]
              );
          }
       }

Answer 1

抛出'flag'变量并使用lock语句。使用此代码可以在您的关键部分中有多个线程（即一个线程即将设置flag = true而另一个线程只评估flag == false为true（btw将来使用！标志）< / p>

lock( obj )
{
    // critical section here
}

obj只需要是对所有线程都可访问的对象的引用。

以下是我对您的代码的修改：

public void goDownLinkType2(string[] splittedString, int index, SuffixNode currentNode)
{
    Boolean writingFlag = false;
    if (index == 2)
    {
        while(writingFlag == false)
        {
            lock( this )
            //while(flag == true)
            //{
                //blocked
            //}
            //if (flag == false)
            {
                //flag = true;
                if (!secondChoiceResults.Contains(currentNode.representingStance.SequenceOfHolds))
                {
                    Console.WriteLine("new addition");
                    secondChoiceResults.Add(currentNode.representingStance.SequenceOfHolds, currentNode.representingStance);
                }
                //flag = false;
                writingFlag = true;
            }
        }


    }
    else
    {
        int nextIndex = index + 1;
        goDownLinkType2(splittedString, nextIndex, (SuffixNode)currentNode.childHolds[splittedString[nextIndex]]);
    }
}

程序无缘无故地挂起

1 个答案: