Question

首先我是论坛的新人，所以请耐心等待我和我的英语。： - ）

我正在编写一个C＃应用程序，它应该将多线程SOAP请求发送到apache后端。到目前为止一切工作正常，但我遇到了问题。应用程序首先读取XML文件从另一个系统首先解析成类，排序并发送到SOAP后端。这里的片段

List<Thread> ThreadsPerOneRecord = new List<Thread>();         
bool ExecuteSingleThreaded = false;
//The variable list is passed as parameter to the function 

foreach (Record prov in list)
{
  XMLResult.AppendText("Type: " + prov.Type + Environment.NewLine);

  Thread t = new Thread(() => Send(prov, c));                                
  t.Start();
  //Here the sleep 
  Thread.Sleep(50);
  ThreadsPerOneRecord.Add(t);               

  #region Code for test single threaded execution
  if (ExecuteSingleThreaded)
  {
    foreach (Thread t2 in ThreadsPerOneRecord)
      t2.Join();
    ThreadsPerOneRecord.Clear();
  }
  #endregion
}

XMLResult.AppendText("Waiting for the threads to finish" + Environment.NewLine);
//Waiting for the threads to finish
foreach (Thread t in ThreadsPerOneRecord)            
  t.Join();

当我将此发送到SOAP Web服务时，除了一个请求之外它工作正常。这些请求彼此混淆。即：

What it should be: 
Record 1 -> SOAP
Record 2 -> SOAP
Record 3 -> SOAP

What it is
Record 1 -> SOAP
Record 2 -> SOAP 
Record 2 -> SOAP 
Record 3 -> nowhere

我已经尝试调试整个代码并使用调试器工作正常。当我插入50毫秒的睡眠时也一样。但没有睡眠就混合了这两个记录......

有人知道为什么会这样吗？不应该每个线程都独立于自身吗？我也检查了集合，数据正确在里面。

由于

Oldfighter

Answer 1

替换

Thread t = new Thread(() => Send(prov, c));
t.Start();

与

Thread t = new Thread(item => Send(item, c));
t.Start(prov);

在你的代码中，lambda表达式实际上看到了对iterator变量的更改（它是每个线程的相同变量，而不是将lambda传递给线程构造函数时捕获的值）。

Answer 2

您的问题是您在Foreach循环中访问修改后的闭包

这是修复：

        List<Thread> ThreadsPerOneRecord = new List<Thread>();
        bool ExecuteSingleThreaded = false;
        //The variable list is passed as parameter to the function  

        foreach (Record prov in list)
        {
            var tempProv = prov;
            XMLResult.AppendText("Type: " + tempProv.Type + Environment.NewLine);

            Thread t = new Thread(() => Send(tempProv, c));
            t.Start();
            //Here the sleep  
            Thread.Sleep(50);
            ThreadsPerOneRecord.Add(t);

            #region Code for test single threaded execution
            if (ExecuteSingleThreaded)
            {
                foreach (Thread t2 in ThreadsPerOneRecord)
                    t2.Join();
                ThreadsPerOneRecord.Clear();
            }
            #endregion
        }

        XMLResult.AppendText("Waiting for the threads to finish" + Environment.NewLine);
        //Waiting for the threads to finish 
        foreach (Thread t in ThreadsPerOneRecord)
            t.Join();

Answer 3

经典的foreach / capture;修复很简单 - 添加一个额外的变量：

foreach (Record tmp in list)
{
  Record prov = tmp;
  XMLResult.AppendText("Type: " + prov.Type + Environment.NewLine);

  Thread t = new Thread(() => Send(prov, c));

否则，所有lambda之间共享“prov”。已经公开提到，正在对c＃5进行评估，但尚未确认。

使用WebRequest的C＃多线程程序

3 个答案: