我有一个简单的tpl数据流,基本上可以完成一些任务。 我注意到,在任何数据块中都存在异常时,它没有被初始父块调用者捕获。 我添加了一些手动代码来检查异常,但似乎不是正确的方法。
if (readBlock.Completion.Exception != null || saveBlockJoinedProcess.Completion.Exception != null || processBlock1.Completion.Exception != null || processBlock2.Completion.Exception != null)
{
throw readBlock.Completion.Exception;
}
我在线查看了建议的方法,但没有发现明显的问题。 因此,我在下面创建了一些示例代码,希望获得有关更好解决方案的指导:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading;
using System.Threading.Tasks;
using System.Threading.Tasks.Dataflow;
namespace TPLDataflow
{
class Program
{
static void Main(string[] args)
{
try
{
//ProcessB();
ProcessA();
}
catch(Exception e)
{
Console.WriteLine("Exception in Process!");
throw new Exception($"exception:{e}");
}
Console.WriteLine("Processing complete!");
Console.ReadLine();
}
private static void ProcessB()
{
Task.WhenAll(Task.Run(() => DoSomething(1, "ProcessB"))).Wait();
}
private static void ProcessA()
{
var random = new Random();
var readBlock = new TransformBlock<int, int>(
x => { try { return DoSomething(x, "readBlock"); } catch (Exception e) { throw e; } },
new ExecutionDataflowBlockOptions { MaxDegreeOfParallelism = 1 }); //1
var braodcastBlock = new BroadcastBlock<int>(i => i); // ⬅ Here
var processBlock1 =
new TransformBlock<int, int>(x => DoSomethingAsync(5, "processBlock1")); //2
var processBlock2 =
new TransformBlock<int, int>(x => DoSomethingAsync(2, "processBlock2")); //3
//var saveBlock =
// new ActionBlock<int>(
// x => Save(x)); //4
var saveBlockJoinedProcess =
new ActionBlock<Tuple<int, int>>(
x => SaveJoined(x.Item1, x.Item2)); //4
var saveBlockJoin = new JoinBlock<int, int>();
readBlock.LinkTo(braodcastBlock, new DataflowLinkOptions { PropagateCompletion = true });
braodcastBlock.LinkTo(processBlock1,
new DataflowLinkOptions { PropagateCompletion = true }); //5
braodcastBlock.LinkTo(processBlock2,
new DataflowLinkOptions { PropagateCompletion = true }); //6
processBlock1.LinkTo(
saveBlockJoin.Target1); //7
processBlock2.LinkTo(
saveBlockJoin.Target2); //8
saveBlockJoin.LinkTo(saveBlockJoinedProcess, new DataflowLinkOptions { PropagateCompletion = true });
readBlock.Post(1); //10
//readBlock.Post(2); //10
Task.WhenAll(
processBlock1.Completion,
processBlock2.Completion)
.ContinueWith(_ => saveBlockJoin.Complete());
readBlock.Complete(); //12
saveBlockJoinedProcess.Completion.Wait(); //13
if (readBlock.Completion.Exception != null || saveBlockJoinedProcess.Completion.Exception != null || processBlock1.Completion.Exception != null || processBlock2.Completion.Exception != null)
{
throw readBlock.Completion.Exception;
}
}
private static int DoSomething(int i, string method)
{
Console.WriteLine($"Do Something, callng method : { method}");
throw new Exception("Fake Exception!");
return i;
}
private static async Task<int> DoSomethingAsync(int i, string method)
{
Console.WriteLine($"Do SomethingAsync");
throw new Exception("Fake Exception!");
await Task.Delay(new TimeSpan(0,0,i));
Console.WriteLine($"Do Something : {i}, callng method : { method}");
return i;
}
private static void Save(int x)
{
Console.WriteLine("Save!");
}
private static void SaveJoined(int x, int y)
{
Thread.Sleep(new TimeSpan(0, 0, 10));
Console.WriteLine("Save Joined!");
}
}
}
答案 0 :(得分:1)
我在线查看了建议的方法,但是没有发现明显的问题。
如果有管道(或多或少),那么通常的方法是使用PropagateCompletion
关闭管道。如果您有更复杂的拓扑,则需要手动完成块。
在您的情况下,您尝试在此处进行传播:
Task.WhenAll(
processBlock1.Completion,
processBlock2.Completion)
.ContinueWith(_ => saveBlockJoin.Complete());
但是此代码不会传播异常。 processBlock1.Completion
和processBlock2.Completion
都完成时,saveBlockJoin
成功完成。
更好的解决方案是使用await
而不是ContinueWith
:
async Task PropagateToSaveBlockJoin()
{
try
{
await Task.WhenAll(processBlock1.Completion, processBlock2.Completion);
saveBlockJoin.Complete();
}
catch (Exception ex)
{
((IDataflowBlock)saveBlockJoin).Fault(ex);
}
}
_ = PropagateToSaveBlockJoin();
使用await
会鼓励您处理异常,您可以通过将异常传递给Fault
来传播异常。
答案 1 :(得分:1)
开箱即用的TPL数据流不支持在管道中向后传播错误,当块具有有限容量时,这尤其令人讨厌。在这种情况下,下游块中的错误可能导致其前面的块无限期地阻塞。我知道的唯一解决方案是使用取消功能,并在任何人失败的情况下取消所有块。这是可以完成的。首先创建一个CancellationTokenSource
:
var cts = new CancellationTokenSource();
然后逐个创建块,将相同的CancellationToken
嵌入所有块的选项中:
var options = new ExecutionDataflowBlockOptions()
{ BoundedCapacity = 10, CancellationToken = cts.Token };
var block1 = new TransformBlock<double, double>(Math.Sqrt, options);
var block2 = new ActionBlock<double>(Console.WriteLine, options);
然后将这些块链接在一起,包括PropagateCompletion
设置:
block1.LinkTo(block2, new DataflowLinkOptions { PropagateCompletion = true });
最后使用扩展方法来在发生异常的情况下触发CancellationTokenSource
的取消:
block1.OnFaultedCancel(cts);
block2.OnFaultedCancel(cts);
OnFaultedCancel
扩展方法如下所示:
public static class DataflowExtensions
{
public static void OnFaultedCancel(this IDataflowBlock dataflowBlock,
CancellationTokenSource cts)
{
dataflowBlock.Completion.ContinueWith(_ => cts.Cancel(), default,
TaskContinuationOptions.OnlyOnFaulted |
TaskContinuationOptions.ExecuteSynchronously, TaskScheduler.Default);
}
}
答案 2 :(得分:0)
乍一看,如果只有一点点(不看您的体系结构)。在我看来,您已经混合了一些较新的结构和较旧的结构。还有一些不必要的代码部分。
例如:
private static void ProcessB()
{
Task.WhenAll(Task.Run(() => DoSomething(1, "ProcessB"))).Wait();
}
使用Wait()方法,如果发生任何异常,则将它们包装在System.AggregateException中。我认为这样比较好:
private static async Task ProcessBAsync()
{
await Task.Run(() => DoSomething(1, "ProcessB"));
}
使用async-await,如果发生异常,则await语句会重新引发包装在System.AggregateException中的第一个异常。这样,您就可以尝试捕获具体的异常类型,并仅处理您真正可以处理的情况。
另一部分是代码的这一部分:
private static void ProcessA()
{
var random = new Random();
var readBlock = new TransformBlock<int, int>(
x =>
{
try { return DoSomething(x, "readBlock"); }
catch (Exception e)
{
throw e;
}
},
new ExecutionDataflowBlockOptions { MaxDegreeOfParallelism = 1 }); //1
为什么只捕获异常才能将其抛出?在这种情况下,try-catch是多余的。
这是这里:
private static void SaveJoined(int x, int y)
{
Thread.Sleep(new TimeSpan(0, 0, 10));
Console.WriteLine("Save Joined!");
}
使用await Task.Delay(....)
更好。使用Task.Delay(...)
,您的应用程序将不会冻结。