我有一个简单的tpl数据流,它基本上完成了一些任务。我注意到,当在任何数据块中出现异常时,它都不会在最初的父块调用者中被捕获。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;
}我在网上看了看什么是建议的方法,但没有看到任何明显的东西。因此,我在下面创建了一些示例代码,并希望获得一些更好的解决方案的指导:
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($"exception:{e}");
}
Console.WriteLine("Processing complete!");
Console.ReadLine();
}
private static void ProcessB()
{
Task.WhenAll(Task.Run(() => DoSomething(1, "ProcessB"))).Wait();
}
private static void ProcessA()
{
var random = new Random();
var readBlock = new TransformBlock<int, int>(x =>
{
try { return DoSomething(x, "readBlock"); }
catch (Exception e) { throw e; }
}); //1
var braodcastBlock = new BroadcastBlock<int>(i => i); // ⬅ Here
var processBlock1 = new TransformBlock<int, int>(x =>
DoSomethingAsync(5, "processBlock1")); //2
var processBlock2 = new TransformBlock<int, int>(x =>
DoSomethingAsync(2, "processBlock2")); //3
//var saveBlock =
// new ActionBlock<int>(
// x => Save(x)); //4
var saveBlockJoinedProcess =
new ActionBlock<Tuple<int, int>>(
x => SaveJoined(x.Item1, x.Item2)); //4
var saveBlockJoin = new JoinBlock<int, int>();
readBlock.LinkTo(braodcastBlock, new DataflowLinkOptions
{ PropagateCompletion = true });
braodcastBlock.LinkTo(processBlock1,
new DataflowLinkOptions { PropagateCompletion = true }); //5
braodcastBlock.LinkTo(processBlock2,
new DataflowLinkOptions { PropagateCompletion = true }); //6
processBlock1.LinkTo(
saveBlockJoin.Target1); //7
processBlock2.LinkTo(
saveBlockJoin.Target2); //8
saveBlockJoin.LinkTo(saveBlockJoinedProcess,
new DataflowLinkOptions { PropagateCompletion = true });
readBlock.Post(1); //10
//readBlock.Post(2); //10
Task.WhenAll(processBlock1.Completion,processBlock2.Completion)
.ContinueWith(_ => saveBlockJoin.Complete());
readBlock.Complete(); //12
saveBlockJoinedProcess.Completion.Wait(); //13
if (readBlock.Completion.Exception != null
|| saveBlockJoinedProcess.Completion.Exception != null
|| processBlock1.Completion.Exception != null
|| processBlock2.Completion.Exception != null)
{
throw readBlock.Completion.Exception;
}
}
private static int DoSomething(int i, string method)
{
Console.WriteLine($"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!");
}
}
}发布于 2019-07-10 12:16:50
我在网上看了一下建议的方法,但没有看到任何明显的东西。
如果您有管道(或多或少),那么常用的方法是使用PropagateCompletion关闭管道。如果您有更复杂的拓扑,那么您将需要手动完成块。
在您的示例中,您尝试在此处进行传播:
Task.WhenAll(
processBlock1.Completion,
processBlock2.Completion)
.ContinueWith(_ => saveBlockJoin.Complete());但此代码不会传播异常。当processBlock1.Completion和processBlock2.Completion都完成时,表示saveBlockJoin已成功完成。
一个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鼓励您处理异常,方法是将异常传递给Fault以传播异常。
发布于 2019-08-13 19:00:47
在开箱即用的TPL数据流中,不支持在管道中向后传播错误,当块具有有限容量时,这尤其令人恼火。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。下面是如何做到这一点的。首先创建一个CancellationTokenSource
var cts = new CancellationTokenSource();然后一个接一个地创建块,在所有块的选项中嵌入相同的CancellationToken:
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);然后将块链接在一起,包括PropagateCompletion设置:
block1.LinkTo(block2, new DataflowLinkOptions { PropagateCompletion = true });最后,在异常情况下,使用扩展方法触发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);
}
}发布于 2019-07-09 19:32:12
乍一看,如果只有一些次要的要点(不看你的架构)。在我看来,你混合了一些新的和一些旧的结构。还有一些代码部分是不必要的。
例如:
private static void ProcessB()
{
Task.WhenAll(Task.Run(() => DoSomething(1, "ProcessB"))).Wait();
}使用Wait()-method,如果发生任何异常,它们将被包装在一个System.AggregateException中。在我看来,这样更好:
private static async Task ProcessBAsync()
{
await Task.Run(() => DoSomething(1, "ProcessB"));
}使用async-await时,如果发生异常,await语句将重新抛出包装在System.AggregateException中的第一个异常。这允许您尝试捕获具体的异常类型,并只处理您真正可以处理的情况。
另一件事是你代码的这一部分:
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为什么捕获异常只是为了重新抛出它?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(....)要好得多。使用Task.Delay(...),您的应用程序不会冻结。
https://stackoverflow.com/questions/56949426
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