TPL 数据流：如何限制整个管道？

Question

我想限制在 Dataflow 管道中 posted 的项目数。项目的数量取决于生产环境。这些对象消耗大量内存（图像），因此我想在管道的最后一个块完成其工作时 post 它们。

我尝试使用 SemaphoreSlim 来限制生产者并在管道的最后一个块中释放它。可以，但是如果过程中抛出异常，程序会一直等待，异常不会被拦截。

这是一个看起来像我们的代码的示例。我怎样才能做到这一点？

static void Main(string[] args)
{
    SemaphoreSlim semaphore = new SemaphoreSlim(1, 2);

    var downloadString = new TransformBlock<string, string>(uri =>
    {
        Console.WriteLine("Downloading '{0}'...", uri);
        return new WebClient().DownloadString(uri);
    });

    var createWordList = new TransformBlock<string, string[]>(text =>
    {
        Console.WriteLine("Creating word list...");

        char[] tokens = text.ToArray();
        for (int i = 0; i < tokens.Length; i++)
        {
            if (!char.IsLetter(tokens[i]))
                tokens[i] = ' ';
        }
        text = new string(tokens);

        return text.Split(new char[] { ' ' },
           StringSplitOptions.RemoveEmptyEntries);
    });

    var filterWordList = new TransformBlock<string[], string[]>(words =>
    {
        Console.WriteLine("Filtering word list...");
        throw new InvalidOperationException("ouch !"); // explicit for test
        return words.Where(word => word.Length > 3).OrderBy(word => word)
           .Distinct().ToArray();
    });

    var findPalindromes = new TransformBlock<string[], string[]>(words =>
    {
        Console.WriteLine("Finding palindromes...");

        var palindromes = new ConcurrentQueue<string>();

        Parallel.ForEach(words, word =>
        {
            string reverse = new string(word.Reverse().ToArray());

            if (Array.BinarySearch<string>(words, reverse) >= 0 &&
                word != reverse)
            {
                palindromes.Enqueue(word);
            }
        });

        return palindromes.ToArray();
    });

    var printPalindrome = new ActionBlock<string[]>(palindromes =>
    {
        try
        {
            foreach (string palindrome in palindromes)
            {
                Console.WriteLine("Found palindrome {0}/{1}",
                   palindrome, new string(palindrome.Reverse().ToArray()));
            }
        }
        finally
        {
            semaphore.Release();
        }
    });

    downloadString.LinkTo(createWordList);
    createWordList.LinkTo(filterWordList);
    filterWordList.LinkTo(findPalindromes);
    findPalindromes.LinkTo(printPalindrome);


    downloadString.Completion.ContinueWith(t =>
    {
        if (t.IsFaulted) 
            ((IDataflowBlock)createWordList).Fault(t.Exception);
        else createWordList.Complete();
    });
    createWordList.Completion.ContinueWith(t =>
    {
        if (t.IsFaulted) 
            ((IDataflowBlock)filterWordList).Fault(t.Exception);
        else filterWordList.Complete();
    });
    filterWordList.Completion.ContinueWith(t =>
    {
        if (t.IsFaulted) 
            ((IDataflowBlock)findPalindromes).Fault(t.Exception); // enter here when an exception throws
        else findPalindromes.Complete();
    });
    findPalindromes.Completion.ContinueWith(t =>
    {
        if (t.IsFaulted)
            ((IDataflowBlock)printPalindrome).Fault(t.Exception); // the fault is propagated here but not catched
        else printPalindrome.Complete();
    });

    try
    {
        for (int i = 0; i < 10; i++)
        {
            Console.WriteLine(i);

            downloadString.Post("http://www.google.com");
            semaphore.Wait(); // waits here when an exception throws
        }

        downloadString.Complete();

        printPalindrome.Completion.Wait();
    }
    catch (AggregateException agg)
    {
        Console.WriteLine("An error has occured : " + agg);
    }
    Console.WriteLine("Done");
    Console.ReadKey();
}

Answer 1

您应该同时等待信号量和完成任务。这样，如果块过早结束（通过异常或取消），那么异常将被重新抛出，否则你将等待你的信号量，直到有空间 post 更多。

您可以使用 Task.WhenAny 和 SemaphoreSlim.WaitAsync:

for (int i = 0; i < 10; i++)
{
    Console.WriteLine(i);
    downloadString.Post("http://www.google.com");

    if (printPalindrome.Completion.IsCompleted)
    {
        break;
    }

    Task.WhenAny(semaphore.WaitAsync(), printPalindrome.Completion).Wait();
}

注意：使用 Task.Wait 仅适用于这种情况，因为它是 Main。通常这应该是一个 async 方法，你应该 await 从 Task.WhenAny.

返回的任务

Answer 2

这就是我处理限制或在任何时候只允许源块中有 10 个项目的方式。您可以将其修改为 1。确保您还限制了管道中的任何其他块，否则，您可以获得 1 的源块和更多的下一个块。

var sourceBlock = new BufferBlock<string>(
    new ExecutionDataflowBlockOptions() { 
        SingleProducerConstrained = true, 
        BoundedCapacity = 10 });

然后生产者这样做：

sourceBlock.SendAsync("value", shutdownToken).Wait(shutdownToken);

如果您使用异步/等待，只需等待 SendAsync 调用。

TPL 数据流：如何限制整个管道？

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