TPL数据流块未按预期处理

Question

我有一组简单的块，它们大多是以串行方式处理的，但我有两个块，我想并行处理(processblock1 & processblock2)。我刚开始玩TPL数据库锁，对它来说太新了。但是，在下面的代码中，我可以看到paraellelblock1被调用为parallelblock2，但从未像预期的那样被调用。我希望他们能同时被踢走。

    class Program
    {
        static void Main(string[] args)
        {
            var readBlock = new TransformBlock<int, int>(x => DoSomething(x, "readBlock"),
                new ExecutionDataflowBlockOptions { MaxDegreeOfParallelism = 1 }); //1

            var processBlock1 =
                new TransformBlock<int, int>(x => DoSomething(x, "processBlock1")); //2
            var processBlock2 =
                new TransformBlock<int, int>(x => DoSomething(x, "processBlock2")); //3

            var saveBlock =
                new ActionBlock<int>(
                x => Save(x)); //4

            readBlock.LinkTo(processBlock1,
                new DataflowLinkOptions { PropagateCompletion = true }); //5

            readBlock.LinkTo(processBlock2,
                new DataflowLinkOptions { PropagateCompletion = true }); //6


            processBlock1.LinkTo(
                saveBlock); //7

            processBlock2.LinkTo(
                saveBlock); //8

            readBlock.Post(1); //10

            Task.WhenAll(
                processBlock1.Completion,
                processBlock2.Completion)
                .ContinueWith(_ => saveBlock.Complete()); //11

            readBlock.Complete(); //12
            saveBlock.Completion.Wait(); //13
            Console.WriteLine("Processing complete!");
            Console.ReadLine();
        }
        private static int DoSomething(int i, string method)
        {
            Console.WriteLine($"Do Something, callng method : { method}");
            return i;
        }
        private static async Task<int> DoSomethingAsync(int i, string method)
        {
            DoSomething(i, method);
            return i;
        }
        private static void Save(int i)
        {
            Console.WriteLine("Save!");
        }


    }

Answer 1

默认情况下，tpl块将只向第一个链接块发送消息。使用BroadcastBlock向许多组件发送消息。

void Main()
{
    var random = new Random();
    var readBlock = new TransformBlock<int, int>(x => {  return DoSomething(x, "readBlock"); },
            new ExecutionDataflowBlockOptions { MaxDegreeOfParallelism = 1 }); //1

    var braodcastBlock = new BroadcastBlock<int>(i => i); // ⬅️ Here

    var processBlock1 =
        new TransformBlock<int, int>(x => DoSomething(x, "processBlock1")); //2
    var processBlock2 =
        new TransformBlock<int, int>(x => DoSomething(x, "processBlock2")); //3

    var saveBlock =
        new ActionBlock<int>(
        x => Save(x)); //4

    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(
        saveBlock); //7

    processBlock2.LinkTo(
        saveBlock); //8

    readBlock.Post(1); //10
    readBlock.Post(2); //10

    Task.WhenAll(
                processBlock1.Completion,
                processBlock2.Completion)
                .ContinueWith(_ => saveBlock.Complete());

    readBlock.Complete(); //12
    saveBlock.Completion.Wait(); //13
    Console.WriteLine("Processing complete!");
}

// Define other methods and classes here

private static int DoSomething(int i, string method)
{
    Console.WriteLine($"Do Something, callng method : { method} {i}");
    return i;
}
private static Task<int> DoSomethingAsync(int i, string method)
{
    DoSomething(i, method);
    return Task.FromResult(i);
}
private static void Save(int i)
{
    Console.WriteLine("Save! " + i);
}

Answer 2

看起来，您只向图表发布了一个项目，而第一个消费它的消费者就赢了。图中没有隐含的“tee”功能--因此不存在可能的并行性。