c# - 将动态创建的 ActionBlock 链接到 BufferBlock

Question

我不确定这是否可行，但如果可行，我可能做得不对。假设我有一个链接到许多消费者(ActionBlocks)的共享缓冲区。每个消费者都应该消费满足用于将其链接到缓冲区的谓词的数据。例如，ActionBlock1 应该消耗满足 x => x % 5 == 0 的数字, ActionBlock2 应该只消耗 x => x % 5 == 1等等

这是我得到的:

private static ITargetBlock<int> BuildPipeline(int NumProductionLines)
{
    var productionQueue = new BufferBlock<int>();

    for (int i = 0; i < NumProductionLines; i++)
    {
        ActionBlock<int> productionLine = new ActionBlock<int>(num => Console.WriteLine("Processed by line {0}: {1}", i + 1, num));

        productionQueue.LinkTo(productionLine, x => x % NumProductionLines == i);
    }

    return productionQueue;
}

然后我调用:

Random rnd = new Random();

ITargetBlock<int> temp = BuildPipeline(5);

while (true)
{
    temp.Post(rnd.Next(255));
}

但是这不起作用。控制台中不显示任何输出。如果我修改 BuildPipeline方法为:

private static ITargetBlock<int> BuildPipeline(int NumProductionLines)
{
    var productionQueue = new BufferBlock<int>();

    ActionBlock<int> productionLine1 = new ActionBlock<int>(num => Console.WriteLine("Processed by line {0}: {1}", 1, num));
    ActionBlock<int> productionLine2 = new ActionBlock<int>(num => Console.WriteLine("Processed by line {0}: {1}", 2, num));
    ActionBlock<int> productionLine3 = new ActionBlock<int>(num => Console.WriteLine("Processed by line {0}: {1}", 3, num));
    ActionBlock<int> productionLine4 = new ActionBlock<int>(num => Console.WriteLine("Processed by line {0}: {1}", 4, num));
    ActionBlock<int> productionLine5 = new ActionBlock<int>(num => Console.WriteLine("Processed by line {0}: {1}", 5, num));

    productionQueue.LinkTo(productionLine1, x => x % 5 == 0);
    productionQueue.LinkTo(productionLine2, x => x % 5 == 1);
    productionQueue.LinkTo(productionLine3, x => x % 5 == 2);
    productionQueue.LinkTo(productionLine4, x => x % 5 == 3);
    productionQueue.LinkTo(productionLine5, x => x % 5 == 4);

    return productionQueue;
}

代码执行预期的操作。

有人可以阐明为什么动态创建和链接操作 block 不起作用吗？

附言如果我在 ITargetBlock<int> temp = BuildPipeline(5); 之后立即进入代码temp 确实显示有 5 个目标链接到缓冲区。而且每个target的Id都不一样。

提前致谢

编辑:添加了 svick 建议的更改，但仍然不好:

private static ITargetBlock<int> BuildPipeline(int NumProductionLines)
{
    var productionQueue = new BufferBlock<int>();
    var opt = new ExecutionDataflowBlockOptions { MaxDegreeOfParallelism = 1 };

    for (int i = 0; i < NumProductionLines; i++)
    {
        ActionBlock<int> productionLine = new ActionBlock<int>(num => Console.WriteLine("Processed by line {0}: {1}", i + 1, num));

        int j = i;
        productionQueue.LinkTo(productionLine, x => x % NumProductionLines == j);
    }

    ActionBlock<int> discardedLine = new ActionBlock<int>(num => Console.WriteLine("Discarded: {0}", num));
    productionQueue.LinkTo(discardedLine);

    return productionQueue;
}

现在只有第二条生产线处理数据(满足 x % 5 == 1 谓词的一条)。并且数据不满足谓词，这意味着我得到以 9 和 7 结尾的数字。

编辑:工作代码如下所示:

private static ITargetBlock<int> BuildPipeline(int NumProductionLines)
{
    var productionQueue = new BufferBlock<int>();
    var opt = new ExecutionDataflowBlockOptions { MaxDegreeOfParallelism = 1 };

    for (int i = 0; i < NumProductionLines; i++)
    {
        int j = i;
        ActionBlock<int> productionLine = new ActionBlock<int>(num => Console.WriteLine("Processed by line {0}: {1}", j + 1, num));

        productionQueue.LinkTo(productionLine, x => x % NumProductionLines == j);
    }

    productionQueue.LinkTo(DataflowBlock.NullTarget<int>());

    return productionQueue;
}

Answer 1

问题是在您的第一个版本中，您对每个目标 block 使用相同的谓词。换句话说，谓词不依赖于 i。

但即使这样做了，您的代码也不会工作，因为 i 变量在谓词之间共享，因此它们都将使用最后一个值。解决方法是将 i 复制到局部变量中并在谓词中使用它。

代码可能是这样的:

private static ITargetBlock<int> BuildPipeline(int NumProductionLines)
{
    var productionQueue = new BufferBlock<int>();

    for (int i = 0; i < NumProductionLines; i++)
    {
        int iCopy = i;

        ActionBlock<int> productionLine = new ActionBlock<int>(
            num => Console.WriteLine("Processed by line {0}: {1}", iCopy + 1, num));

        productionQueue.LinkTo(
            productionLine, x => x % NumProductionLines == iCopy);
    }

    return productionQueue;
}

如果您问为什么您的代码不至少处理 x % 5 == 1 个数字，那是因为随机生成器可能会生成一个与该谓词不匹配的数字, 所以没有一个 ActionBlock 会接受它。因此，该号码将留在源 block 的输出队列中，其他号码将无法通过。

如果在您的实际代码中可能会发生类似情况，并且您想丢弃所有不符合任何谓词的数字，您可以将源代码块链接到 a block that does nothing并在将其链接到所有有用的 block 之后接受任何内容:

productionQueue.LinkTo(DataflowBlock.NullTarget<int>());