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First Posted 1 Jul 2008
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Action Extensions

By | 1 Jul 2008 | Article
Parallelization of multicast delegates

Introduction

Ennis Ray Lynch, Jr. gave me an idea the other day in the Lounge. The basic idea was running multicast delegates in parallel.

Background

The code to run delegates in parallel is rather trivial.

static void Run(Action[] aa)
{
  List<IAsyncResult> waits = new List<IAsyncResult>();

  foreach (Action a in aa)
  {
    waits.Add(a.BeginInvoke(null, null));
  }

  foreach (IAsyncResult ar in waits)
  {
    ar.AsyncWaitHandle.WaitOne();
  }
}

To run a multicast delegate in parallel follows the same pattern.

static void Run(Action t)
{
  List<IAsyncResult> waits = new List<IAsyncResult>();

  foreach (Action a in t.GetInvocationList())
  {
    waits.Add(a.BeginInvoke(null, null));
  }

  foreach (IAsyncResult ar in waits)
  {
    ar.AsyncWaitHandle.WaitOne();
  }
}

Using the Code

Following the above pattern, we simply create stubs for the generic Action delegate (only 1 shown for clarity). If needed, you can replace with your own delegate type. There is one important aspect to keep in mind; the delegate MUST return void. Why, you may ask? The answer is simple. There is no easy way to consume multiple return values in C#.

static class ActionExtensions
{
  class WaitList : IDisposable
  {
    readonly List<IAsyncResult> waits = new List<IAsyncResult>();

    public void Add(IAsyncResult ar)
    {
      waits.Add(ar);
    }

    public void Dispose()
    {
      foreach (var ar in waits)
      {
        ar.AsyncWaitHandle.WaitOne();
      }
    }
  }

  public static Action MakeParallel(this Action t)
  {
    return () =>
    {
      using (var w = new WaitList())
      {
        foreach (Action a in t.GetInvocationList())
        {
          w.Add(a.BeginInvoke(null, null));
        }
      }
    };
  }
}

The usage is also trivial. Simply call the MakeParallel extension method.

class Program
{
  static void Main(string[] args)
  {
    Action<int> f = null;

    for (int i = 0; i < 8; i++)
    {
      f += Thread.Sleep;
    }

    Stopwatch ws = Stopwatch.StartNew();

    f(250);

    Console.WriteLine("ser: {0:f3}", ws.Elapsed.TotalMilliseconds);

    f = f.MakeParallel();

    ws = Stopwatch.StartNew();

    f(250);

    Console.WriteLine("par: {0:f3}", ws.Elapsed.TotalMilliseconds);

    Console.ReadLine();
  }
}

The output should show the parallel version running at half (or quarter) the time of the serial version.

Points of Interest

From what I can determine, BeginInvoke utilizes the number of logical CPUs. I have however not been able to test this.

History

  • 2nd July, 2008 - Initial version

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

About the Author

leppie

Software Developer

South Africa South Africa

Member

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GeneralNice and simple. PinmvpSacha Barber23:15 6 Jan '11  
GeneralNice, but you should call EndInvoke() PinmemberJelle Hissink9:48 6 Jul '08  
GeneralRe: Nice, but you should call EndInvoke() Pinmemberleppie10:20 6 Jul '08  
GeneralSweet PinmemberNick Butler22:28 3 Jul '08  
GeneralRe: Sweet Pinmemberleppie7:21 4 Jul '08  
GeneralNice work Pinmemberkin3tik5:42 2 Jul '08  
Dude brilliant stuff ... great to see you still among the living also man
 
Banshee for windows YAY !!!
http://sourceforge.net/projects/banshee32

GeneralNice PinmvpPete O'Hanlon22:33 1 Jul '08  
GeneralRe: Nice Pinmemberleppie22:48 1 Jul '08  

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