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What is the best way to modify a list in a 'foreach' loop?

Asked 2009-04-17T10:48:53.803
109

A new feature in C# / .NET 4.0 is that you can change your enumerable in a foreach without getting the exception. See Paul Jackson's blog entry An Interesting Side-Effect of Concurrency: Removing Items from a Collection While Enumerating for information on this change.

What is the best way to do the following?

foreach(var item in Enumerable)
{
    foreach(var item2 in item.Enumerable)
    {
        item.Add(new item2)
    }
}

Usually I use an IList as a cache/buffer until the end of the foreach, but is there better way?

c# .net list enumeration enumerable
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1 Answer

3

Here's how you can do that (quick and dirty solution. If you really need this kind of behavior, you should either reconsider your design or override all IList<T> members and aggregate the source list):

using System;
using System.Collections.Generic;

namespace ConsoleApplication3
{
    public class ModifiableList<T> : List<T>
    {
        private readonly IList<T> pendingAdditions = new List<T>();
        private int activeEnumerators = 0;

        public ModifiableList(IEnumerable<T> collection) : base(collection)
        {
        }

        public ModifiableList()
        {
        }

        public new void Add(T t)
        {
            if(activeEnumerators == 0)
                base.Add(t);
            else
                pendingAdditions.Add(t);
        }

        public new IEnumerator<T> GetEnumerator()
        {
            ++activeEnumerators;

            foreach(T t in ((IList<T>)this))
                yield return t;

            --activeEnumerators;

            AddRange(pendingAdditions);
            pendingAdditions.Clear();
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            ModifiableList<int> ints = new ModifiableList<int>(new int[] { 2, 4, 6, 8 });

            foreach(int i in ints)
                ints.Add(i * 2);

            foreach(int i in ints)
                Console.WriteLine(i * 2);
        }
    }
}
answered 2009-04-17T11:02:43.223

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