Get next N elements from enumerable

Question

Context: C# 3.0, .Net 3.5
Suppose I have a method that generates random numbers (forever):

private static IEnumerable RandomNumberGenerator(         


        

Answer 1

It seems like we'd prefer for an IEnumerable to have a fixed position counter so that we can do

var group1 = items.Take(10);
var group2 = items.Take(10);
var group3 = items.Take(10);
var group4 = items.Take(10);

and get successive slices rather than getting the first 10 items each time. We can do that with a new implementation of IEnumerable which keeps one instance of its Enumerator and returns it on every call of GetEnumerator:

public class StickyEnumerable : IEnumerable, IDisposable
{
    private IEnumerator innerEnumerator;

    public StickyEnumerable( IEnumerable items )
    {
        innerEnumerator = items.GetEnumerator();
    }

    public IEnumerator GetEnumerator()
    {
        return innerEnumerator;
    }

    System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
    {
        return innerEnumerator;
    }

    public void Dispose()
    {
        if (innerEnumerator != null)
        {
            innerEnumerator.Dispose();
        }
    }
}

Given that class, we could implement Slice with

public static IEnumerable> Slices(this IEnumerable items, int size)
{
    using (StickyEnumerable sticky = new StickyEnumerable(items))
    {
        IEnumerable slice;
        do
        {
            slice = sticky.Take(size).ToList();
            yield return slice;
        } while (slice.Count() == size);
    }
    yield break;
}

That works in this case, but StickyEnumerable is generally a dangerous class to have around if the consuming code isn't expecting it. For example,

using (var sticky = new StickyEnumerable(Enumerable.Range(1, 10)))
{
    var first = sticky.Take(2);
    var second = sticky.Take(2);
    foreach (int i in second)
    {
        Console.WriteLine(i);
    }
    foreach (int i in first)
    {
        Console.WriteLine(i);
    }
}

prints

1
2
3
4

rather than

3
4
1
2