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

I got this solution for the same problem:

int[] ints = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
IEnumerable> chunks = Chunk(ints, 2, t => t.Dump());
//won't enumerate, so won't do anything unless you force it:
chunks.ToList();

IEnumerable Chunk(IEnumerable src, int n, Func, T> action){
  IEnumerable head;
  IEnumerable tail = src;
  while (tail.Any())
  {
    head = tail.Take(n);
    tail = tail.Skip(n);
    yield return action(head);
  }
}

if you just want the chunks returned, not do anything with them, use chunks = Chunk(ints, 2, t => t). What I would really like is to have to have t=>t as default action, but I haven't found out how to do that yet.