How do I implement IEnumerable<T>
If you choose to use a generic collection, such as List<MyObject>
instead of ArrayList
, you'll find that the List<MyObject>
will provide both generic and non-generic enumerators that you can use.
using System.Collections;
class MyObjects : IEnumerable<MyObject>
{
List<MyObject> mylist = new List<MyObject>();
public MyObject this[int index]
{
get { return mylist[index]; }
set { mylist.Insert(index, value); }
}
public IEnumerator<MyObject> GetEnumerator()
{
return mylist.GetEnumerator();
}
IEnumerator IEnumerable.GetEnumerator()
{
return this.GetEnumerator();
}
}
You probably do not want an explicit implementation of IEnumerable<T>
(which is what you've shown).
The usual pattern is to use IEnumerable<T>
's GetEnumerator
in the explicit implementation of IEnumerable
:
class FooCollection : IEnumerable<Foo>, IEnumerable
{
SomeCollection<Foo> foos;
// Explicit for IEnumerable because weakly typed collections are Bad
System.Collections.IEnumerator IEnumerable.GetEnumerator()
{
// uses the strongly typed IEnumerable<T> implementation
return this.GetEnumerator();
}
// Normal implementation for IEnumerable<T>
IEnumerator<Foo> GetEnumerator()
{
foreach (Foo foo in this.foos)
{
yield return foo;
//nb: if SomeCollection is not strongly-typed use a cast:
// yield return (Foo)foo;
// Or better yet, switch to an internal collection which is
// strongly-typed. Such as List<T> or T[], your choice.
}
// or, as pointed out: return this.foos.GetEnumerator();
}
}
Why do you do it manually? yield return
automates the entire process of handling iterators. (I also wrote about it on my blog, including a look at the compiler generated code).
If you really want to do it yourself, you have to return a generic enumerator too. You won't be able to use an ArrayList
any more since that's non-generic. Change it to a List<MyObject>
instead. That of course assumes that you only have objects of type MyObject
(or derived types) in your collection.