C# Interface IEnumerable Any() without specifying generic types

Question

I have casted

var info = property.Info;
object data = info.GetValue(obj);

...

var enumerable = (IEnumerable)data;

if (enumerable.Any())  ///Does not compile
{
}

if (enumerable.GetEnumerator().Current != null) // Run time error
{
}

and I would like to see if this enumerable has any elements, via using Linq Query Any(). But unfortunately, even with using Linq, I can't.

How would I do this without specifying the generic type.

score 25 · Accepted Answer · answered Oct 02 '18 at 00:47

While you can't do this directly, you could do it via Cast:

if (enumerable.Cast<object>().Any())

That should always work, as any IEnumerable can be wrapped as an IEnumerable<object>. It will end up boxing the first element if it's actually an IEnumerable<int> or similar, but it should work fine. Unlike most LINQ methods, Cast and OfType target IEnumerable rather than IEnumerable<T>.

You could write your own subset of extension methods like the LINQ ones but operating on the non-generic IEnumerable type if you wanted to, of course. Implementing LINQ to Objects isn't terribly hard - you could use my Edulinq project as a starting point, for example.

There are cases where you could implement Any(IEnumerable) slightly more efficiently than using Cast - for example, taking a shortcut if the target implements the non-generic ICollection interface. At that point, you wouldn't need to create an iterator or take the first element. In most cases that won't make much performance difference, but it's the kind of thing you could do if you were optimizing.

user2864740 · Answer 2 · 2018-10-02T00:47:26.957

2

One method is to use foreach, as noted in IEnumerable "Remarks". It also provides details on the additional methods off of the result of GetEnumerator.

bool hasAny = false;
foreach (object i in (IEnumerable)(new int[1] /* IEnumerable of any type */)) {
    hasAny = true;
    break;
}

(Which is itself easily transferable to an Extension method.)

edited Oct 02 '18 at 00:47

answered Oct 02 '18 at 00:45

user2864740

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is that really the only approach? Suprising. Seems like a big hack – Ilan Keshet Oct 02 '18 at 00:47
@IlanKeshet Not the "only" approach. It is also not really a "hack", given what the IEnumerable interface exposes. – user2864740 Oct 02 '18 at 00:47
Surely not a Hack, quite a valid way to enumerate through the `IEnumerable` and check various conditions – Mrinal Kamboj Oct 02 '18 at 09:42

score 2 · Answer 3 · answered Oct 10 '18 at 15:15

Your attempt to use GetEnumerator().Current tried to get the current value of an enumerator that had not yet been moved to the first position yet. It would also have given the wrong result if the first item existed or was null. What you could have done (and what the Any() in Enumerable does) is see if it was possible to move to that first item or not; i.e. is there a first item to move to:

internal static class UntypedLinq
{
    public static bool Any(this IEnumerable source)
    {
        if (source == null) throw new ArgumentNullException(nameof(source));
        IEnumerator ator = source.GetEnumerator();
        // Unfortunately unlike IEnumerator<T>, IEnumerator does not implement
        // IDisposable. (A design flaw fixed when IEnumerator<T> was added).
        // We need to test whether disposal is required or not.
        if (ator is IDisposable disp)
        {
            using(disp)
            {
                return ator.MoveNext();
            }
        }

        return ator.MoveNext();
    }

    // Not completely necessary. Causes any typed enumerables to be handled by the existing Any
    // in Linq via a short method that will be inlined.
    public static bool Any<T>(this IEnumerable<T> source) => Enumerable.Any(source);
}

C# Interface IEnumerable Any() without specifying generic types

3 Answers3