getting type T from IEnumerable<T>

is there a way to retrieve type T from IEnumerable<T> through reflection?

e.g.

i have a variable IEnumerable<Child> info; i want to retrieve Child's type through reflection

IEnumerable<T> myEnumerable;
Type type = myEnumerable.GetType().GetGenericArguments()[0]; 

Thusly,

IEnumerable<string> strings = new List<string>();
Console.WriteLine(strings.GetType().GetGenericArguments()[0]);

prints System.String.

See MSDN for Type.GetGenericArguments.

Edit: I believe this will address the concerns in the comments:

// returns an enumeration of T where o : IEnumerable<T>
public IEnumerable<Type> GetGenericIEnumerables(object o) {
    return o.GetType()
            .GetInterfaces()
            .Where(t => t.IsGenericType
                && t.GetGenericTypeDefinition() == typeof(IEnumerable<>))
            .Select(t => t.GetGenericArguments()[0]);
}

Some objects implement more than one generic IEnumerable so it is necessary to return an enumeration of them.

Edit: Although, I have to say, it's a terrible idea for a class to implement IEnumerable<T> for more than one T.

I'd just make an extension method. This worked with everything I threw at it.

public static Type GetItemType<T>(this IEnumerable<T> enumerable)
{
    return typeof(T);
}

I had a similar problem. The selected answer works for actual instances. In my case I had only a type (from a PropertyInfo).

The selected answer fails when the type itself is typeof(IEnumerable<T>) not an implementation of IEnumerable<T>.

For this case the following works:

public static Type GetAnyElementType(Type type)
{
   // Type is Array
   // short-circuit if you expect lots of arrays 
   if (type.IsArray)
      return type.GetElementType();

   // type is IEnumerable<T>;
   if (type.IsGenericType && type.GetGenericTypeDefinition() == typeof (IEnumerable<>))
      return type.GetGenericArguments()[0];

   // type implements/extends IEnumerable<T>;
   var enumType = type.GetInterfaces()
                           .Where(t => t.IsGenericType && 
                                  t.GetGenericTypeDefinition() == typeof(IEnumerable<>))
                           .Select(t => t.GenericTypeArguments[0]).FirstOrDefault();
   return enumType ?? type;
}

If you know the IEnumerable<T> (via generics), then just typeof(T) should work. Otherwise (for object, or the non-generic IEnumerable), check the interfaces implemented:

        object obj = new string[] { "abc", "def" };
        Type type = null;
        foreach (Type iType in obj.GetType().GetInterfaces())
        {
            if (iType.IsGenericType && iType.GetGenericTypeDefinition()
                == typeof(IEnumerable<>))
            {
                type = iType.GetGenericArguments()[0];
                break;
            }
        }
        if (type != null) Console.WriteLine(type);

Thank you very much for the discussion. I used it as a basis for the solution below, which works well for all cases that are of interest to me (IEnumerable, derived classes, etc). Thought I should share here in case anyone needs it also:

  Type GetItemType(object someCollection)
  {
    var type = someCollection.GetType();
    var ienum = type.GetInterface(typeof(IEnumerable<>).Name);
    return ienum != null
      ? ienum.GetGenericArguments()[0]
      : null;
  }

public static Type GetInnerGenericType(this Type type)
{
  // Attempt to get the inner generic type
  Type innerType = type.GetGenericArguments().FirstOrDefault();

  // Recursively call this function until no inner type is found
  return innerType is null ? type : innerType.GetInnerGenericType();
}

This is a recursive function that will go depth first down the list of generic types until it gets a concrete type definition with no inner generic types.

I tested this method with this type: ICollection<IEnumerable<ICollection<ICollection<IEnumerable<IList<ICollection<IEnumerable<T>>>>>>>>

which should return T

Just use typeof(T)

EDIT: Or use .GetType().GetGenericParameter() on an instantiated object if you don't have T.

An alternative for simpler situations where it's either going to be an IEnumerable<T> or T - note use of GenericTypeArguments instead of GetGenericArguments().

Type inputType = o.GetType();
Type genericType;
if ((inputType.Name.StartsWith("IEnumerable"))
    && ((genericType = inputType.GenericTypeArguments.FirstOrDefault()) != null)) {

    return genericType;
} else {
    return inputType;
}

I know this is a bit old, but I believe this method will cover all the problems and challenges stated in the comments. Credit to Eli Algranti for inspiring my work.

/// <summary>Finds the type of the element of a type. Returns null if this type does not enumerate.</summary>
/// <param name="type">The type to check.</param>
/// <returns>The element type, if found; otherwise, <see langword="null"/>.</returns>
public static Type FindElementType(this Type type)
{
   if (type.IsArray)
      return type.GetElementType();

   // type is IEnumerable<T>;
   if (ImplIEnumT(type))
      return type.GetGenericArguments().First();

   // type implements/extends IEnumerable<T>;
   var enumType = type.GetInterfaces().Where(ImplIEnumT).Select(t => t.GetGenericArguments().First()).FirstOrDefault();
   if (enumType != null)
      return enumType;

   // type is IEnumerable
   if (IsIEnum(type) || type.GetInterfaces().Any(IsIEnum))
      return typeof(object);

   return null;

   bool IsIEnum(Type t) => t == typeof(System.Collections.IEnumerable);
   bool ImplIEnumT(Type t) => t.IsGenericType && t.GetGenericTypeDefinition() == typeof(IEnumerable<>);
}

This is an improvement on Eli Algranti's solution in that it will also work where the IEnumerable<> type is at any level in the inheritance tree.

This solution will obtain the element type from any Type. If the type is not an IEnumerable<>, it will return the type passed in. For objects, use GetType. For types, use typeof, then call this extension method on the result.

public static Type GetGenericElementType(this Type type)
{
    // Short-circuit for Array types
    if (typeof(Array).IsAssignableFrom(type))
    {
        return type.GetElementType();
    }

    while (true)
    {
        // Type is IEnumerable<T>
        if (type.IsGenericType && type.GetGenericTypeDefinition() == typeof(IEnumerable<>))
        {
            return type.GetGenericArguments().First();
        }

        // Type implements/extends IEnumerable<T>
        Type elementType = (from subType in type.GetInterfaces()
            let retType = subType.GetGenericElementType()
            where retType != subType
            select retType).FirstOrDefault();

        if (elementType != null)
        {
            return elementType;
        }

        if (type.BaseType == null)
        {
            return type;
        }

        type = type.BaseType;
    }
}

typeof(IEnumerable<Foo>).GetGenericArguments()[0] will return the first generic argument - in this case typeof(Foo).

this is how I usually do it (via extension method):

public static Type GetIEnumerableUnderlyingType<T>(this T iEnumerable)
    {
        return typeof(T).GetTypeInfo().GetGenericArguments()[(typeof(T)).GetTypeInfo().GetGenericArguments().Length - 1];
    }

Here's my unreadable Linq query expression version ..

public static Type GetEnumerableType(this Type t) {
	return !typeof(IEnumerable).IsAssignableFrom(t) ? null : (
	from it in (new[] { t }).Concat(t.GetInterfaces())
	where it.IsGenericType
	where typeof(IEnumerable<>)==it.GetGenericTypeDefinition()
	from x in it.GetGenericArguments() // x represents the unknown
	let b = it.IsConstructedGenericType // b stand for boolean
	select b ? x : x.BaseType).FirstOrDefault()??typeof(object);
}

Note the method also takes non-generic IEnumerable into account, it returns object in this case, because it takes a Type rather than a concrete instance as the argument. By the way, for x represents the unknown, I found this video insteresting, though it is irrelevant ..