How to get the type of T from a member of a generic class or method?

Question

Let say I have a generic member in a class or method, so:

public class Foo
{
    public List Bar { get; set; }

    public void Baz()
    {         


        

Answer 1

(note: I'm assuming that all you know is object or IList or similar, and that the list could be any type at runtime)

If you know it is a List<T>, then:

Type type = abc.GetType().GetGenericArguments()[0];

Another option is to look at the indexer:

Type type = abc.GetType().GetProperty("Item").PropertyType;

Using new TypeInfo:

using System.Reflection;
// ...
var type = abc.GetType().GetTypeInfo().GenericTypeArguments[0];

Answer 2

If I understand correctly, your list has the same type parameter as the container class itself. If this is the case, then:

Type typeParameterType = typeof(T);

If you are in the lucky situation of having object as a type parameter, see Marc's answer.

Answer 3

With the following extension method you can get away without reflection:

public static Type GetListType<T>(this List<T> _)
{
    return typeof(T);
}

Or more general:

public static Type GetEnumeratedType<T>(this IEnumerable<T> _)
{
    return typeof(T);
}

Usage:

List<string>        list    = new List<string> { "a", "b", "c" };
IEnumerable<string> strings = list;
IEnumerable<object> objects = list;

Type listType    = list.GetListType();           // string
Type stringsType = strings.GetEnumeratedType();  // string
Type objectsType = objects.GetEnumeratedType();  // BEWARE: object

Answer 4

Consider this: I use it to export 20 typed list by same way:

private void Generate<T>()
{
    T item = (T)Activator.CreateInstance(typeof(T));

    ((T)item as DemomigrItemList).Initialize();

    Type type = ((T)item as DemomigrItemList).AsEnumerable().FirstOrDefault().GetType();
    if (type == null) return;
    if (type != typeof(account)) //account is listitem in List<account>
    {
        ((T)item as DemomigrItemList).CreateCSV(type);
    }
}