Let say I have a generic member in a class or method, so:
public class Foo
{
public List Bar { get; set; }
public void Baz()
{
Using 3dGrabber's solution:
public static T GetEnumeratedType<T>(this IEnumerable<T> _)
{
return default(T);
}
//and now
var list = new Dictionary<string, int>();
var stronglyTypedVar = list.GetEnumeratedType();
If you dont need the whole Type variable and just want to check the type you can easily create a temp variable and use is operator.
T checkType = default(T);
if (checkType is MyClass)
{}
Try
list.GetType().GetGenericArguments()
That's work for me. Where myList is some unknown kind of list.
IEnumerable myEnum = myList as IEnumerable;
Type entryType = myEnum.AsQueryable().ElementType;
You can get the type of "T" from any collection type that implements IEnumerable<T> with the following:
public static Type GetCollectionItemType(Type collectionType)
{
var types = collectionType.GetInterfaces()
.Where(x => x.IsGenericType
&& x.GetGenericTypeDefinition() == typeof(IEnumerable<>))
.ToArray();
// Only support collections that implement IEnumerable<T> once.
return types.Length == 1 ? types[0].GetGenericArguments()[0] : null;
}
Note that it doesn't support collection types that implement IEnumerable<T> twice, e.g.
public class WierdCustomType : IEnumerable<int>, IEnumerable<string> { ... }
I suppose you could return an array of types if you needed to support this...
Also, you might also want to cache the result per collection type if you're doing this a lot (e.g. in a loop).
This is how i did it
internal static Type GetElementType(this Type type)
{
//use type.GenericTypeArguments if exist
if (type.GenericTypeArguments.Any())
return type.GenericTypeArguments.First();
return type.GetRuntimeProperty("Item").PropertyType);
}
Then call it like this
var item = Activator.CreateInstance(iListType.GetElementType());
OR
var item = Activator.CreateInstance(Bar.GetType().GetElementType());