741
votes

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

public class Foo<T>
{
    public List<T> Bar { get; set; }
    
    public void Baz()
    {
        // get type of T
    }   
}

When I instantiate the class, the T becomes MyTypeObject1, so the class has a generic list property: List<MyTypeObject1>. The same applies to a generic method in a non-generic class:

public class Foo
{
    public void Bar<T>()
    {
        var baz = new List<T>();
        
        // get type of T
    }
}

I would like to know what type of objects the list of my class contains. So what type of T does the list property called Bar or the local variable baz contain?

I cannot do Bar[0].GetType(), because the list might contain zero elements. How can I do it?

16

16 Answers

775
votes

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.

545
votes

(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];
54
votes

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
33
votes

Try

list.GetType().GetGenericArguments()
16
votes

The following works for me. Where myList is some unknown kind of list.

IEnumerable myEnum = myList as IEnumerable;
Type entryType = myEnum.AsQueryable().ElementType;
14
votes

If you don’t need the whole Type variable and just want to check the type, you can easily create a temporary variable and use the is operator.

T checkType = default(T);

if (checkType is MyClass)
{}
9
votes

You can use this one for the return type of a generic list:

public string ListType<T>(T value)
{
    var valueType = value.GetType().GenericTypeArguments[0].FullName;
    return valueType;
}
8
votes

Consider this:

I use it to export 20 typed lists by the 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);
    }
}
7
votes

I use this extension method to accomplish something similar:

public static string GetFriendlyTypeName(this Type t)
{
    var typeName = t.Name.StripStartingWith("`");
    var genericArgs = t.GetGenericArguments();
    if (genericArgs.Length > 0)
    {
        typeName += "<";
        foreach (var genericArg in genericArgs)
        {
            typeName += genericArg.GetFriendlyTypeName() + ", ";
        }
        typeName = typeName.TrimEnd(',', ' ') + ">";
    }
    return typeName;
}

public static string StripStartingWith(this string s, string stripAfter)
{
    if (s == null)
    {
        return null;
    }
    var indexOf = s.IndexOf(stripAfter, StringComparison.Ordinal);
    if (indexOf > -1)
    {
        return s.Substring(0, indexOf);
    }
    return s;
}

You use it like this:

[TestMethod]
public void GetFriendlyTypeName_ShouldHandleReallyComplexTypes()
{
    typeof(Dictionary<string, Dictionary<string, object>>).GetFriendlyTypeName()
        .ShouldEqual("Dictionary<String, Dictionary<String, Object>>");
}

This isn't quite what you're looking for, but it's helpful in demonstrating the techniques involved.

6
votes

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).

6
votes

The GetGenericArgument() method has to be set on the Base Type of your instance (whose class is a generic class myClass<T>). Otherwise, it returns a type[0].

Example:

Myclass<T> instance = new Myclass<T>();
Type[] listTypes = typeof(instance).BaseType.GetGenericArguments();
1
votes

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();
0
votes
public bool IsCollection<T>(T value){
  var valueType = value.GetType();
  return valueType.IsArray() || typeof(IEnumerable<object>).IsAssignableFrom(valueType) || typeof(IEnumerable<T>).IsAssignableFrom(valuetype);
}
0
votes

If you want to know a property's underlying type, try this:

propInfo.PropertyType.UnderlyingSystemType.GenericTypeArguments[0]
-1
votes

This is how I did it:

internal static Type GetElementType(this Type type)
{
    // Use type.GenericTypeArguments if it exists
    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());
-9
votes

Type:

type = list.AsEnumerable().SingleOrDefault().GetType();