Return the same instance type as the input type in generics

Question

Tried this code:

public static T NotEmpty<T, X> (T instance, string paramName = null) where T : IEnumerable<X>
{
    if (instance != null)
        if (instance.Any ())
            return instance;

    throw new Exception ();
}

static void Main (string[] args)
{
    var list = new List<int> () { 1, 2, 3 };
    var t = NotEmpty (list);
}

EDIT: So I want to pass any type that implements IEnumerable like List<int> and return instance of same type ( LIst<int> ).

error: var t = NotEmpty (list); The type arguments for method NoName.Program.NotEmpty<T,X>(T, string) cannot be inferred from the usage. Try specifying the type arguments explicitly.

Answer 1

I believe you don't need two generic parameter. You can replace T as IEnumerable<T>

public static IEnumerable<T> NotEmpty<T> (IEnumerable<T> instance, string paramName = null)
{
    if (instance != null && instance.Any())
        return instance;

    throw new Exception ();
}

Demo.