Passing arguments to C# generic new() of templated type

In order to create an instance of a generic type in a function you must constrain it with the "new" flag.

public static string GetAllItems<T>(...) where T : new()

However that will only work when you want to call the constructor which has no parameters. Not the case here. Instead you'll have to provide another parameter which allows for the creation of object based on parameters. The easiest is a function.

public static string GetAllItems<T>(..., Func<ListItem,T> del) {
  ...
  List<T> tabListItems = new List<T>();
  foreach (ListItem listItem in listCollection) 
  {
    tabListItems.Add(del(listItem));
  }
  ...
}

You can then call it like so

GetAllItems<Foo>(..., l => new Foo(l));

in .Net 3.5 and after you could use the activator class:

(T)Activator.CreateInstance(typeof(T), args)

Since nobody bothered to post the 'Reflection' answer (which I personally think is the best answer), here goes:

public static string GetAllItems<T>(...) where T : new()
{
   ...
   List<T> tabListItems = new List<T>();
   foreach (ListItem listItem in listCollection) 
   {
       Type classType = typeof(T);
       ConstructorInfo classConstructor = classType.GetConstructor(new Type[] { listItem.GetType() });
       T classInstance = (T)classConstructor.Invoke(new object[] { listItem });

       tabListItems.Add(classInstance);
   } 
   ...
}

Edit: This answer is deprecated due to .NET 3.5's Activator.CreateInstance, however it is still useful in older .NET versions.