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Home/ Questions/Q 1110755
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T02:26:33+00:00 2026-05-17T02:26:33+00:00

This is probably me just remembering things completely backwards, but I’d like to know

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This is probably me just remembering things completely backwards, but I’d like to know more about what I’m doing wrong…

I have declared a class to be nothing more than a direct inheritance from a generic list (done to simplify naming), something like this:

public class FooList : List<Foo> {}

now in another method completely separate from this class, I am trying to return an instance of this class, however I want to filter the class based on a criterion, so I’m using a lambda expression:

var list = new FooList(); // imagine this fills it with different items
var filtered = list.FindAll(c => c.Something == "filter criteria");

now according to the FindAll method, this SHOULD return a List[Foo]. However, I want to return this object as a FooList, not a List[Foo]. Do I have to create a new instance of FooList and copy the items from the List[Foo]?

If so, why? why can’t I convert a List to a FooList directly, since they are the same object?

If this CAN be done, how do I do it?

many thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-17T02:26:34+00:00Added an answer on May 17, 2026 at 2:26 am

    They are not the same thing. A FooList is a List<foo> but a List<foo>, which is being returned by the FindAll() function inherited from List<foo>, is not a FooList. You will need to construct a new FooList.

    You could do something like create a Constructor for FooList that takes a IEnumerable<foo> object like this:

    class FooList : List<foo>
    {
       public FooList(IEnumerable<foo> l)
            : base(l)
       {  }
    }
    

    Or you could also do something like this:

     var list = new FooList();
     var filtered = list.FindAll(c => c.Something == "filter criteria");
     var newList = new FooList();
     newList.AddRange(filtered);
     return newList;
    

    However as mentioned in some of the other answers, if you are not adding any functionality then you should just create an alias with the using keyword.

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