out parameter without a method

Question

You may please look at this area

http://msdn.microsoft.com/en-us/library/9eekhta0.aspx

please note

public interface IEnumerable<out T> : IEnumerable

We know that ban "out" parameter can only be  used to pass values out of a method

Now if we code some thing as

public class My_Good_Class<T>: IEnumerable<out T>
{// .....etc}

How are we using the out parameter withot using a method ??

Answer 1

Question: What exactly are you trying to do? We have "out IEnumerable<T>" as parameter but never "IEnumerable<out T>", let alone assign it as a type.

"out" is decorator to a parameter, telling the compiler what to do with the parameter, but it's neither a type or a part of a type. You may spend some time to make your mind clear about the concepts involved, then try to ask again.

Answer 2

That is not a out Parameter.

This is a out Generic modifier. Similar Syntax, totally different kind of beast. It deals with a very advanced form of Polymorphy called "Covarriance":

http://msdn.microsoft.com/en-us/library/dd469487.aspx

Basically it goes:

Because the generics type was marked with out (is covariant) and because you could store a String in an Object Reference, you can store a Sample<String> Instance in a Sample<Object> Reference.

Sounds logical, but is only a recent addition.

---

Let's talk about MVVM: http://social.msdn.microsoft.com/Forums/en-US/wpf/thread/b1a8bf14-4acd-4d77-9df8-bdb95b02dbe2 Please mark post as helpfull and answers respectively.

Answer 3

You're right. Never implemented things using this feature and not aware that it exist.