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Message: [QUOTE="SteveCriscuoli, post: 8544748"] I originally posted this in microsoft.public.xml ... my apologies to those to who read both groups ... Someone please tell me if I'm missing something here. The XML Comment generator in VS2005 seems to be woefully incomplete. I created a test solution with 2 classes in different namespaces XTest.Derived.DerivedClass XTest.Base.BaseClass and yes, DerivedClass is dreived from BaseClass. I gave them each 2 constructors, DerivedClass received a method, a property, and a private field. below is what I got: <?xml version="1.0" ?> - <doc> - <assembly> <name>XTest</name> </assembly> - <members> - <member name="T:XTest.Derived.DerivedClass"> <summary>This is my derived class.</summary> </member> - <member name="T:XTest.Base.BaseClass"> <summary>comments for the entire class</summary> </member> - <member name="M:XTest.Base.BaseClass.#ctor"> <summary>default constructor comments.</summary> </member> - <member name="M:XTest.Base.BaseClass.#ctor(System.String)"> <summary>alternate constructor comments</summary> <param name="s">first parameter</param> </member> - <member name="F:XTest.Derived.DerivedClass.ia"> <summary>my array</summary> </member> - <member name="M:XTest.Derived.DerivedClass.#ctor"> <summary>default DC constructor</summary> </member> - <member name="M:XTest.Derived.DerivedClass.#ctor(System.String)"> <summary>alternate constructor for DC</summary> <param name="s">same silly string</param> </member> - <member name="M:XTest.Derived.DerivedClass.getVal"> <summary>returns the value 19</summary> <returns>the value 19</returns> </member> - <member name="P:XTest.Derived.DerivedClass.Ia"> <summary>accessor and mutator for ia.</summary> </member> </members> </doc> You'll notice there is no relation in here between the two classes. How come? How can anyone generate a sufficiently complete API document if you can't show class inheritance? I've combed through the MSDN site and can't even find a tag I could add to my documentation to 'help things along' which I shouldn't have to do in the first place. Someone please enlighten me. Steve PS - I'll send you the whole solution if you send me an email [/QUOTE]

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