Can you change the value of the Name property of a Type at runtime?

Question

Is it possible to change/override the Name property of a Type?

Given the class definition:

class Sample
{}

Can you change the value that typeof(Sample).Name returns?

I'm using an custom serialization library, which literally use typeof(T).Name in their source code:

writer.WriteStartElement(typeof(T).Name);
writer.WriteValue(item);
writer.WriteEndElement();

Answer 1

No, you cannot change the .Name of a Type at runtime. However, most serialization libraries allow you to have some control of the processing of names, either by providing a custom "binder" (etc), or by annotating the type with attributes to indicate the preferred name to use (note: most libraries that allow attributes also allow the name to be provided via runtime configuration of the serialization library).

An important question, then, is: what is the serialization library that is being used here?

If the serialization library doesn't support this, and cannot be changed, then the only alternative (short of renaming Sample ) to all of this is to create a type (either by hand, or via TypeBuilder at runtime) that looks like the original type, but with a different code, and similarly: create code that translates between the two types.

Answer 2

I don't know any way to accomplish such a task. CLR does similar thing with marshalled by ref objects, but I doubt that it will allow you to "rename" your class or its property(field) in such a way.

Your best bet is to either conform to the old code or(the better one)is to replace or change the serialization routines to be less rigid.

Answer 3

As for XML serialization using the XmlSerialization library in the BCL, you can override the xml Element name to be generated using the following attribute:

[XmlElement("SomeOtherName")]
public class Sample{...}

Answer 4

From your question, you mentioned you want to control the name during serialize

If you want to change the name during the serialization, you can use the following attributes

[DataContract(Name = "NewName")] 
[XmlRoot("NewName")] 
class Sample
{
}