How to use default serialization in a custom JsonConverter

Question

I have a complex object graph that I am serializing/deserializing with Json.NET. Some of the objects derive from an abstract class, so in order for the deserialization to work properly, I needed to create a custom `JsonConverter`. Its only role is to select the appropriate concrete implementation of the abstract class at deserialization-time and allow Json.NET to continue on its way. My problem comes when I want to serialize. I don't need to do anything custom at all. I want to get exactly the same behavior as I would get using `JsonConvert.SerializeObject` with no custom `JsonConverter`. However, since I'm using the custom JsonConverter class for my deserialization needs, I'm forced to supply a `WriteJson` implementation. Since WriteJson is abstract, I can't just call `base.WriteJson`, but I want to do essentially that. So my question is, what do I put in that method to get the plain-Jane, default behavior? In other words: ``` public override void WriteJson(JsonWriter writer, object…

Accepted Answer

In your custom `JsonConverter`, override `CanWrite` and return false:

```
public override bool CanWrite { get { return false; } }

public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
    throw new NotImplementedException();
}
```

Then you can just throw an exception from `WriteJson`, since it won't get called.

(Similarly, to get default behavior during deserialization, override `CanRead` and return `false`.)

Note that the same approach can be used for `JsonConverter<T>` (introduced in Json.NET 11.0.1) since it is just a subclass of `JsonConverter` that introduces type-safe versions of `ReadJson()` and `WriteJson()`.

How to use default serialization in a custom JsonConverter

Solution

Problem