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Home/ Questions/Q 3273914
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T18:58:55+00:00 2026-05-17T18:58:55+00:00

I have the following: public class MyClass : SuperClass { [JsonProperty] public virtual string

  • 0

I have the following:

public class MyClass : SuperClass {    
  [JsonProperty]
  public virtual string Id { get; set; }
}

public abstract class SuperClass { 
  public int GetHashCode() {
  //do things here
  }
}

I cannot alter SuperClass. When I go to serialize to Json using JsonNet I’ll do something like this:

    JsonSerializerSettings serializer = new JsonSerializerSettings {
        //serializer settings
    };

    var jsonNetResult = new JsonNetResult {
        Data = myClass,
        SerializerSettings = serializer
    };

    return jsonNetResult;

Obviously it will not serialize GetHashCode(). If I go:

    var jsonNetResult = new JsonNetResult {
        Data = myClass.GetHashCode(),
        SerializerSettings = serializer
    };

It will correctly serialize the value, is there some serializer setting I can use to tell it to include GetHashCode()?

Edit: I should add that right now I’m creating a property with only get to accomplish this, i.e.

[JsonProperty]
public virtual int GetHashCodeJson { get { return GetHashCode(); }
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-17T18:58:56+00:00Added an answer on May 17, 2026 at 6:58 pm

    This is not so much an issue with JSON.Net as with .net serialization in general.

    You need to serialize objects by their properties and you are asking to serialize the return value of a method. So there is not a way to do this with the syntax you want.

    That you are able to do this:

    Data = myClass.GetHashCode()
    

    Only means that the return value of the method (an int) can be serialized and not that the serializer cares at all about what method that value is coming from.

    If you think about it, there is not a lot of sense to saying that a value is the serialized return value of a method because how do you deserialize that then? You would never be able to write the value back to the method because its a return value only, not a 2-way relationship like a property with {get;set;}.

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