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How to serialize a class with an interface?

Asked 2011-01-25T15:30:18.607
40

I have never done much with serialization, but am trying to use Google's gson to serialize a Java object to a file. Here is an example of my issue:

public interface Animal {
    public String getName();
}


 public class Cat implements Animal {

    private String mName = "Cat";
    private String mHabbit = "Playing with yarn";

    public String getName() {
        return mName;
    }

    public void setName(String pName) {
        mName = pName;
    }

    public String getHabbit() {
        return mHabbit;
    }

    public void setHabbit(String pHabbit) {
        mHabbit = pHabbit;
    }

}

public class Exhibit {

    private String mDescription;
    private Animal mAnimal;

    public Exhibit() {
        mDescription = "This is a public exhibit.";
    }

    public String getDescription() {
        return mDescription;
    }

    public void setDescription(String pDescription) {
        mDescription = pDescription;
    }

    public Animal getAnimal() {
        return mAnimal;
    }

    public void setAnimal(Animal pAnimal) {
        mAnimal = pAnimal;
    }

}

public class GsonTest {

public static void main(String[] argv) {
    Exhibit exhibit = new Exhibit();
    exhibit.setAnimal(new Cat());
    Gson gson = new Gson();
    String jsonString = gson.toJson(exhibit);
    System.out.println(jsonString);
    Exhibit deserializedExhibit = gson.fromJson(jsonString, Exhibit.class);
    System.out.println(deserializedExhibit);
}
}

So this serializes nicely -- but understandably drops the type information on the Animal:

{"mDescription":"This is a public exhibit.","mAnimal":{"mName":"Cat","mHabbit":"Playing with yarn"}}

This causes real problems for deserialization, though:

Exception in thread "main" java.lang.RuntimeException: No-args constructor for interface com.atg.lp.gson.Animal does not exist. Register an
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2 Answers

92

Here is a generic solution that works for all cases where only interface is known statically.

  1. Create serialiser/deserialiser:

    final class InterfaceAdapter<T> implements JsonSerializer<T>, JsonDeserializer<T> {
        public JsonElement serialize(T object, Type interfaceType, JsonSerializationContext context) {
            final JsonObject wrapper = new JsonObject();
            wrapper.addProperty("type", object.getClass().getName());
            wrapper.add("data", context.serialize(object));
            return wrapper;
        }
    
        public T deserialize(JsonElement elem, Type interfaceType, JsonDeserializationContext context) throws JsonParseException {
            final JsonObject wrapper = (JsonObject) elem;
            final JsonElement typeName = get(wrapper, "type");
            final JsonElement data = get(wrapper, "data");
            final Type actualType = typeForName(typeName); 
            return context.deserialize(data, actualType);
        }
    
        private Type typeForName(final JsonElement typeElem) {
            try {
                return Class.forName(typeElem.getAsString());
            } catch (ClassNotFoundException e) {
                throw new JsonParseException(e);
            }
        }
    
        private JsonElement get(final JsonObject wrapper, String memberName) {
            final JsonElement elem = wrapper.get(memberName);
            if (elem == null) throw new JsonParseException("no '" + memberName + "' member found in what was expected to be an interface wrapper");
            return elem;
        }
    }
    
  2. make Gson use it for the interface type of your choice:

    Gson gson = new GsonBuilder().registerTypeAdapter(Animal.class, new InterfaceAdapter<Animal>())
                                 .create();
    
answered 2012-03-03T21:37:02.547
14

Put the animal as transient, it will then not be serialized.

Or you can serialize it yourself by implementing defaultWriteObject(...) and defaultReadObject(...) (I think thats what they were called...)

EDIT See the part about "Writing an Instance Creator" here.

Gson cant deserialize an interface since it doesnt know which implementing class will be used, so you need to provide an instance creator for your Animal and set a default or similar.

answered 2011-01-25T15:33:06.167

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