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How do I implement type TypeAdapterFactory in Gson?

The main method of create is generic. Why?

The registration method registerTypeAdapterFactory() does not receive type a type argument. So, how does Gson know which classes are processed by the factory?

Should I implement one factory for multiple classes, or can I implement one for many classes?

If I implement one factory for multiple classes, then what should I return in case of out-of-domain type argument?

1 Answer 1

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When you register a regular type adapter (GsonBuilder.registerTypeAdapter), it only generates a type adapter for THAT specific class. For example:

public abstract class Animal { abstract void speak(); }
public class Dog extends Animal {
   private final String speech = "woof";
   public void speak() {
       System.out.println(speech);
   }
}

// in some gson related method
gsonBuilder.registerTypeAdapter(Animal.class, myTypeAdapterObject);
Gson g = gsonBuilder.create();
Dog dog = new Dog();
System.out.println(g.toJson(dog));

If you did this, then Gson will not use your myTypeAdapterObject, it will use the default type adapter for Object.

So, how can you make a type adapter object that can convert ANY Animal subclass to Json? Create a TypeAdapterFactory! The factory can match using the generic type and the TypeToken class. You should return null if your TypeAdapterFactory doesn't know how to handle an object of that type.

The other thing TypeAdapterFactory can be used for is that you can't CHAIN adapters any other way. By default, Gson doesn't pass your Gson instance into the read or write methods of TypeAdapter. So if you have an object like:

public class MyOuterClass {
    private MyInnerClass inner;
}

There is no way to write your TypeAdapter<MyOuterClass> that knows how to use your TypeAdapter<MyInnerClass> without using the TypeAdapterFactory. The TypeAdapterFactory.create method DOES pass the Gson instance, which allows you to teach your TypeAdapter<MyOuterClass> how to serialize the MyInnerClass field.


Generally, here is a good standard way to begin to write an implementation of a TypeAdapterFactory:

public enum FooAdapterFactory implements TypeAdapterFactory {
    INSTANCE; // Josh Bloch's Enum singleton pattern

    @SuppressWarnings("unchecked")
    @Override
    public <T> TypeAdapter<T> create(Gson gson, TypeToken<T> type) {
        if (!Foo.class.isAssignableFrom(type.getRawType())) return null;

        // Note: You have access to the `gson` object here; you can access other deserializers using gson.getAdapter and pass them into your constructor
        return (TypeAdapter<T>) new FooAdapter();
    }

    private static class FooAdapter extends TypeAdapter<Foo> {
        @Override
        public void write(JsonWriter out, Foo value) {
            // your code
        }

        @Override
        public Foo read(JsonReader in) throws IOException {
            // your code
        }
    }
}
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  • 1
    So, will Gson call my adapter factory for hundreds of classes met during serialization?
    – Suzan Cioc
    Commented Mar 10, 2014 at 18:09
  • 3
    yes, but Gson keeps a lookup table of class to AdapterFactory so it only tries your class once. if create returns null it never tries for that class again.
    – durron597
    Commented Mar 10, 2014 at 18:22
  • 3
    @SuzanCioc Umm yeah... I figured that out looking at the source code :-D
    – durron597
    Commented Mar 10, 2014 at 18:30
  • 1
    If FooAdapter is the only Adapter that can be provided then why use an TypeAdapterFactory? Why not just use a TypeAdapter?
    – Subby
    Commented Apr 6, 2016 at 21:59
  • 1
    Because a TypeAdapter<T> can only serialize one specific type, T. If you want to serialize an interface and all implementors, you need to create a reference in the Gson lookup table for all requested classes that implement that interface.
    – durron597
    Commented Apr 6, 2016 at 23:34

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