Deserializing ImmutableList using Gson

问题

I'm using quite a few immutable collections and I'm curious how to deserialize them using Gson. As nobody answered and I've found the solution myself, I'm simplifying the question and presenting my own answer.

I had two problems:

• How to write a single Deserializer working for all ImmutableList<XXX>?
• How to register it for all ImmutableList<XXX>?

回答1:

Update: There's https://github.com/acebaggins/gson-serializers which covers many guava collections:

• ImmutableList
• ImmutableSet
• ImmutableSortedSet
• ImmutableMap
• ImmutableSortedMap

---

How to write a single Deserializer working for all ImmutableList?

The idea is simple, transform the passed Type representing an ImmutableList<T> into a Type representing List<T>, use the build-in Gson's capability to create a List and convert it to an ImmutableList.

class MyJsonDeserializer implements JsonDeserializer<ImmutableList<?>> {
    @Override
    public ImmutableList<?> deserialize(JsonElement json, Type type, JsonDeserializationContext context) throws JsonParseException {
        final Type type2 = ParameterizedTypeImpl.make(List.class, ((ParameterizedType) type).getActualTypeArguments(), null);
        final List<?> list = context.deserialize(json, type2);
        return ImmutableList.copyOf(list);
    }
}

There are multiple ParameterizedTypeImpl classes in Java libraries I use, but none of them intended for public usage. I tested it with sun.reflect.generics.reflectiveObjects.ParameterizedTypeImpl.

How to register it for all ImmutableList?

That part is trivial, the first argument to register is java.lang.reflect.Type which mislead me to using ParameterizedType, where simply using Class does the job:

final Gson gson = new GsonBuilder()
    .registerTypeAdapter(ImmutableList.class, myJsonDeserializer)
    .create();

回答2:

One more implementation without ParameterizedTypeImpl

@Override
public ImmutableList<?> deserialize(final JsonElement json, final Type type, final JsonDeserializationContext context) throws JsonParseException {
    @SuppressWarnings("unchecked")
    final TypeToken<ImmutableList<?>> immutableListToken = (TypeToken<ImmutableList<?>>) TypeToken.of(type);
    final TypeToken<? super ImmutableList<?>> listToken = immutableListToken.getSupertype(List.class);
    final List<?> list = context.deserialize(json, listToken.getType());
    return ImmutableList.copyOf(list);
}

回答3:

@maaartinus already covered the second question, so I'll post a complementary Guava-based solution to the first question which doesn't require ParametrizedTypeImpl

public final class ImmutableListDeserializer implements JsonDeserializer<ImmutableList<?>> {
  @Override
  public ImmutableList<?> deserialize(final JsonElement json, final Type type,
                                      final JsonDeserializationContext context)
      throws JsonParseException {
    final Type[] typeArguments = ((ParameterizedType) type).getActualTypeArguments();
    final Type parameterizedType = listOf(typeArguments[0]).getType();
    final List<?> list = context.deserialize(json, parameterizedType);
    return ImmutableList.copyOf(list);
  }

  private static <E> TypeToken<List<E>> listOf(final Type arg) {
    return new TypeToken<List<E>>() {}
        .where(new TypeParameter<E>() {}, (TypeToken<E>) TypeToken.of(arg));   
  }
}

来源：https://stackoverflow.com/questions/7706772/deserializing-immutablelist-using-gson