/*
* Copyright (C) 2008 Google Inc.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package com.smartandroid.sa.json;
import java.lang.reflect.Type;
/**
* <p>
* Interface representing a custom deserializer for Json. You should write a
* custom deserializer, if you are not happy with the default deserialization
* done by Gson. You will also need to register this deserializer through
* {@link GsonBuilder#registerTypeAdapter(Type, Object)}.
* </p>
*
* <p>
* Let us look at example where defining a deserializer will be useful. The
* {@code Id} class defined below has two fields: {@code clazz} and
* {@code value}.
* </p>
*
* <pre>
* public class Id<T> {
* private final Class<T> clazz;
* private final long value;
*
* public Id(Class<T> clazz, long value) {
* this.clazz = clazz;
* this.value = value;
* }
*
* public long getValue() {
* return value;
* }
* }
* </pre>
*
* <p>
* The default deserialization of {@code Id(com.foo.MyObject.class, 20L)} will
* require the Json string to be
* <code>{"clazz":com.foo.MyObject,"value":20}</code>. Suppose, you already know
* the type of the field that the {@code Id} will be deserialized into, and
* hence just want to deserialize it from a Json string {@code 20}. You can
* achieve that by writing a custom deserializer:
* </p>
*
* <pre>
* class IdDeserializer implements JsonDeserializer<Id>() {
* public Id fromJson(JsonElement json, Type typeOfT, JsonDeserializationContext context)
* throws JsonParseException {
* return (Id) new Id((Class)typeOfT, id.getValue());
* }
* </pre>
*
* <p>
* You will also need to register {@code IdDeserializer} with Gson as follows:
* </p>
*
* <pre>
* Gson gson = new GsonBuilder().registerTypeAdapter(Id.class,
* new IdDeserializer()).create();
* </pre>
*
* @author Inderjeet Singh
* @author Joel Leitch
*
* @param <T>
* type for which the deserializer is being registered. It is
* possible that a deserializer may be asked to deserialize a
* specific generic type of the T.
*/
public interface JsonDeserializer<T> {
/**
* Gson invokes this call-back method during deserialization when it
* encounters a field of the specified type.
* <p>
* In the implementation of this call-back method, you should consider
* invoking
* {@link JsonDeserializationContext#deserialize(JsonElement, Type)} method
* to create objects for any non-trivial field of the returned object.
* However, you should never invoke it on the the same type passing
* {@code json} since that will cause an infinite loop (Gson will call your
* call-back method again).
*
* @param json
* The Json data being deserialized
* @param typeOfT
* The type of the Object to deserialize to
* @return a deserialized object of the specified type typeOfT which is a
* subclass of {@code T}
* @throws JsonParseException
* if json is not in the expected format of {@code typeofT}
*/
public T deserialize(JsonElement json, Type typeOfT,
JsonDeserializationContext context) throws JsonParseException;
}