code: /*
   * 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 tools.com.google.gson;
  
  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&lt;T&gt; {
   *   private final Class&lt;T&gt; clazz;
   *   private final long value;
   *   public Id(Class&lt;T&gt; 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&lt;Id&gt;() {
   *   public Id deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context)
   *       throws JsonParseException {
   *     return 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>
   *
   * <p>New applications should prefer {@link TypeAdapter}, whose streaming API
   * is more efficient than this interface's tree API.
   *
   * @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;
  }
  