jsondeserializer.java

地图生成的一个文件

/*
 * 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.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 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;
}