+++ /dev/null
-/*
- * 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<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 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;
-}