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diff --git a/src/com/google/gson/JsonDeserializer.java b/src/com/google/gson/JsonDeserializer.java
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+/*
+ * 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 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;
+}