[BUG] 嵌套 MAP 的子类在反序列化时无法正确生成实例

[huangluyu]

FastJson 2.0.26

当出现下面的嵌套Map时，无法正确反序列化成对应的类型

public class A extends HashMap<String, B> {

}

public class B extends HashMap<String, String> {

}

测试如下

@Test
void test_nest_map_subclass() {
    A a = new A();
    a.put("a", new B());
    a.get("a").put("b", "1");

    String jsonString = JSON.toJSONString(a);
    assert jsonString.equals("{\"a\":{\"b\":\"1\"}}");

    A a1 = JSON.parseObject(jsonString, A.class);
    // java.lang.ClassCastException: com.alibaba.fastjson2.JSONObject cannot be cast to B
    assert a1.get("a").get("b").equals("1");
}

[wenshao]

https://oss.sonatype.org/content/repositories/snapshots/com/alibaba/fastjson2/fastjson2/2.0.38-SNAPSHOT/
问题已修复，请用2.0.38-SNAPSHOT版本帮忙验证

[huangluyu]

提问中的 case 已测试通过。但如果 case 升级成如下时，仍然无法实现：

public class A extends HashMap<String, B> {

}

public class B extends HashMap<String, String> {

    public String get(Integer number) {
        return get(String.valueOf(number));
    }
}

@Test
void test_nest_map_subclass() {
    A a = new A();
    a.put("a", new B());
    a.get("a").put("b", "1");

    String jsonString = JSON.toJSONString(a);
    assert jsonString.equals("{\"a\":{\"b\":\"1\"}}");

    A a1 = JSON.parseObject(jsonString, A.class);
    // java.lang.ClassCastException: com.alibaba.fastjson2.JSONObject cannot be cast to B
    assert a1.get("a").get(1).equals("1");
}

https://oss.sonatype.org/content/repositories/snapshots/com/alibaba/fastjson2/fastjson2/2.0.38-SNAPSHOT/ 问题已修复，请用2.0.38-SNAPSHOT版本帮忙验证

[wenshao]

你的例子是不是写错了，是不是应该是这样

  @Test
    public void test_nest_map_subclass() {
        A a = new A();
        a.put("a", new B());
        a.get("a").put("1", "b");

        String jsonString = JSON.toJSONString(a);
        assertEquals(jsonString, "{\"a\":{\"1\":\"b\"}}");

        A a1 = JSON.parseObject(jsonString, A.class);
        assertEquals(a1.get("a").get("1"), "b");
    }

    public static class A
            extends HashMap<String, B> {
    }

    public static class B
            extends HashMap<String, String> {
        public String get(Integer number) {
            return get(String.valueOf(number));
        }
    }

[wenshao]

https://github.com/alibaba/fastjson2/releases/tag/2.0.38
请用新版本