[QUESTION] fastjson1.x DefaultJSONParser.getContext().object 如何迁移为FastJson2等价代码？

### 请描述您的问题
请问如何将fastjson1.x的应用里的自定义序列化的代码替换为等价的fastjson2代码？
FastJson1: com.alibaba.fastjson.parser.DefaultJSONParser.getContext().object
Fastjson2: JSONReader的context里如何拿当前上下文里的Bean Instance?
1. DTO：
``` java
@Data
public class UserDto{
@JSONField(deserializeUsing = AgeParamConverter.class)
private Integer age;
@JSONField(deserializeUsing = AgeCodeParamConverter.class)
private AgeCodeEnum ageCodeEnum;
}
```

2. 自定义反序列化：
``` java
public class AgeParamConverter extends IntegerCodec {
@Override
public Integer deserialze(DefaultJSONParser parser, Type type, Object fieldName) {
UserDto user = (UserDto) parser.getContext().object;
Integer age = super.deserialze(parser, type, fieldName);
if (user == null){
return age;
}

if (user.getAgeCodeEnum() == null){
user.setAgeCodeEnum(getAgeCodeEnum(age)
}
return age;
}
```

Start by comparing the fastjson1.x DefaultJSONParser.getContext().object entry point with the fastjson2 JSONReader context APIs, then inspect existing custom deserialization examples in the repository. Done means documenting or demonstrating a confirmed fastjson2 equivalent for retrieving the current UserDto instance during AgeParamConverter deserialization.

Written by the indexing model from the issue text.