How can I properly serialize and deserialize JSON in Java with generic collections?

General Discussions
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1

I’m implementing JSON serialization for objects, but I’m facing an issue with integrating generic collections.

My AwesomeList<T> extends JSONSerializable, but during deserialization, I can’t instantiate T due to Java’s type restrictions. What’s the best way to handle java serialize json while maintaining flexibility with generics?

2

Gson provides TypeToken to preserve generic type information during deserialization.

import com.google.gson.*;
import com.google.gson.reflect.TypeToken;
import java.lang.reflect.Type;
import java.util.List;

class AwesomeList<T> {
    List<T> items;
}

public class GsonGenericExample {
    public static void main(String[] args) {
        Gson gson = new Gson();
        String json = "[\"Hello\", \"World\"]";
        
        Type type = new TypeToken<AwesomeList<String>>(){}.getType();
        AwesomeList<String> list = gson.fromJson(json, type);

        System.out.println(list.items); // Output: [Hello, World]
    }
}

Gson automatically handles generics using TypeToken.

3

You can also use Jackson with TypeReference. It solves the type erasure problem, similar to Gson.

import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.ObjectMapper;
import java.util.List;

class AwesomeList<T> {
    public List<T> items;
}

public class JacksonGenericExample {
    public static void main(String[] args) throws Exception {
        ObjectMapper mapper = new ObjectMapper();
        String json = "[\"Hello\", \"World\"]";

        AwesomeList<String> list = mapper.readValue(json, new TypeReference<AwesomeList<String>>() {});
        
        System.out.println(list.items); // Output: [Hello, World]
    }
}

Jackson efficiently maintains type safety for generics.

4

If you need full control over serialization, implement a custom deserializer.

class AwesomeList<T> {
    List<T> items;
}

class AwesomeListDeserializer<T> implements JsonDeserializer<AwesomeList<T>> {
    private final Class<T> clazz;
    
    public AwesomeListDeserializer(Class<T> clazz) { this.clazz = clazz; }

    @Override
    public AwesomeList<T> deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) 
            throws JsonParseException {
        AwesomeList<T> list = new AwesomeList<>();
        list.items = context.deserialize(json, TypeToken.getParameterized(List.class, clazz).getType());
        return list;
    }
}

It works when T needs special handling during deserialization.