How to deserialize generic List with Jackson?

Question

I've been using Jackson to serialize/deserialize objects for years and have always found it needlessly complicated to use TypeReference<T> to deserialize List etc. I created a simple helper function:

public static <T> TypeReference<List<T>> list() {
    return new TypeReference<List<T>>(){}
}

With intended use:

List<Foo> foos = objectMapper.readValue(json, list());

And it works! Kind of. When inspecting through the debugger, rather than a list of Foo, it is rather a list of LinkedHashMap. I understand that ObjectMapper deserializes into LinkedHashMap for type Object and I read the explanation for that here:

Jackson and generic type reference

However, why is it able to assign List<LinkedHasMap> to a List<Foo>? At the very least shouldn't that be some sort of ClassCastException?

Also, is there anyway to do this with Java's type system?

NOTE: the following method declaration has the same issue, which makes sense because the additional argument is not needed for T to be determined:

public static <T> TypeReference<List<T>> listOf(Class<T> ignored) {
    return new TypeReference<List<T>>(){}
}

Does this answer your question? [Jackson and generic type reference](https://stackoverflow.com/questions/6846244/jackson-and-generic-type-reference) — Savior, Apr 11 '20 at 01:24
Or here: https://stackoverflow.com/questions/34578452/how-to-use-jacksons-typereference-with-generics — Savior, Apr 11 '20 at 01:27
There's no difference between a `List` and a `List` at runtime, because of type erasure. — Savior, Apr 11 '20 at 02:49
See also [Gson deserialize list and class, how does erasure work here?](https://stackoverflow.com/a/61134913/5221149) Different library, exact same issue. — Andreas, Apr 11 '20 at 03:21

score 4 · Accepted Answer · answered Apr 11 '20 at 09:31

It works like this because of type erasure in Java. Please, read about it before you start reading next part of this answer:

As you probably know right now, after reading above articles, your method after compilation looks like this:

static <T> TypeReference<List> listOf(Class<T> ignored) {
    return new TypeReference<List>(){};
}

Jackson will try to find out the most appropriate type for it which will be java.util.LinkedHashMap for a JSON Object. To create irrefutable type you need to use com.fasterxml.jackson.databind.type.TypeFactory class. See below example:

import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.JavaType;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.type.TypeFactory;

import java.io.File;
import java.util.List;

public class JsonTypeApp {

    public static void main(String[] args) throws Exception {
        File jsonFile = new File("./resource/test.json").getAbsoluteFile();

        ObjectMapper mapper = new ObjectMapper();

        System.out.println("Try with 'TypeFactory'");
        List<Id> ids = mapper.readValue(jsonFile, CollectionsTypeFactory.listOf(Id.class));
        System.out.println(ids);
        Id id1 = ids.get(0);
        System.out.println(id1);

        System.out.println("Try with 'TypeReference<List<T>>'");
        List<Id> maps = mapper.readValue(jsonFile, CollectionsTypeFactory.erasedListOf(Id.class));
        System.out.println(maps);
        Id maps1 = maps.get(0);
        System.out.println(maps1);
    }
}

class CollectionsTypeFactory {
    static JavaType listOf(Class clazz) {
        return TypeFactory.defaultInstance().constructCollectionType(List.class, clazz);
    }

    static <T> TypeReference<List> erasedListOf(Class<T> ignored) {
        return new TypeReference<List>(){};
    }
}

class Id {
    private int id;

    // getters, setters, toString
}

Above example, for below JSON payload:

[
  {
    "id": 1
  },
  {
    "id": 22
  },
  {
    "id": 333
  }
]

prints:

Try with 'TypeFactory'
[{1}, {22}, {333}]
{1}
Try with 'TypeReference<List<T>>'
[{id=1}, {id=22}, {id=333}]
Exception in thread "main" java.lang.ClassCastException: java.util.LinkedHashMap cannot be cast to com.example.Id
    at com.example.JsonTypeApp.main(JsonTypeApp.java:27)

How to deserialize generic List with Jackson?

1 Answers1

Linked