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DefaultScalaModule breaks deserialization of java classes #644

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@kudrevatykh

Description

@kudrevatykh

I found problem that adding DefaultScalaModule to object mapper breaks Deserialization of classes that successfuly deserialized without scala module

package test;

import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonValue;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.module.scala.DefaultScalaModule;
import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;

public class ScalaModuleTest {


    public record M(String name, String domain) {
        @JsonCreator(mode = JsonCreator.Mode.DELEGATING)
        public static M fromString(String uri) {
            String[] parts = uri.split("@");
            if (parts.length > 2 || parts.length == 0) {
                throw new RuntimeException(String.format("Invalid SIP URI: %s", uri));
            }

            String[] usernameParts = parts[0].split("\\.");
            return new M(usernameParts[usernameParts.length - 1], parts.length == 2 ? parts[1] : null);
        }

        @Override
        @JsonValue
        public String toString() {
            return String.format("%s@%s", name, domain);
        }
    }

    public record Obj(M val) {

    }

    @Test
    public void testScalaModule() throws JsonProcessingException {
        var json = "{\"val\": \"a@b\"}";
        var mapper = new ObjectMapper().registerModule(new DefaultScalaModule());
        var val = mapper.readValue(json, Obj.class);
        Assertions.assertEquals(new M("a", "b"), val.val);

    }
}

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