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jdereg / json-io

Convert Java to JSON. Convert JSON to Java. Pretty print JSON. Java JSON serializer. Deep copy Java object graphs.
Apache License 2.0
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Deserialization of records doesn't work in Java15 #151

Closed reuschling closed 2 years ago

reuschling commented 3 years ago

If you want to deserialize a record, you get an IllegalAccessException: Can not set final ... field.

I assume the problem will exist also in the new Java16 which comes out next week, where records won't be a preview feature anymore.

Example code:

    public static record TestRecord(String str, boolean bool)
    {

    }

...

        TestRecord rec = new TestRecord("check", false);

        String strJson = JsonWriter.objectToJson(rec);
        System.out.println(strJson);

        TestRecord clone = (TestRecord) JsonReader.jsonToJava(strJson);
        System.out.println(clone);

Gson also has this issue: https://github.com/google/gson/issues/1794. According this, the solution would be to use the constructor in favour of setting the final fields via reflection.

jdereg commented 2 years ago

json-io has the assign instantiator customization that lets you write the code that creates the object, which allows you to set up the object in regular code - your code is called when the troubled class is encountered. This is dicussed in the user-guide, linked here: https://github.com/jdereg/json-io/blob/master/user-guide.md under the Customization section. Also, recent changes have been added to support JDK's through 17.