Thou shalt be binary compatible

2012-11-23  |   |  Java  

I learned some new tricks today thanks to Gunnar around backward compatibilities in Java.

There is compatibility and compatibility

In Bean Validation, we need to fix a mistake I made. One easy solution is to create a sub-interface and return that sub-interface.

//API
public interface Contract {
    public Result testMe();
    public Result testMeMore();

    public static interface Result {
        void doIt();
    }
}

//client code
Contract contract = new ContractImpl();
contract.testMeMore().doIt();

In our example, I need to add a new method to results coming from testMeMore(). I thought of this approach:

//API
public interface Contract {
    public Result testMe();
    public ResultMore testMeMore();

    public static interface Result {
        void doIt();
    }

    public static interface ResultMore extends Result {
        void doItAgain();
    }
}

//client code
Contract contract = new ContractImpl();
contract.testMeMore().doIt();

This approach is source compatible: if I recompile the client code, things will work perfectly. But it's not backward compatible at the binary level.

If you don't recompile the client code and simply update the API jar, you will get a nasty exception

Exception in thread "main" java.lang.NoSuchMethodError: com.jboss.test.Contract.testMeMore()Lcom/jboss/test/Contract$Result;

That's because the contract is now com.jboss.test.Contract.testMeMore()Lcom/jboss/test/Contract$ResultMore and even if Result is a super interface, Java does not let go with it.

And since application deployed in Java EE 6 are supposed to work out of the box for Java EE 7, we can't do that. Note that testing binary compatibilities is not trivial.

Use the erasure hack, Luke

The what? It turns out the Java designers already had this problem when they introduced the generics type system. You can solve the problem by using intersection types.

//API
public interface Contract {
    public Result testMe();
    public <T extends Result & ResultMore> T testMeMore();

    public static interface Result {
        void doIt();
    }

    public static interface ResultMore extends Result {
        void doItAgain();
    }
}

//client code
Contract contract = new ContractImpl();
contract.testMeMore().doIt();

Because generic types are erased by their most left upper bounds which is Result in our case, things work out smoothly.

By the way, you can reproduce this ad nauseam.

//API
public interface Contract {
    public Result testMe();
    public <T extends Result & ResultMore & ResultUltimate> T testMeMore();

    public static interface Result {
        void doIt();
    }

    public static interface ResultMore extends Result {
        void doItAgain();
    }

    public static interface ResultUltimate extends Result {
        void doItForEver();
    }

}

//client code
Contract contract = new ContractImpl();
contract.testMeMore().doItAgain();

Conclusion

Now the big question is, should we use this trick in Bean Validation to solve this problem. What's your take on it?


Name: Emmanuel Bernard
Bio tags: French, Open Source actor, Hibernate, (No)SQL, JCP, JBoss, Snowboard, Economy
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Geoloc: Paris, France

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