Why UnaryFunction

Question

When I read Effective Java item 27, the type casting between UnaryFunction<Object> and UnaryFunction<T> confused me.

interface UnaryFunction<T> {
    T apply(T t);
}

public class Main {
    private static final UnaryFunction<Object>  IDENTITY = new UnaryFunction<Object>() {
        public Object apply(Object t) {
            return t;
        }
    };

    @SuppressWarnings("unchecked")
    public static <T> UnaryFunction<T> identityFunction() {
        return  (UnaryFunction<T>) IDENTITY;
    }

    public static void main(String ... args) {
        UnaryFunction<A> identityA = Main.identityFunction();
        A a = identityA.apply(new A());
    }

}

class A {}

Why UnaryFunction<Object> can be casted to UnaryFunction<T>?

I know the generic type will be erased after complier. So (UnaryFunction<T>) IDENTITY will eventually be (UnaryFunction<Object>) IDENTITY, this will be working in Runtime.

But directly casting UnaryFunction<Object> to UnaryFunction<A> is not allowed by compiler.

UnaryFunction<A> identityA = (UnaryFunction<A>)IDENTITY;
//incompatible types: UnaryFunction<java.lang.Object> cannot be converted to UnaryFunction<A>

And there is no inheritance relationship between UnaryFunction<Object> and UnaryFunction<T>. So why UnaryFunction<Object> can be casted to UnaryFunction<T>?

@user2357112 it doesn't compile even if explicitly state A extends Object — Alexander Kulyakhtin, Sep 10 '16 at 06:59
@AlexanderKulyakhtin: Inheritance isn't enough. `A` isn't `Object`, so a `UnaryFunction — user2357112, Sep 10 '16 at 07:00
@user2357112 Do you mean T **might be** `Object`, so the compiler allow `UnaryFunction — Robin Han, Sep 10 '16 at 07:04
@RobinHan: Pretty much. If you put in a type bound like `T extends Integer`, so `T` definitely isn't `Object`, the compiler will reject the cast. — user2357112, Sep 10 '16 at 07:25
@user2357112 if I change the method definition to `public static UnaryFunction identityFunction()` and `UnaryFunction identityA = Main.identityFunction();`, the compile is still pass. So I don't think it's related to the bound. — Robin Han, Sep 10 '16 at 07:32
@RobinHan: [The compilation fails.](http://ideone.com/cPTIzj) You must have screwed up your test. — user2357112, Sep 10 '16 at 07:34
@user2357112 Sorry my fault. The IDE doesn't mark it as red. — Robin Han, Sep 10 '16 at 07:40
This is indeed a strange beast. [Compilation succeeds](http://ideone.com/KpWTmt) but when I inline method `identityFunction` it fails. This runs: `Exception in thread "main" java.lang.ClassCastException: java.lang.Object cannot be cast to A` — Mark Jeronimus, Sep 10 '16 at 14:42
`@SuppressWarnings("unchecked")` is you telling the compiler "It may not look like it, but I promise I'm doing the right thing." So it believes you. It's your own fault if you lied to Java. — Louis Wasserman, Sep 11 '16 at 17:22

score 5 · Accepted Answer · edited May 23 '17 at 12:32

All generic types, without a bound (... extends ..., is considered Object due to type erasure. So this cast might be valid for some values of T. @SuppressWarnings("unchecked") tells the compiler you are doing the right thing so the warning is suppressed.

Casting to a specific type is problematic. Let's assume you are dealing with List<Object> which you cast to List<A> where A is a class. In this case list might have elements which are a supertype of A so this cast is unsound hence the compiler does not allow it. In case of the function (UnaryFunction<Object>) it is implied that this can return an Object. Say your code was IDENTITY = t -> new Object(); in which case casting to UnaryFunction<A> would be unsound as it returns an Object. In the case it is UnaryFunction<T> there is some type T which satisfies the cast, that is, when T is an Object.

For some background reading on this see: Liskov substitution principle which deals with subtyping of functions.

1 Answers1