Error compiling lambda with bounded generic

Question

The following is a contrived, self-contained example I created to demonstrate a problem I'm having with a much more complicated program:

public class MyTest {
    public static void main(String[] args) {
        SubObject[] array = new SubObject[5];
        Iterator<? extends SuperObject> iterator 
                = Arrays.asList((SuperObject[]) array).iterator();
        Iterable<? extends SuperObject> iterable = () -> iterator;
    }
}

class SuperObject {};

class SubObject extends SuperObject{};

This does not compile:

MyTest.java:9: error: incompatible types: bad return type in lambda expression
        Iterable<? extends SuperObject> iterable = () -> iterator;
                                                         ^
    Iterator<CAP#1> cannot be converted to Iterator<SuperObject>
  where CAP#1 is a fresh type-variable:
    CAP#1 extends SuperObject from capture of ? extends SuperObject

I know I could simplify the generics, replacing <? extends SuperObject> with <SuperObject>. But this would defeat my purpose in the larger program. Why this error?

May I just ask why having Iterator<SuperObject> and Iterable<SuperObject> would defeat the purpose of your program? — Ousmane D.
– Ousmane D., Commented Jan 21, 2018 at 5:11
Consider a super container class and a set of sub container classes that extend the super container. The super container defines an abstract iterable() method that returns an Iterable<Widget>. Each sub container contains an array of a class that extends Widget, and implements iterable() to return an Iterable over the array. If the array in a sub container is implemented using ArrayList<ExtOfWidget>, I can't seem to create an Iterable that can be returned by iterable(), i.e. an Iterable<Widget>. Can't cast ArrayList<ExtOfWidget> to ArrayList<Widget>. — Steve Brandli
– Steve Brandli, Commented Jan 21, 2018 at 5:19
To continue: However it all works if iterable() returns an Iterable<? extends Widget>, at least except for the problem I present in this question. — Steve Brandli
– Steve Brandli, Commented Jan 21, 2018 at 5:21
Your example is a bit convoluted. You can demonstrate the same issue like this: Iterator<?> iterator = Collections.emptyIterator(); Iterable<?> iterable = () -> iterator; — shmosel
– shmosel, Commented Jan 21, 2018 at 5:53
Either way, the example serves to demonstrate the error, but not why it's actually a problem. Can you elaborate on your sample use case above so we can understand what you're really trying to do? — shmosel
– shmosel, Commented Jan 21, 2018 at 5:59

Steve Brandli · Accepted Answer · 2018-01-21 12:50:40Z

2

Thanks to @Radiodef's comment, I was able to figure this out. The trick, as explained in https://docs.oracle.com/javase/tutorial/java/generics/capture.html, is to use a helper function:

public class MyTest4 {
    public static void main(String[] args) {
        SubObject[] array = new SubObject[5];
        Iterator<? extends SuperObject> iterator
                = Arrays.asList((SuperObject[]) array).iterator();
        Iterable<? extends SuperObject> iterable = getIterable(iterator);
    }

    static <T> Iterable<T> getIterable(Iterator<T> iterator) {
        return () -> iterator;
    }

}

class SuperObject {};

class SubObject extends SuperObject{};

answered Jan 21, 2018 at 12:50

Steve Brandli

5864 silver badges14 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Andrey Tyukin Over a year ago

Interesting observation. If one could write "lambdas with type-arguments" in Java, one could do the same thing inline:

Iterator<? extends Blah> a = ...;  Iterable<? extends Blah> b = (<T>(Iterator<T> iterator) -> (Iterable<T>)(() -> iterator)).apply(a);

. This would be akin to "inline syntax for natural transformations from Iterator<-> to Iterable<->". Unfortunately, this isn't valid (yet). +1

Holger Over a year ago

Note that the type cast in the first statement is obsolete; you can simply do Iterator<? extends SuperObject> iterator = Arrays.asList(array).iterator();, but that does not imply that you should. As the same tutorial states: “Using a wildcard as a return type should be avoided because it forces programmers using the code to deal with wildcards”. In your example, no-one is forcing you, you are rather forcing yourself, but keep that in mind for whatever real life scenario you simplified to this example.

Holger Over a year ago

And just for records, you can do Iterable<? extends SuperObject> iterable = Arrays.<SuperObject>asList(array)::iterator; which would even be a semantically correct Iterable, is it isn’t just a one-time iteration only iterable. Still, there is no real reason to force having a wildcard in the variable…

Collectives™ on Stack Overflow

Error compiling lambda with bounded generic

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related