The short answer is that generics are invariant, so this will not work.
The long answer takes a while to understand. It starts simple:
Dog woof = new Dog();
Animal animal = woof;
Works just fine, since a Dog is an Animal. On the other hand:
List< Animal > fauna = new ArrayList<>();
List< Dog > dogs = new ArrayList<>();
fauna = dogs;
will fail to compile, because generics are invariant; basically a List<Dog> is not a List<Animal>.
How come? Well if the assignment would have been possible, what is stopping you from doing:
fauna.add(new Cat());
dogs.get(0); // what is this now?
A compiler could be smarter here, actually. What if your Lists are immutable? After creation, you can't put anything into them. In such a case, fauna = dogs, should be allowed, but java does not do this (scala does), even with the newly added Immutable collections in java-9.
When Lists are immutable, they are said to be Producers, meaning they don't take the generic type as input. For example:
interface Sink<T> {
T nextElement();
}
Since Sink never takes T as input, it is a Producer of Ts (not a Consumer), thus it could be possible to say:
Sink<Object> objects ...
Sink<String> strings ...
objects = strings;
Since Sink has no option to add elements, we can't break anything, but java does not care and prohibits this. kotlinc (just like scalac) allows it.
In java this shortcoming is solved with a "bounded type":
List<? extends Animal> animals = new ArrayList<>();
animals = dogs;
The good thing is that you still can't do: animals.add(new Cat()). You know exactly what that list holds - some types of Animals, so when you read from it, you always, for a fact, know that you will get an Animal. But because List<? extends Animal> is assignable to List<Dog> for example, addition is prohibited, otherwise:
animals.add(new Cat()); // if this worked
dogs.get(0); // what is this now?
This "addition is prohibited" is not exactly correct, since it's always possible to do:
private static <T> void topLevelCapture(List<T> list) {
T t = list.get(0);
list.add(t);
}
topLevelCapture(animals);
Why this works is explained here, what matters is that this does not break anything.
What if you wanted to say that you have a group of animals, like a List<List...>? May be the first thing you want to do is List<List<Animal>>:
List<List<Animal>> groups = new ArrayList<>();
List<List<Dog>> dogs = new ArrayList<>();
groups = dogs;
this would obviously not work. But what if we added bounded types?
List<List<? extends Animal>> groups = new ArrayList<>();
List<List<Dog>> dogs = new ArrayList<>();
groups = dogs;
even if List<Dog> is a List<? extends Animal> the generics of these are not (generics are invariant). Again, if this would have been allowed, you could do:
groups.add(<list of cats>);
dogs.get(0); // obvious problems
The only way to make it work would be via:
List<? extends List<? extends Animal>> groups = new ArrayList<>();
List<List<Dog>> dogs = new ArrayList<>();
groups = dogs;
that is, we found a super type of List<Dog> in List<? extends Animal> and we also need the bounded type ? extends List... so that the outer lists themselves are assignable.
This huge intro was to show that:
Map<Integer, Map<Integer, String>> map = new HashMap<>();
Map<?, ?> broader = new HashMap<>();
broader = map;
would compile because there are no restrictions what-so-ever here, the broader map basically is a map "of anything".
If you read what I had to say above, you probably know why this is not allowed:
Map<Integer, Map<Integer, String>> map = new HashMap<>();
Map<?, Map<?, ?>> lessBroader = new HashMap<>();
lessBroader = map;
if it would have been allowed, you could do:
Map<Double, Float> newMap = new HashMap<>(); // this is a Map<?, ?> after all
lessBroader.add(12, newMap);
map.get(12); // hmm...
If maps were immutable and the compiler would care, this could have been avoided and the assignment could have made to work just fine.
return someMap; //compilation fails– BucksawMap<?, ? extends Map<?, ?>>– ReinMap. Also i didn't find the answer in that 'List posts' – Bucksaw