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Incompatible return type while implementing interface method with generic wildcard return type [duplicate]

This compiles (1.6)
List<? extends Object> l = new ArrayList<Date>();
But this does not
List<List<? extends Object>> ll = new ArrayList<List<Date>>();
with the error of
Type mismatch: cannot convert from ArrayList<List<Date>> to List<List<? extends Object>>
Could someone explain why?
Thanks
EDIT: edited for being consequent

Well the explanations are correct, but I think it'd be a nice thing to add the actual working solution as well ;)
List<? extends List<? extends Object>>
Will work just fine, but obviously the use of such a collection is quite limited by the usual limitations of generic Collections (but then the same is true for the simpler List< ? extends Date >)

Because it would break type safety:
List<List<Object>> lo = new ArrayList<List<Object>>();
List<List<? extends Object>> ll = lo;
List<String> ls = new ArrayList<String>();
ll.add(ls);
lo.get(0).add(new Object());
String s = ls.get(0); // assigns a plain Object instance to a String reference

Suppose D is subtype of B, G<T> is a generic type
B x = new D(); // OK
G<B> y = new G<D>(); // FAIL
Now, G<Date> is a subtype of G<?>, therefore
G<?> x = new G<Date>(); // OK
G<G<?>> y = new G<G<Date>>(); // FAIL

When assigning to a variable (List<T>) with a non-wildcard generic type T, the object being assigned must have exactly T as its generic type (including all generic type parameters of T, wildcard and non-wildcard). In your case T is List<? extends Object>, which is not the same type as List<Date>.
What you can do, because List<Date> is assignable to List<? extends Object>, is use the wildcard type:
List<? extends List<? extends Object>> a = new ArrayList<List<Date>>();

<? extends Object>
means that the wildcard could be substituted only for those objects, that are subclass of Object class.
List<List<? extends Object>> ll = new ArrayList<List<Object>>();
gives you error of type mismatch because you are trying to assign a ArrayList of List of object of java class Object to a List that contains the List of any type of objects that are subclass of java class Object.
For more ref, have a look at the Wildcard documentation

Java nested generic casting (? super Type) [duplicate]

How come one must use the generic type Map<?, ? extends List<?>> instead of a simpler Map<?, List<?>> for the following test() method?
public static void main(String[] args) {
Map<Integer, List<String>> mappy =
new HashMap<Integer, List<String>>();
test(mappy);
}
public static void test(Map<?, ? extends List<?>> m) {}
// Doesn't compile
// public static void test(Map<?, List<?>> m) {}
Noting that the following works, and that the three methods have the same erased type anyways.
public static <E> void test(Map<?, List<E>> m) {}

Fundamentally, List<List<?>> and List<? extends List<?>> have distinct type arguments.
It's actually the case that one is a subtype of the other, but first let's learn more about what they mean individually.
Understanding semantic differences
Generally speaking, the wildcard ? represents some "missing information". It means "there was a type argument here once, but we don't know what it is anymore". And because we don't know what it is, restrictions are imposed on how we can use anything that refers to that particular type argument.
For the moment, let's simplify the example by using List instead of Map.
A List<List<?>> holds any kind of List with any type argument. So i.e.:
List<List<?>> theAnyList = new ArrayList<List<?>>();
// we can do this
theAnyList.add( new ArrayList<String>() );
theAnyList.add( new LinkedList<Integer>() );
List<?> typeInfoLost = theAnyList.get(0);
// but we are prevented from doing this
typeInfoLost.add( new Integer(1) );
We can put any List in theAnyList, but by doing so we have lost knowledge of their elements.
When we use ? extends, the List holds some specific subtype of List, but we don't know what it is anymore. So i.e.:
List<? extends List<Float>> theNotSureList =
new ArrayList<ArrayList<Float>>();
// we can still use its elements
// because we know they store Float
List<Float> aFloatList = theNotSureList.get(0);
aFloatList.add( new Float(1.0f) );
// but we are prevented from doing this
theNotSureList.add( new LinkedList<Float>() );
It's no longer safe to add anything to the theNotSureList, because we don't know the actual type of its elements. (Was it originally a List<LinkedList<Float>>? Or a List<Vector<Float>>? We don't know.)
We can put these together and have a List<? extends List<?>>. We don't know what type of List it has in it anymore, and we don't know the element type of those Lists either. So i.e.:
List<? extends List<?>> theReallyNotSureList;
// these are fine
theReallyNotSureList = theAnyList;
theReallyNotSureList = theNotSureList;
// but we are prevented from doing this
theReallyNotSureList.add( new Vector<Float>() );
// as well as this
theReallyNotSureList.get(0).add( "a String" );
We've lost information both about theReallyNotSureList, as well as the element type of the Lists inside it.
(But you may note that we can assign any kind of List holding Lists to it...)
So to break it down:
// ┌ applies to the "outer" List
// ▼
List<? extends List<?>>
// ▲
// └ applies to the "inner" List
The Map works the same way, it just has more type parameters:
// ┌ Map K argument
// │ ┌ Map V argument
// ▼ ▼
Map<?, ? extends List<?>>
// ▲
// └ List E argument
Why ? extends is necessary
You may know that "concrete" generic types have invariance, that is, List<Dog> is not a subtype of List<Animal> even if class Dog extends Animal. Instead, the wildcard is how we have covariance, that is, List<Dog> is a subtype of List<? extends Animal>.
// Dog is a subtype of Animal
class Animal {}
class Dog extends Animal {}
// List<Dog> is a subtype of List<? extends Animal>
List<? extends Animal> a = new ArrayList<Dog>();
// all parameterized Lists are subtypes of List<?>
List<?> b = a;
So applying these ideas to a nested List:
List<String> is a subtype of List<?> but List<List<String>> is not a subtype of List<List<?>>. As shown before, this prevents us from compromising type safety by adding wrong elements to the List.
List<List<String>> is a subtype of List<? extends List<?>>, because the bounded wildcard allows covariance. That is, ? extends allows the fact that List<String> is a subtype of List<?> to be considered.
List<? extends List<?>> is in fact a shared supertype:
List<? extends List<?>>
╱ ╲
List<List<?>> List<List<String>>
In review
Map<Integer, List<String>> accepts only List<String> as a value.
Map<?, List<?>> accepts any List as a value.
Map<Integer, List<String>> and Map<?, List<?>> are distinct types which have separate semantics.
One cannot be converted to the other, to prevent us from doing modifications in an unsafe way.
Map<?, ? extends List<?>> is a shared supertype which imposes safe restrictions:
Map<?, ? extends List<?>>
╱ ╲
Map<?, List<?>> Map<Integer, List<String>>
How the generic method works
By using a type parameter on the method, we can assert that List has some concrete type.
static <E> void test(Map<?, List<E>> m) {}
This particular declaration requires that all Lists in the Map have the same element type. We don't know what that type actually is, but we can use it in an abstract manner. This allows us to perform "blind" operations.
For example, this kind of declaration might be useful for some kind of accumulation:
static <E> List<E> test(Map<?, List<E>> m) {
List<E> result = new ArrayList<E>();
for(List<E> value : m.values()) {
result.addAll(value);
}
return result;
}
We can't call put on m because we don't know what its key type is anymore. However, we can manipulate its values because we understand they are all List with the same element type.
Just for kicks
Another option which the question does not discuss is to have both a bounded wildcard and a generic type for the List:
static <E> void test(Map<?, ? extends List<E>> m) {}
We would be able to call it with something like a Map<Integer, ArrayList<String>>. This is the most permissive declaration, if we only cared about the type of E.
We can also use bounds to nest type parameters:
static <K, E, L extends List<E>> void(Map<K, L> m) {
for(K key : m.keySet()) {
L list = m.get(key);
for(E element : list) {
// ...
}
}
}
This is both permissive about what we can pass to it, as well as permissive about how we can manipulate m and everything in it.
See also
"Java Generics: What is PECS?" for the difference between ? extends and ? super.
JLS 4.10.2. Subtyping among Class and Interface Types and JLS 4.5.1. Type Arguments of Parameterized Types for entry points to the technical details of this answer.

This is because the subclassing rules for generics are slightly different from what you may expect. In particular if you have:
class A{}
class B extends A{}
then
List<B> is not a subclass of List<A>
It's explained in details here and the usage of the wildcard (the "?" character) is explained here.

What is wrong with my method generic parameters

I want to populate a List with generic maps, but my code does not compile. I have prepared the most simplified example for the problem. In the comments above problematic lines I have put the error the line below produces.
void populateList(List<? extends Map<String,?>> list) {
list.clear();
HashMap<String, ?> map;
map = new HashMap<String,String>();
//The method put(String, capture#2-of ?) in the type HashMap<String,capture#2-of ?> is not applicable for the arguments (String, String)
map.put("key", "value"); // this line does not compile
// The method add(capture#3-of ? extends Map<String,?>) in the type List<capture#3-of ? extends Map<String,?>> is not applicable for the arguments (HashMap<String,capture#5-of ?>)
list.add(map); //This line does not compile
}
Why is this so? Is there something I do not understand?
EDIT 1
According to one of the answers below in which he pointed out that ? stands for unknown type and not a descendant of Object. This is a valid point. And also, inside the method I know the type which go into map so I have modified my simple code accordingly.
void populateList(List<? extends Map<String,?>> list) {
list.clear();
HashMap<String, String> map; //known types
map = new HashMap<String,String>();
map.put("key", "value"); // this line now compiles
// The method add(capture#3-of ? extends Map<String,?>) in the type List<capture#3-of ? extends Map<String,?>> is not applicable for the arguments (HashMap<String,capture#5-of ?>)
list.add(map); //This line STILL does not compile. Why is that?
}
The reason I am asking this is because a method form android SDK expects such list and as it seems one cannot populate such lists. How does one do that? Typecast?
EDIT 2
Since there several proposals to change my signature I will add that I cannot do that. Basicaly, I would like to populate lists for SimpleExpandablaListAdapter.
void test() {
ExpandableListView expandableListView.setAdapter(new ArrayAdapterRetailStore(this, R.layout.list_item_retail_store, retailStores));
List<? extends Map<String, ?>> groupData= new ArrayList<HashMap<String,String>>();
populateGroup(groupData)
// child data ommited for simplicity
expandableListView.setAdapter( new SimpleExpandableListAdapter(
this,
groupdata,
R.layout.list_group,
new String[] {"GroupKey"},
new int[] {R.id.tvGroupText},
childData,
R.layout.list_item_child,
new String[] {"ChildKey"},
new int[] {R.id.tvChilText}));
}
// I want populateGroupData() to be generic
void populateGroupData(List<? extends Map<String,?>> groupData) {
groupData.clear();
HashMap<String,String> map;
map = new HashMap<String,String>();
map.put("key", "value");
groupData.add(map); // does not compile
}

From the documentation
When the actual type parameter is ?, it stands for some unknown type. Any parameter we pass to add would have to be a subtype of this unknown type. Since we don't know what type that is, we cannot pass anything in. The sole exception is null, which is a member of every type.
so, you can add only
list.add(null);
Please read this tutorial on Generics Wildcards

here is the working code
//also works with void populateList(List<Map<String,?>> list) {
void populateList(List<? super Map<String,?>> list) {
list.clear();
Map<String, String> map;
map = new HashMap<String,String>();
map.put("key", "value"); // this line now compiles
list.add(map); //This line compiles
}
and why it works:
List<? super Map<String,?>> list or simply List<Map<String,?>> list
// => this ensure you that the list can contains a Map<String,?>.
Map<String, String> map is a Map<String,?>
// =>that can ber inserted to the list, so you don't need any cast
Edit:
The common mistake is that the wildcard "? extends Map" will limit the function call to a list that is "at least" typed with map. This is not what you want, because you could pass a List<TreeMap<String,?>>which can not contain a HashMap for example. Additionnaly you couldn't call your method with a List<Object>
-> To illustrate generics limitation i have added 2 examples with super and extends
void exampleWithExtends(List<? extends Map<String,?>> list) {
}
void exampleWithSuper(List<? super Map<String,?>> list) {
}
void funWithGenerics(){
exampleWithExtends(new ArrayList<TreeMap<String,String>>());
exampleWithExtends(new ArrayList<Map<String,?>>());//works in both cases
//exampleWithExtends(new ArrayList<Object>()); /does not compile
//exampleWithSuper(new ArrayList<TreeMap<String,String>>()); //does not compile
exampleWithSuper(new ArrayList<Map<String,?>>());//works in both cases
exampleWithSuper(new ArrayList<Object>());
}

There is no way you can write Map<String, String> map = getMap("abc"); without a cast
The problem has more to do with easymock and the types returned/expected by the expect and andReturn methods, which I'm not familiar with. You could write
Map<String, String> expected = new HashMap<String, String> ();
Map<?, ?> actual = getMap("someKey");
boolean ok = actual.equals(pageMaps);
//or in a junit like syntax
assertEquals(expected, actual);
Not sure if that can be mixed with your mocking stuff. This would maybe work:
EasyMock.expect((Map<String, String>) config.getMap("sillyMap")).andReturn(pageMaps);
Also note that you can't add anything to a generic collection with a wildcard. So this:
Map<?, ?> map = ...
map.put(a, b);
won't compile, unless a and b are null

Java is type-safe! At least at this point :)
This will do the trick:
HashMap<String, String> map = new HashMap<String,String>();
map.put("key", "value");
((List<Map<String,String>>)groupData).add(map);

Detail about the "super" wildcard in java generics

I have a question regarding generics:
Map<? super String, ? super String> mappa1 = new HashMap<Object,Object>();
with super it's possible to instantiate a HashMap<Object,Object> for a <? super String>.
However then you can add only objects which extends String ( in this case only String itself).
Why don't they forbid by compilation error as well as happens with the extends wildcard.
I mean if once created a Map <Object, Object> it's possible only to add Strings.. why not forcing to create a Map<String, String> in the first place? (like it happens with the extends wildcard)
Again I know the difference between super and extends concerning generics. I would like just to know the details I have aboved-mentioned.
Thanks in advance.

Let's use List instead of Map for brevity.
Essentially, practical meaning of extends and super can be defined as follows:
List<? extends T> means "a List you can get T from"
List<? super T> means "a List you can put T into"
Now you can see that there is nothing special about extends - behavior of extends and super is completely symmetric:
List<? extends Object> a = new ArrayList<String>(); // Valid, you can get an Object from List<String>
List<? extends String> b = new ArrayList<Object>(); // Invalid, there is no guarantee that List<Object> contains only Strings
List<? super String> a = new ArrayList<Object>(); // Valid, you can put a String into List<Object>
List<? super Object> b = new ArrayList<String>(); // Invalid, you cannot put arbitrary Object into List<String>

I think you are thrown off because you picked a collection type. Collections are rarely used as consumers and thus a lower bound (? super X) is not put on their element types. A more appropriate example is predicate.
Consider a method such as <E> List<E> filter(List<? extends E> p, Predicate<? super E> p). It will take a list l and a predicate p and return a new list containing all elements of l which satisfy p.
You could pass in a List<Integer> and a Predicate<Number> which is satisfied by all multiples of 2.5. The Predicate<Number> would become a Predicate<? super Integer>. If it did not, you could not invoke filter as follows.
List<Integer> x = filter(Arrays.asList(1,5,8,10), Predicates.multipleOf(2.5));

Map<? super String, ? super String> mappa1 = new HashMap<Object,Object>();
Since Java Generics are based on type erasure, with this line you didn't create a MashMap<Object,Object>. You just created an instance of the HashMap class; the type parameters get lost immediately after this line of code and all that stays is the type of your mappa1 variable, which doesn't even mention Object. The type of the new expression is assignment-compatible with the type of mappa1 so the compiler allows the assignment.
In general, the type parameters used with new are irrelevant and to address this issue, Java 7 has introduced the diamond operator <>. All that really matters is the type of mappa1, which is is Map<? super String, ? super String>; as far as the rest of your code is concerned, this is the type of the instantiated map.

The problem you're describing doesn't exist.
It is because your reference is declared as Map<? super String, ? super String>. But your actual object can hold any object since it's HashMap<Object,Object>
Map<? super String, ? super String> mappa1 = new HashMap<Object,Object>();
map1.put("", "");
//you can put only string to map1
//but you can do this
Map map2 = map1;
map2.put(23, 234);
the same can be described by a better example:
String a = "a".
a.length(); // legal.
Object b = a;
b.length() // compilation error

Java generics, nested collection of wildcard

This compiles (1.6)
List<? extends Object> l = new ArrayList<Date>();
But this does not
List<List<? extends Object>> ll = new ArrayList<List<Date>>();
with the error of
Type mismatch: cannot convert from ArrayList<List<Date>> to List<List<? extends Object>>
Could someone explain why?
Thanks
EDIT: edited for being consequent

Well the explanations are correct, but I think it'd be a nice thing to add the actual working solution as well ;)
List<? extends List<? extends Object>>
Will work just fine, but obviously the use of such a collection is quite limited by the usual limitations of generic Collections (but then the same is true for the simpler List< ? extends Date >)

Because it would break type safety:
List<List<Object>> lo = new ArrayList<List<Object>>();
List<List<? extends Object>> ll = lo;
List<String> ls = new ArrayList<String>();
ll.add(ls);
lo.get(0).add(new Object());
String s = ls.get(0); // assigns a plain Object instance to a String reference

Suppose D is subtype of B, G<T> is a generic type
B x = new D(); // OK
G<B> y = new G<D>(); // FAIL
Now, G<Date> is a subtype of G<?>, therefore
G<?> x = new G<Date>(); // OK
G<G<?>> y = new G<G<Date>>(); // FAIL

When assigning to a variable (List<T>) with a non-wildcard generic type T, the object being assigned must have exactly T as its generic type (including all generic type parameters of T, wildcard and non-wildcard). In your case T is List<? extends Object>, which is not the same type as List<Date>.
What you can do, because List<Date> is assignable to List<? extends Object>, is use the wildcard type:
List<? extends List<? extends Object>> a = new ArrayList<List<Date>>();

<? extends Object>
means that the wildcard could be substituted only for those objects, that are subclass of Object class.
List<List<? extends Object>> ll = new ArrayList<List<Object>>();
gives you error of type mismatch because you are trying to assign a ArrayList of List of object of java class Object to a List that contains the List of any type of objects that are subclass of java class Object.
For more ref, have a look at the Wildcard documentation