Multiple Wildcards on a Generic Methods Makes Java Compiler (And Me!) Very Confused

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Multiple wildcards on a generic methods makes Java compiler (and me!) very confused

As Appendix B indicates, this has nothing to do with multiple wildcards, but rather, misunderstanding what List<List<?>> really means.

Let's first remind ourselves what it means that Java generics is invariant:

  1. An Integer is a Number
  2. A List<Integer> is NOT a List<Number>
  3. A List<Integer> IS a List<? extends Number>

We now simply apply the same argument to our nested list situation (see appendix for more details):

  1. A List<String> is (captureable by) a List<?>
  2. A List<List<String>> is NOT (captureable by) a List<List<?>>
  3. A List<List<String>> IS (captureable by) a List<? extends List<?>>

With this understanding, all of the snippets in the question can be explained. The confusion arises in (falsely) believing that a type like List<List<?>> can capture types like List<List<String>>, List<List<Integer>>, etc. This is NOT true.

That is, a List<List<?>>:

  • is NOT a list whose elements are lists of some one unknown type.
    • ... that would be a List<? extends List<?>>
  • Instead, it's a list whose elements are lists of ANY type.

Snippets

Here's a snippet to illustrate the above points:

List<List<?>> lolAny = new ArrayList<List<?>>();

lolAny.add(new ArrayList<Integer>());
lolAny.add(new ArrayList<String>());

// lolAny = new ArrayList<List<String>>(); // DOES NOT COMPILE!!

List<? extends List<?>> lolSome;

lolSome = new ArrayList<List<String>>();
lolSome = new ArrayList<List<Integer>>();

More snippets

Here's yet another example with bounded nested wildcard:

List<List<? extends Number>> lolAnyNum = new ArrayList<List<? extends Number>>();
    
lolAnyNum.add(new ArrayList<Integer>());
lolAnyNum.add(new ArrayList<Float>());
// lolAnyNum.add(new ArrayList<String>());     // DOES NOT COMPILE!!

// lolAnyNum = new ArrayList<List<Integer>>(); // DOES NOT COMPILE!!

List<? extends List<? extends Number>> lolSomeNum;
    
lolSomeNum = new ArrayList<List<Integer>>();
lolSomeNum = new ArrayList<List<Float>>();
// lolSomeNum = new ArrayList<List<String>>(); // DOES NOT COMPILE!!

Back to the question

To go back to the snippets in the question, the following behaves as expected (as seen on ideone.com):

public class LOLUnknowns1d {
    static void nowDefinitelyIllegal(List<? extends List<?>> lol, List<?> list) {
        lol.add(list); // DOES NOT COMPILE!!!
            // The method add(capture#1-of ? extends List<?>) in the
            // type List<capture#1-of ? extends List<?>> is not 
            // applicable for the arguments (List<capture#3-of ?>)
    }
    public static void main(String[] args) {
        List<Object> list = null;
        List<List<String>> lolString = null;
        List<List<Integer>> lolInteger = null;

        // these casts are valid
        nowDefinitelyIllegal(lolString, list);
        nowDefinitelyIllegal(lolInteger, list);
    }
}

lol.add(list); is illegal because we may have a List<List<String>> lol and a List<Object> list. In fact, if we comment out the offending statement, the code compiles and that's exactly what we have with the first invocation in main.

All of the probablyIllegal methods in the question, aren't illegal. They are all perfectly legal and typesafe. There is absolutely no bug in the compiler. It is doing exactly what it's supposed to do.


References

  • Angelika Langer's Java Generics FAQ
    • Which super-subtype relationships exist among instantiations of generic types?
    • Can I create an object whose type is a wildcard parameterized type?
  • JLS 5.1.10 Capture Conversion

Related questions

  • Any simple way to explain why I cannot do List<Animal> animals = new ArrayList<Dog>()?
  • Java nested wildcard generic won’t compile

Appendix: The rules of capture conversion

(This was brought up in the first revision of the answer; it's a worthy supplement to the type invariant argument.)

5.1.10 Capture Conversion

Let G name a generic type declaration with n formal type parameters A1…An with corresponding bounds U1…Un. There exists a capture conversion from G<T1…Tn> to G<S1…Sn>, where, for 1 <= i <= n:

  1. If Ti is a wildcard type argument of the form ? then …
  2. If Ti is a wildcard type argument of the form ? extends Bi, then …
  3. If Ti is a wildcard type argument of the form ? super Bi, then …
  4. Otherwise, Si = Ti.

Capture conversion is not applied recursively.

This section can be confusing, especially with regards to the non-recursive application of the capture conversion (hereby CC), but the key is that not all ? can CC; it depends on where it appears. There is no recursive application in rule 4, but when rules 2 or 3 applies, then the respective Bi may itself be the result of a CC.

Let's work through a few simple examples:

  • List<?> can CC List<String>
    • The ? can CC by rule 1
  • List<? extends Number> can CC List<Integer>
    • The ? can CC by rule 2
    • In applying rule 2, Bi is simply Number
  • List<? extends Number> can NOT CC List<String>
    • The ? can CC by rule 2, but compile time error occurs due to incompatible types

Now let's try some nesting:

  • List<List<?>> can NOT CC List<List<String>>
    • Rule 4 applies, and CC is not recursive, so the ? can NOT CC
  • List<? extends List<?>> can CC List<List<String>>
    • The first ? can CC by rule 2
    • In applying rule 2, Bi is now a List<?>, which can CC List<String>
    • Both ? can CC
  • List<? extends List<? extends Number>> can CC List<List<Integer>>
    • The first ? can CC by rule 2
    • In applying rule 2, Bi is now a List<? extends Number>, which can CC List<Integer>
    • Both ? can CC
  • List<? extends List<? extends Number>> can NOT CC List<List<Integer>>
    • The first ? can CC by rule 2
    • In applying rule 2, Bi is now a List<? extends Number>, which can CC, but gives a compile time error when applied to List<Integer>
    • Both ? can CC

To further illustrate why some ? can CC and others can't, consider the following rule: you can NOT directly instantiate a wildcard type. That is, the following gives a compile time error:

    // WildSnippet1
    new HashMap<?,?>();         // DOES NOT COMPILE!!!
    new HashMap<List<?>, ?>();  // DOES NOT COMPILE!!!
    new HashMap<?, Set<?>>();   // DOES NOT COMPILE!!!

However, the following compiles just fine:

    // WildSnippet2
    new HashMap<List<?>,Set<?>>();            // compiles fine!
    new HashMap<Map<?,?>, Map<?,Map<?,?>>>(); // compiles fine!

The reason WildSnippet2 compiles is because, as explained above, none of the ? can CC. In WildSnippet1, either the K or the V (or both) of the HashMap<K,V> can CC, which makes the direct instantiation through new illegal.

Generic method vs wildcard - compilation error

Confusions do come when you deal with multi-level wildcard syntax. Let's understand what those types exactly mean in there:

  • List<List<?>> is a concrete parameterized type. It is a heterogenous collection of different types of List<E>. Since List<?> represent a family of all the instantiation of List, you can't really pass an ArrayList<List<String>> to List<List<?>>. Because, nothing stops you from adding a List<Integer> to it inside the method, and that will crash at runtime, had compiler allowed it.

  • List<? extends List<?>> is a wildcard parameterized type. It represents a family of different types of List<E>. Basically, it might be a List<ArrayList<String>>, List<LinkedList<Date>>, so on. It can be a list of any type that extend from a List<?>. So, it will be safe to pass a ArrayList<List<String>> to it, the reason being, you won't be allowed to add anything, but null to the list. Adding anything to the list will be a compile time error.

  • As for List<List<T>>, it is again a concrete parameterized type. And since you're dealing with a generic method now, the type parameter will be inferred to be the type that is passed for it. So, for an ArrayList<List<String>>, type T will be inferred as T. A generic method deals with the types that are declared with it. So, there is only a single type T here. All the lists you get out of List<List<T>> will certainly be a List<T> for any type T. So, it's a homogenous collection of that type of List. Inside the method, you can not add any arbitrary List<E> to the List<List<T>>, because the compiler doesn't know whether that type E is compatible with T or not. So, it is safe invocation.

Related:

  • Multiple wildcards on a generic methods makes Java compiler (and me!) very confused
  • Java HashMap nested generics with wildcards
  • What are multi-level wild cards? Confusion in syntax
  • When to use generic methods and when to use wild-card?

Java: Wildcard Types Mismatch Results in Compilation Error

Use the following:

IManager<IAnotherRandomType<?>> test3Manager =
        Factory.<IAnotherRandomType<?>>createManager(test3);

This is just a case of the compiler's type inference falling on its face, so it's necessary explicitly provide the type argument for T.

More technically:

test3 is declared to have the type IAnotherRandomType<?>, where ? is a wildcard capture - a sort of one-use type parameter representing some specific unknown type. That's what the compiler's referring to when it says capture#1-of ?. When you pass test3 into createManager, T gets inferred as IAnotherRandomType<capture#1-of ?>.

Meanwhile, test3Manager is declared to have the type IManager<IAnotherRandomType<?>>, which has a nested wildcard - it does not behave like a type parameter but rather represents any type.

Since generics aren't covariant, the compiler can't convert from IManager<IAnotherRandomType<capture#1-of ?>> to IManager<IAnotherRandomType<?>>.

More reading on nested wildcards:

  • Multiple wildcards on a generic methods makes Java compiler (and me!) very confused
  • Java Generic List<List<? extends Number>>
  • Which super-subset relationships exist among wildcards?

Java Generics: unbounded wildcard generalization not working past second level

Note that List<B> is not a subtype of List<A> even if B is a subtype of A. Therefore, List<Class<? extends Integer>> is not a subtype of List<Class<?>> even though you can assign a Class<? extends Integer> object to a Class<?> variable. Just consider that in bar, it would be legal to invoke a.add(Object.class), as Class<Object> is a subtype of Class<?>.

You therefore want to extend the bar argument type to List<? extends Class<?>>.

Passing any concrete class for a wildcard in a generic class gives error. Why?

Use wildcard for Entry object also, wildcard represents an unknown type. In the above code you defined Entry<Integer, ?> entry with key Integer and value unknown type, But in the same way you have to say for entry objects in List

static void bucketSort(List<? extends Entry<Integer, ?>> list, int n)
{
    //some code
}

Multiple wildcards on a generic methods makes Java compiler (and me!) very confused

As Appendix B indicates, this has nothing to do with multiple wildcards, but rather, misunderstanding what List<List<?>> really means.

Let's first remind ourselves what it means that Java generics is invariant:

  1. An Integer is a Number
  2. A List<Integer> is NOT a List<Number>
  3. A List<Integer> IS a List<? extends Number>

We now simply apply the same argument to our nested list situation (see appendix for more details):

  1. A List<String> is (captureable by) a List<?>
  2. A List<List<String>> is NOT (captureable by) a List<List<?>>
  3. A List<List<String>> IS (captureable by) a List<? extends List<?>>

With this understanding, all of the snippets in the question can be explained. The confusion arises in (falsely) believing that a type like List<List<?>> can capture types like List<List<String>>, List<List<Integer>>, etc. This is NOT true.

That is, a List<List<?>>:

  • is NOT a list whose elements are lists of some one unknown type.
    • ... that would be a List<? extends List<?>>
  • Instead, it's a list whose elements are lists of ANY type.

Snippets

Here's a snippet to illustrate the above points:

List<List<?>> lolAny = new ArrayList<List<?>>();

lolAny.add(new ArrayList<Integer>());
lolAny.add(new ArrayList<String>());

// lolAny = new ArrayList<List<String>>(); // DOES NOT COMPILE!!

List<? extends List<?>> lolSome;

lolSome = new ArrayList<List<String>>();
lolSome = new ArrayList<List<Integer>>();

More snippets

Here's yet another example with bounded nested wildcard:

List<List<? extends Number>> lolAnyNum = new ArrayList<List<? extends Number>>();
    
lolAnyNum.add(new ArrayList<Integer>());
lolAnyNum.add(new ArrayList<Float>());
// lolAnyNum.add(new ArrayList<String>());     // DOES NOT COMPILE!!

// lolAnyNum = new ArrayList<List<Integer>>(); // DOES NOT COMPILE!!

List<? extends List<? extends Number>> lolSomeNum;
    
lolSomeNum = new ArrayList<List<Integer>>();
lolSomeNum = new ArrayList<List<Float>>();
// lolSomeNum = new ArrayList<List<String>>(); // DOES NOT COMPILE!!

Back to the question

To go back to the snippets in the question, the following behaves as expected (as seen on ideone.com):

public class LOLUnknowns1d {
    static void nowDefinitelyIllegal(List<? extends List<?>> lol, List<?> list) {
        lol.add(list); // DOES NOT COMPILE!!!
            // The method add(capture#1-of ? extends List<?>) in the
            // type List<capture#1-of ? extends List<?>> is not 
            // applicable for the arguments (List<capture#3-of ?>)
    }
    public static void main(String[] args) {
        List<Object> list = null;
        List<List<String>> lolString = null;
        List<List<Integer>> lolInteger = null;

        // these casts are valid
        nowDefinitelyIllegal(lolString, list);
        nowDefinitelyIllegal(lolInteger, list);
    }
}

lol.add(list); is illegal because we may have a List<List<String>> lol and a List<Object> list. In fact, if we comment out the offending statement, the code compiles and that's exactly what we have with the first invocation in main.

All of the probablyIllegal methods in the question, aren't illegal. They are all perfectly legal and typesafe. There is absolutely no bug in the compiler. It is doing exactly what it's supposed to do.


References

  • Angelika Langer's Java Generics FAQ
    • Which super-subtype relationships exist among instantiations of generic types?
    • Can I create an object whose type is a wildcard parameterized type?
  • JLS 5.1.10 Capture Conversion

Related questions

  • Any simple way to explain why I cannot do List<Animal> animals = new ArrayList<Dog>()?
  • Java nested wildcard generic won’t compile

Appendix: The rules of capture conversion

(This was brought up in the first revision of the answer; it's a worthy supplement to the type invariant argument.)

5.1.10 Capture Conversion

Let G name a generic type declaration with n formal type parameters A1…An with corresponding bounds U1…Un. There exists a capture conversion from G<T1…Tn> to G<S1…Sn>, where, for 1 <= i <= n:

  1. If Ti is a wildcard type argument of the form ? then …
  2. If Ti is a wildcard type argument of the form ? extends Bi, then …
  3. If Ti is a wildcard type argument of the form ? super Bi, then …
  4. Otherwise, Si = Ti.

Capture conversion is not applied recursively.

This section can be confusing, especially with regards to the non-recursive application of the capture conversion (hereby CC), but the key is that not all ? can CC; it depends on where it appears. There is no recursive application in rule 4, but when rules 2 or 3 applies, then the respective Bi may itself be the result of a CC.

Let's work through a few simple examples:

  • List<?> can CC List<String>
    • The ? can CC by rule 1
  • List<? extends Number> can CC List<Integer>
    • The ? can CC by rule 2
    • In applying rule 2, Bi is simply Number
  • List<? extends Number> can NOT CC List<String>
    • The ? can CC by rule 2, but compile time error occurs due to incompatible types

Now let's try some nesting:

  • List<List<?>> can NOT CC List<List<String>>
    • Rule 4 applies, and CC is not recursive, so the ? can NOT CC
  • List<? extends List<?>> can CC List<List<String>>
    • The first ? can CC by rule 2
    • In applying rule 2, Bi is now a List<?>, which can CC List<String>
    • Both ? can CC
  • List<? extends List<? extends Number>> can CC List<List<Integer>>
    • The first ? can CC by rule 2
    • In applying rule 2, Bi is now a List<? extends Number>, which can CC List<Integer>
    • Both ? can CC
  • List<? extends List<? extends Number>> can NOT CC List<List<Integer>>
    • The first ? can CC by rule 2
    • In applying rule 2, Bi is now a List<? extends Number>, which can CC, but gives a compile time error when applied to List<Integer>
    • Both ? can CC

To further illustrate why some ? can CC and others can't, consider the following rule: you can NOT directly instantiate a wildcard type. That is, the following gives a compile time error:

    // WildSnippet1
    new HashMap<?,?>();         // DOES NOT COMPILE!!!
    new HashMap<List<?>, ?>();  // DOES NOT COMPILE!!!
    new HashMap<?, Set<?>>();   // DOES NOT COMPILE!!!

However, the following compiles just fine:

    // WildSnippet2
    new HashMap<List<?>,Set<?>>();            // compiles fine!
    new HashMap<Map<?,?>, Map<?,Map<?,?>>>(); // compiles fine!

The reason WildSnippet2 compiles is because, as explained above, none of the ? can CC. In WildSnippet1, either the K or the V (or both) of the HashMap<K,V> can CC, which makes the direct instantiation through new illegal.

Double wildcard parameterization (nested wildcards) with Java generics

Use the following instead:

Set<? extends Class<?>> subTypes = reflections.getSubTypesOf(type);

The problem is that nested wildcards don't perform type capture. Declaring a Set<Class<?>> means "a set of classes of any type", while what's being returned is a Set<Class<? extends capture#19-of ?>>, which means "a set of classes of any type extending from some specific unknown type". In this case, that "specific unknown type" is derived from the type argument to T, which is inferred from type to be an unbounded wildcard capture (the ? in Class<?>).

For example, pretend that "specific unknown type" is Number:

Class<Number> type = ...
Set<Class<?>> subTypes = reflections.getSubTypesOf(type);

Here, getSubTypesOf returns a Set<Class<? extends Number>>. That type is not assignable to Set<Class<?>> because generic types aren't covariant. But, wildcard capture helps us express covariance, allowing types like Set<? extends Class<?>>. The caveat is that we can't add anything but null to such a set, since we don't know its specific type.

Related posts:

  • Java Generic List<List<? extends Number>>
  • Multiple wildcards on a generic methods makes Java compiler (and me!) very confused
  • What is PECS (Producer Extends Consumer Super)?

Similar posts:

  • Java: Wildcard Types Mismatch Results in Compilation Error
  • Issue with declaration of Map<String,Class<? extends Serializable>>
  • Bounded-wildcard related compiler error

Nested wildcards

It is important to understand the implication of the wildcard types.

You already understood that you can assign your Map<Integer, Map<Integer, String>> to Map<?, ?> as Map<?, ?> implies arbitrary types, unknown to whoever might have a reference of the declared type Map<?, ?>. So you can assign any map to Map<?, ?>.

In contrast, if you have a Map<?, Map<?, ?>> it has an unknown key type but the value type is not unknown. It’s Map<?,?> the type, recall the information above, that can be assigned with any map.

So, the following code is legal:

Map<?, Map<?, ?>> map=new HashMap<>();
map.put(null, Collections.<String,String>singletonMap("foo", "bar"));
map.put(null, Collections.<Double,Integer>singletonMap(42.0, 1000));
map.put(null, Collections.<Object,Boolean>singletonMap(false, true));

Here, we are putting a null key as we can’t put anything else for keys but arbitrary typed maps as values as that’s what a value type of Map<?, ?> implies: can be assigned from arbitrary maps. Note that by iterating over the entries we can also set other entries having non-null keys to arbitrary maps then.

So I’m quite sure that you don’t want to assign your Map<Integer, Map<Integer, String>> to a Map<?, Map<?, ?>> and discover arbitrary maps not being Map<Integer, String> as values afterwards and that you are quite happy that the compiler doesn’t allow this.

What you actually want to do is to assign your map to a type which has both, key and value type, unknown but still telling that your values are maps:

Map<Integer, Map<Integer, String>> someMap = new HashMap<>();
Map<?, ? extends Map<?, ?>> map=someMap;

In the generic type system Map<Integer, String> is a sub-type of Map<?, ?> so you can assign it to Map<?, ?> as well as ? extends Map<?, ?>. This sub-type relationship is not different than the relationship of String to Object. You can assign any String to a variable of type Object but if you have a Map<?,String> you can’t assign it to Map<?,Object> but only to Map<?, ? extends Object> for the same reason: the map shall continue to contain Strings as values rather than receiving arbitrary objects.

Note that you can workaround this limitation. You can say:

Map<Integer, Map<Integer, String>> someMap = new HashMap<>();
Map<?, Map<?, ?>> map=Collections.unmodifiableMap(someMap);

Since the map returned by unmodifiableMap does not allow any modifications, it allows widening the key and value types. The contained values are of the specified type (i.e. Map<?, ?>) when you query the map, but attempts to put in arbitrary map values, while not rejected by the compiler, will be rejected at runtime.


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