Develop Reference

Adding Object to Generic List with two types

I read a couple of posts such as here but I was unable to find the solution for my problem.
Why I am unable to add d? It is a subtype of Object...
Type of d: A<B<X>>
List<A<B<? extends Object>>> rv=new LinkedList<>();
rv.add(d); //not working
EDIT
I tried to simplify the problem. When I do:
A<B<?>> abcv=new A<B<String>>();
I get the error: Type mismatch: cannot convert from A<B<String>> to A<B<?>>
However, String is compatible with "?" - so why is it not working? I want to add elements to a list where the last type can by anything, something like this:
List<A<B<?>>> rv=new LinkedList<>();
rv.add(new A<B<X>>());
rv.add(new A<B<String>>());
rv.add(new A<B<Integer>>());

List<SubClaz> is not a subtype of List<SuperClaz> in Java. That's why the wildcards are used: List<SubClaz> is a subtype of List<? extends SuperClaz>.
Now for your A<B<?>> abcv=new A<B<String>>(); example:
By adding the wildcard, you're making B<String> a subtype of B<?>, but since these are also wrapped by another type A, we're back to the first problem:
A<B<String>> is not a subtype of A<B<?>>
(Notice B<?> is the SuperClaz and B<String> is the SubClaz in this case).
You can fix this the same way; by adding another wildcard:
A<B<String>>() is a subtype of A<? extends B<?>>.
Keep in mind that this doesn't allow you to read or manipulate the list as you want. Search for covariance and contravariance for more detail. Here is a good one: http://bayou.io/draft/Capturing_Wildcards.html

d should have type A<B<? extends Object>> or compatible.

List<A<B<?>>> rv = new LinkedList<>();
? in generic means any type.
For example:
You can assign any type of generic list to List<?> but you can not add any type of object into list because compiler does not know what type it is because of wildcard(?)
List<?> genericList1 = new ArrayList<String>();
List<?> genericList2 = new ArrayList<Integer>();
List<?> genericList3 = new ArrayList<X>();
/**
* Compiler will allow to assign any type of generic list to List<?>
* but it will not allow to add.
*/
genericList1.add("xyz"); // It will give compiler error
The <?> wildcard allows a list of ANY type to be assigned, but the add() method is not valid on that list because you could put the wrong kind of thing into the collection. Compiler does not know which type you would pass because of wildcard(?).
If you want to add any type of object to your list than you can take list like this.
List<Object> rv = new LinkedList<>();
rv.add(new A<B<X>>());
rv.add(new A<B<String>>());
rv.add(new A<B<Integer>>());
Now compiler knows that rv is the list which accept only Object type, so compiler will not complain anything.

Why is it possible to cast generic class?

Java generics are invariant so it's impossible to do such cast:
List<Object> li = (List<Object>)new ArrayList<Integer>();
But in the following code in line 4 I can cast from List<Integer> to List<T>, where T could be any type. Why is that type of cast allowed?
I know it generates warning about unchecked cast, but the point is that this cast is possible inside parametrized method, but not in normal code. Keeping in mind that generics are invariant why is it allowed? In normal code when having List<Integer> I can only cast it to List<Integer> which doesn't make sense and other casts are illegal. So what is the point of allowing such cast as in line 4?
I know that generic types are removed at compile time and it ends with List xlist = (List)list, but before removing those types it is obvious that this cast should not be allowed unless it is accepted only for the case when someone passes Integer as el which doesn't make much sense.
class Test {
public static <T> void t(List<Integer> list, T el) {
List<T> xlist = (List<T>)list; //OK
xlist.add(el);
}
public static void main(String[] args) {
List<Integer> list = new ArrayList<>();
t(list, "a");
t(list, "b");
//prints [a, b] even if List type is Integer
System.out.println(list);
}
}

In Java, it is a compile error to do an explicit cast that is known at compile time to be always incorrect or always correct. A cast from List<Integer> to List<Object> is known at compile time to be always incorrect, therefore it is not allowed. A cast from List<Integer> to List<T> is not known at compile time to be always incorrect -- it would be correct if T were Integer, and T is not known at compile time.

I guess the answer is that this cast is possible only because Integer can be passed as argument el and in that case that cast will be correct. In all other cases of types other than Integer it will be illegal cast. But as I see in generics, if there is at least one type which would give correct cast (Integer) then the compiler doesn't give an error, but warning, even if all other cases are invalid.

Object is a super class of the class Integer. List<Object> is not a super class of the class List<Integer> , but furthermore if you would like a List<Object> to reference to a List<Integer> you can use joker sign (?).
List<?> list1 = new List<Object>();
List<?> list2 = new List<Integer>();
list1 = list2;

Casting Java ArrayList<Custom Class> to ArrayList<Object>

I am attempting to convert an ArrayList of class SomeClass to an ArrayList of class Object. This new ArrayList of Object will then be passed to a function. I currently have done the following:
// convert ArrayList<SomeClass> to generic ArrayList<Object>
Object[] objectArray = someClassList.toArray();
ArrayList<Object> objects = new ArrayList<Object>();
for (int i = 0; i < objectArray.length; i++) {
objects.add(objectArray[i]);
}
someFunction(objects);
public void someFunction(ArrayList<Object> objects) {
// do something with objects
}
Is there a more efficient or "standard" way of doing this? Is what I am doing "wrong" in the first place?
The purpose of converting it to ArrayList of class Object is that I have created an external library to process ArrayList of generic Objects.

If you are able to change the function's signature to taking an ArrayList<? extends Object> objects or an ArrayList<?> objects (or even better, List instead of ArrayList), you will be able to pass your ArrayList<SomeClass> directly.
The reason an ArrayList<SomeClass> is not an ArrayList<Object> is that an ArrayList<Object> would accept that you add() any kind of Object into it, which is not something you can do with an ArrayList<SomeClass>. On the other hand, an ArrayList<? extends Object> will allow you to retrieve elements from the list, but not add elements, so ArrayList<SomeClass> can safely be assigned to it.

Since you created the external library, I think it would be easier to modify the function signature to accept lists of any type. This can be accomplished using the unbounded wildcard ?:
public static void someFunction(List<?> objects) {
// whatever
}
Then you don't need to make any conversions to call it:
public static void main(String[] args) {
List<String> words = new ArrayList<>();
someFunction(words);
}
Also, unless you have a good reason not to, it would be better to accept any List in someFunction instead of limiting your input to ArrayLists. This makes your code more flexible and easier to change in the future.

A simple way to convert a List<SubFoo> to a List<Foo> is to use Collections.unmodifiableList(listOfSubFoos), which is perfectly type-safe and actually enforces that you can't do anything bad with it (like adding a DifferentSubFoo).

It is possible to transform the type parameters of a type in arbitrary ways with two casts:
ArrayList<SomeClass> l1 = ...;
ArrayList<Object> l2 = (ArrayList<Object>) (Object) l1;
But, as Aasmund Eldhuset also says in his answer: This is probably not a good idea! It is better to give a more suitable type to l2 instead, like ArrayList<?>.
This code gives you an compile warning saying Type safetyThat: Unchecked cast from Object to ArrayList<Object> for a reason. If for example a String is added to l2 and then someone reads l1 and expects a SomeClass they will get a very unexpected ClassCastException.

Java Generics: How does method inference work when wildcard is being used in the method parameters?

Supposedy i have the following:
class x {
public static void main(String [] args) {
List <?> a = new LinkedList<Object>();
List <? extends Object> b = new LinkedList<Object>();
List <? super Object> c = new LinkedList<Object>();
abc(a, "Hello"); // (1) Error
abc(b, "Hello"); // (2) Error
abc(c, "Hello"); // (3) ok
def(b); // (4) ok
// Showing inference at work
Integer[] a = {10, 20, 30}; // (5)
T is inferred to be ? extends Object
Method signature: ppp(? extends Object, ? extends Object[])
Method call signature: ppp(String, Integer[]);
ppp("Hello", a); // ok
}
static <T> void abc(List<T> a, T b) {}
static <T> void def(List<T> a) {}
static <T> void ppp(T t1, T[] t2){}
}
To begin with, look at clause 5 showing inference at work. Now clause 5 section is a working section.
If that is what it is, then why does clause (1) & (2) have errors?
From my view, all these 3 methods calling have the same inference generated since no actual type parameters is used on the abc method call.
method parameter <T> abc (List <T> a, T b>)
inferred <Object> abc (List <Object>, Object) // (4)
Please bear in mind, method abc() and def() is my method. Compiler doesn't know what i want to do with the List in this method. I might just print the list size or might not even do anything at all as shown above. So there is no get or set involved here.
CONTINUATION -->
This is getting very confusing for me.
class y {
public static void main(String [] args) {
List <Integer> a = new LinkedList<Integer>();
List <Object> b = new LinkedList<Object>();
ppp("Hello", new Integer(1)); // (20) ok
qqq("Hello", a); // (21) error
qqq("Hello", b); // (22) ok
}
static <T> void ppp(T t1, T t2) {}
static <T> void qqq(T t1, List <T> t2) {}
}
Note that clause 21 is the same as clause 20 except 2nd parameter is being made to be a List instead of Integer.
Clause 20 is ok cos' T is inferred to be Object.
Clause 22 is ok. Same reason as clause 20.
Clause 21 failed? T could also be inferred to be Object too - would work too?

The hard thing about the wildcard is to realize ? extends Foo does not mean "anything that extends Foo", but instead it means "some specific type that extends Foo". And since you are outside that definition, you have no way to know which specific sub-type of Foo it is.
Update:
As I said, it's complicated. Here are some comments on your code.
// a list of some specific type, and you don't know what type that is.
// it's a sub-type ob Object, yes, which means that you can do
// Object foo = a.get(0); , but the compiler has no way of knowing
// whether it's a String so you can't pass in a String
List <?> a = new LinkedList<Object>();
// same here. '?' and '? extends Object' are equivalent
List <? extends Object> b = new LinkedList<Object>();
// this is a list of Objects and superclasses thereof.
// since there are no superclasses of Object, this is equivalent to
// List<Object>. And through inheritance, a String is an Object, so
// you can pass it in.
List <? super Object> c = new LinkedList<Object>();
Update 2:
The problem here is that you are dealing with fixed, but unresolveable variables.
// you can pass in a List<String> and a String,
// but if you pass in a List<?>, the compiler has no idea what
// '?' is and just can't substitute 'String'.
// 'T' doesn't help you here, because 'T' can't match both
// '?' and 'String'.
static <T> void abc(List<T> a, T b) {}
// this works because 'T' substitutes '?' and doesn't have to worry
// about a 2nd parameter
static <T> void def(List<T> a) {}
Read this question, it might shed some light on the problem:
What is PECS (Producer Extends Consumer Super)?

You've set up a bit of a straw man by creating a LinkedList<Object> in each case. That can make it difficult to see the problem. What you have to remember is that when the compiler gets to those method invocations, it doesn't know that you created a LinkedList<Object>. It could be a LinkedList<Integer>, for example.
So let's look at your code with more interesting initializations:
List<Integer> integers = new LinkedList<Integer>();
List <?> a = integers;
List <? extends Object> b = integers;
List <? super Object> c = new LinkedList<Object>();
//INVALID. T maps to a type that could be Object OR anything else. "Hello"
//would only be type-assignable to T if T represented String, Object, CharSequence,
//Serializable, or Comparable
abc(a, "Hello");
//INVALID. T maps to a type that could be Object OR anything else. "Hello"
//would only be type-assignable to T if T represented String, Object, CharSequence,
//Serializable, or Comparable
abc(b, "Hello");
//VALID. T maps to an unknown super type of Object (which can only be Object itself)
//since String is already type-assignable to Object, it is of course guaranteed to be
//type-assignable to any of Object's super types.
abc(c, "Hello");
Integer i1 = integers.get(0);
Integer i2 = integers.get(1);
It doesn't take much to see that if the implementation of abc was this:
//a perfectly valid implementation
static <T> void abc(List<T> a, T b) {
a.add(b);
}
That you would get a ClassCastException when initializing i1.
From my view, all these 3 methods calling has the following inference generated since no actual type parameters is used on the abc static method call.
method parameter <T> abc (List <T> a, T b>)
inferred <Object> abc (List <Object>, Object) // (4)
This is categorically wrong. It is not inferred that T is Object in any of your examples, not even in the case of ? super Object. T is resolved to the capture of a, and unless you can assign a String to that capture (as is the case when it's ? super Object) you will have a type error.
Edit #1
Regarding your update (I've replaced your generic array with a List<T> since generic arrays needlessly cloud the issue):
// Showing inference at work
List<Integer> a = Arrays.asList(10, 20, 30); // (5)
T is inferred to be ? extends Object
Method signature: ppp(? extends Object, List<? extends Object>)
Method call signature: ppp(String, List<Integer>);
ppp("Hello", a); // ok
This is not correct. The crucial mistake you're making is here:
Method signature: ppp(? extends Object, List<? extends Object>)
This is not at all what the capture engine does or should translate your invocation into. It resolves T as <? extends Object> but as one specific capture of <? extends Object>. Let's call it capture-1-of<? extends Object>. Thus your method must be like this:
Method signature: ppp(capture-1-of<? extends Object>, List<capture-1-of<? extends Object>>)
This means that there is a binding between the two parameters...they must resolve to the same capture. In general it is very difficult to tell the compiler that two things are the same capture. In fact, even this is not a valid invocation of ppp (even though they are clearly the same capture):
List<? extends Integer> myList;
ppp(myList.get(0), myList);
One way we could invoke ppp is through a generic intermediary:
public static <T> void pppCaller(List<T> items) {
ppp(items.get(0), items);
}
pppCaller(myList);
The only sure-fire way you could invoke ppp with a wildcarded list would be to invoke it like this:
List<? extends Integer> myList = new ArrayList<Integer>();
ppp(null, myList);
That's because the null is the only thing that you can assign to anything. On the other hand, if you had this method:
private static <T> void qqq(T item1, T item2) {}
You could indeed invoke it like this:
List<? extends Integer> myList;
qqq(myList.get(0), myList.get(1));
Because in this case, the inference can generalize T to Object. Since List<? extends Integer> is not covariant with List<Object>, it cannot do the same for ppp().
However, what most people do to get around this is to relax their method signature. Instead, declare ppp as the following:
public static <T> ppp(T item, List<? super T> items) {
}
This follows the guidelines that Sean put in his post of "PECS"
If (your method) produces, use extends, if it consumes, use super.
Edit #2
Regarding your latest edit:
public static void main(String [] args) {
List <Integer> a = new LinkedList<Integer>();
qqq("Hello", a); // (21) error
}
static <T> void qqq(T t1, List <T> t2) {}
Object is not a valid inference for T. I think this is something fundamental you're missing, so I'll say it clear:
A List<Integer> is NOT type-assignable to List<Object>
Not at all. If it were, you could do something like this which obviously violates type safety:
List<Integer> myInts = new ArrayList<Integer>();
List<Object> myObjects = myInts; //doesn't compile!
myObjects.add("someString");
Integer firstInt = myInts.get(0); //ClassCastException!
So T cannot be inferred as Object, since it would require assigning a List<Integer> to a variable of type List<Object>.

A wildcard would then needed to induce subtype covariance
I'd rather say "try to simulate" since even after using wild-cards you can't get the same functionality you get for arrays.
Then the question is why clause (3) works and not clause(2) or (1)?
Consider the first declaration:
List <?> a = new LinkedList<Object>();
This declaration effectively says, I really don't know (or care) what kind of element the collection a contains. This effectively shuts you off from "mutating" the collection since you might end up adding elements of type which are not really compatible with a. You can have List<?> a = new ArrayList<String>() but you still won't be able to put anything in it. Basically, in case an add is allowed, the compiler can't guarantee the type safety of the collection.
List <? extends Object> b = new LinkedList<Object>();
Here you say b is a collection which contains elements which extend an Object. What kind of element, you don't know. This again as per the previous discussion doesn't allow you to add anything since you could end up compromising type safety.
List <? super Object> c = new LinkedList<Object>();
Here you say, c is a collection which contains elements of type Object and it's super-classes or in other words, at least an Object. Since each reference type in Java is assignment compatible with Object, it works in the third case.

Integer[] is a subtype of Object[], however List<Integer> is not a subtype of List<Object>. This is quite confusing; arrays are more primitive and should be avoided.
If a method parameter type is T[], it accepts all S[] where S is a subtype of T.
To do this with List, the type should be List<? extends T>. It accepts all List<S> where S is a subtype of T.

List<?> a means that a holds a specific type, which is unknown. Consider this more complete example:
List<Float> floats = new ArrayList<Float>();
List<?> a = floats; /* Alias "floats" as "a" */
abc(a, "Hello"); /* This won't compile... */
float f = floats.get(0); /* .. if it did, you'd get a ClassCastException */
static <T> abc(List<T> a, T b) {
a.add(b); /* Absolutely no problem here. */
}
List<? extends Object> means essentially the same thing as List<?>, and would cause the same error.
List<? super Object> means that the list holds a specific, but unknown super-type of Object, and the parameterized method can accept any object that is-a Object for the second parameter. While the method invocation is type-safe, attempting to assign an unsafe value to c will cause an error:
List<Number> numbers = new ArrayList<Number>();
List<? super Object> a = numbers; /* This won't compile... */
abc(a, "Hello");
Number n = numbers.get(0); /* ...if it did, you'd get a ClassCastException */

Collection in Java

Quick Question...
Can collections in Java hold more than one type? Or do they all have to be the same type?
thanks

Simple answer
Yes.
More detailed answer
You can either use generic collection, without <T> value, for example:
ArrayList a = new ArrayList();
a.add(2);
a.add("String");
Using collections without <T> is a bad habit and most IDEs / compilers give a warning here. You can circumvent it by using a collection of Object, i.e.:
ArrayList<Object> a = new ArrayList<Object>();
Or you can find some common interface or supertype that these element must have in, for example ArrayList<Number> - and you can store various objects that have common Number superclass, i.e. BigDecimal, BigInteger, Byte, Double, Float, Integer, Long, Short:
ArrayList<Number> a = new ArrayList<Number>();
a.add(2); // integer
a.add(42L); // long
a.add(123.45d); // double
System.out.println(a.toString()); // => [2, 42, 123.45]
Note that it essentially means that a elements are of Number class — i.e. you can't ask to execute subclass-specific methods (for example, Double#isInfinite(), which doesn't exist in Number superclass), although you can typecast in run-time if you somehow know it's safe to typecast:
a.get(2).isInfinite() // compile-time error
((Double) a.get(2)).isInfinite() // => false
((Double) a.get(1)).isInfinite() // run-time error (ClassCastException)
Run-time typecasting is also generally frowned upon, as it effectively circumvents proper compile-time type safety.
Also note that it's impossible to assign (or use) ArrayList<Number> in place of ArrayList<Integer> and vice-versa, i.e. this will fail to compile:
public void printNumbers(ArrayList<Number> list) {
list.forEach(System.out::println);
}
ArrayList<Integer> a = new ArrayList<Integer>();
printNumbers(a); // "incompatible types"
as well as this:
public void printIntegers(ArrayList<Integer> list) {
list.forEach(System.out::println);
}
ArrayList<Number> a = new ArrayList<Number>();
printIntegers(a); // "incompatible types"
To declare a variable to be able to accept both ArrayList<Number> or any of its subclasses, one can use ArrayList<? extends Number> or ArrayList<? super Number> syntax. extends is generally used when you're going to consume (i.e. read) from the object in your method, super is used when you're going to produce (i.e. write). Given that printout is consuming, it's safe to use extends:
public void printNumbers(ArrayList<? extends Number> list) {
list.forEach(System.out::println);
}
ArrayList<Integer> listInt = new ArrayList<Integer>();
printNumbers(listInt); // works
ArrayList<Double> listDbl = new ArrayList<Double>();
printNumbers(listDbl); // also works
There is a good answer in
Difference between <? super T> and <? extends T> in Java for more in-depth explanation.

If you want them to hold any more than one type, use Collection<Object>. However, you won't know what you're getting without doing some if (x instanceof MyType) calls, which are rather inefficient.

They have to be of the same Supertype. So if you have objects of type A, then a Collection<A> can store objects of type A and of every subtype of A.
If you want to allow arbitrary types, then use Collection<Object>, otherwise take the most general appropriate super-class.
However, you will then have to manually cast from the most general type (Object) to the specific type you have in mind. You can use the typeof operator to find out what the type is.

Every Collection classes can contains heterogeneous objects except TreeSet and TreeMap. Since TreeSet and TreeMap stores elements according to some sorting order. so, if objects are of different type it will not be able to sort it because comparison between the objects will not be possible for sorting.

Yes they can but they should not (that's why generics have been put in place since 5th version of jdk) in general store different types, as this is the straight way to errors.

Yes collections in java can hold more than one type as below. But it will throw an exception if done using the following way.
ArrayList al = new ArrayList();
al.add(1);
al.add("name");
al.add(1.2f);
Iterator itr =al.iterator();
while(itr.hasNext())
{
System.out.println(itr.next());
}
Hence it's better to mention the type that you're using. To get rid of the exception the above program can be modified as below.
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(1);
al.add(2);
al.add(3);
Iterator itr =al.iterator();
while(itr.hasNext())
{
System.out.println(itr.next());
}
ArrayList<String> al1 = new ArrayList<String>();
al1.add("Words");
al1.add("Names");
al1.add("Characters");
Iterator itr1 =al1.iterator();
while(itr1.hasNext())
{
System.out.println(itr1.next());
}
You can also use more than these types.

Yes,
My mistake the correct code is this one and
ArrayList<Elements>()=new ArrayList();
or
ArrayList<E>()=new ArrayList();
should be the correct declaration if you want to use Generics in Collection.
class Test
{
public static void main(String[] args)
{
// For Generic class of List
ArrayList<E> arrL1 = new ArrayList<E>();
arrL1.add("stackoverflow");
arrL1.add(1);
Iterator itr1=list.iterator();
while(itr1.hasNext())
{
System.out.println(itr1.next());
}
// for Particular datatype in List
ArrayList<String> list=new ArrayList<String>(); // Creating arraylist
list.add("Ravi"); // Adding object in arraylist
list.add("Vijay");
list.add("Ravi");
list.add("Ajay");
// transversing the values
Iterator itr=list.iterator();
while(itr.hasNext())
{
System.out.println(itr.next());
}
}
}
Output 1
stackoverflow
1
Output 2
Ravi
Vijay
Ravi
Ajay

I believe you can also use Collection<?>.

Yes, you can have more than one datatype in ArrayList of Collection.
class Test
{
public static void main(String[] args)
{
// For Generic class of List
ArrayList<> arrL1 = new ArrayList<>();
arrL1.add("stackoverflow");
arrL1.add(1);
// for Particular datatype in List
ArrayList<String> list=new ArrayList<String>(); // Creating arraylist
list.add("Ravi"); // Adding object in arraylist
list.add("Vijay");
list.add("Ravi");
list.add("Ajay");
// transversing the values
Iterator itr=list.iterator();
while(itr.hasNext())
{
System.out.println(itr.next());
}
}
}
Output 1:
stackoverflow
1
Output 2:
Ravi
Vijay
Ravi
Ajay