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Difference between Bounded Type parameter (T extends) and Upper Bound Wildcard (? extends)

I know that there was a similar question already posted, although I think mine is somewhat different...
Suppose you have two methods:
// Bounded type parameter
private static <T extends Number> void processList(List<T> someList) {
}
// Upper bound wildcard
private static void processList2(List<? extends Number> someList) {
// ...
}
As far as I know, both methods accepts arguments, that are List of type Number or List of subtype of Number.
But what's the difference between the two methods after all?

There are several differences between the two syntaxes during compile time :
With the first syntax, you can add elements to someList but with the second, you can't. This is commonly known as PECS and less commonly known as the PUT and GET prinicple.
With the first syntax, you have a handle to the type parameter T so you can use it to do things such as define local variables within the method of type T, cast a reference to the type T, call methods that are available in the class represented by T, etc. But with the second syntax, you don't have a handle to the type so you can't do any of this.
The first method can actually be called from the second method to
capture the wildcard. This is the most common way to capture a
wildcard via a helper method.
private static <T extends Number> void processList(List<T> someList) {
T n = someList.get(0);
someList.add(1,n); //addition allowed.
}
private static void processList2(List<? extends Number> someList) {
Number n = someList.get(0);
//someList.add(1,n);//Compilation error. Addition not allowed.
processList(someList);//Helper method for capturing the wildcard
}
Note that since generics are compile time sugar, these differences at a broader level are only limited to the compilation.

I can think of the below differences :
a) Modifying your list inside the method, consider the below code :
// Bounded type parameter
private static <T extends Number> void processList(List<T> someList)
{
T t = someList.get(0);
if ( t.getClass() == Integer.class )
{
Integer myNum = new Integer(4);
someList.add((T) myNum);
}
}
// Upper bound wildcard
private static void processList2(List<? extends Number> someList)
{
Object o = someList.get(0);
if ( o instanceof Integer )
{
Integer myNum = new Integer(4);
someList.add(myNum); // Compile time error !!
}
}
With wildcard, you cannot add elements to the list! The compiler tells you that it doesn't know what is myNum. But in the first method, you could add an Integer by first checking if T is Integer, with no compile time error.
b) The first method is called generic method. It follows the syntax that is defined for a generic method.
The upper bounds specified in the method definition are used to restrict the parameter types.
The second one is NOT necessarily called a generic method, it is a normal method that happens to accept a generic parameter.
The wildcard ? with extends keyword is used as a means of relaxing the types that the method can accept.

The difference is on the compiler side.
On the first one you can use the type (to cast something or use it as a bound to call another method for example) while on the second one, you cannot use it.

If you want to use the type information then go with bounded. With the wildcard, the argument will appear as a generic Object and you won't be able to call methods based on that type.
public static <T extends Object> ListIterator<T> createListIterator(ListIterator<T> o)
{
return new ListIteratorAdaptor<T>(o);
}
https://docs.oracle.com/javase/tutorial/java/generics/bounded.html

There are following three types of Wildcard usually used with Generic in JAVA. Each one is explained as below with example.
Upper-bounded Wildcard:
? extends T : In Upper bounded wildcard only T or its subtypes will be supported.
For example we have an Animal class and have Dog , Cat as its subtypes. So following generic methods will only
accept parameters of type Data<Animal>, Data<Dog> and Data<Cat>
public static void add(Data<? extends Animal> animalData) {
}
Lower-bounded Wildcard:
? super T : In Lower-bounded wildcard only T or its super types will be supported.
Same example we used for defining Lower-bounded Wildcard. Lets say we have Animal class as super or parent class
and Dog as its child class. Now below method use Lower-bounded Wildcard and will only accept parameters of type
Data<Animal>, Data<Dog> and Data<Object>
public static void add(Data<? super Dog> animalData) {
}
Unbounded Wildcard:
? : Unbounded wildcard supports all types. So our above example method can take parameters of type
Data<Animal>, Data<Dog> , Data<Object> and Data<Cat>
public static void add(Data<?> animalData) {
}

Differences between these generic statements [duplicate]

I'm a newbie in Generic and my question is: what difference between two functions:
function 1:
public static <E> void funct1 (List<E> list1) {
}
function 2:
public static void funct2(List<?> list) {
}

The first signature says: list1 is a List of Es.
The second signature says: list is a List of instances of some type, but we don't know the type.
The difference becomes obvious when we try to change the method so it takes a second argument, which should be added to the list inside the method:
import java.util.List;
public class Experiment {
public static <E> void funct1(final List<E> list1, final E something) {
list1.add(something);
}
public static void funct2(final List<?> list, final Object something) {
list.add(something); // does not compile
}
}
The first one works nicely. And you can't change the second argument into anything that will actually compile.
Actually I just found an even nicer demonstration of the difference:
public class Experiment {
public static <E> void funct1(final List<E> list) {
list.add(list.get(0));
}
public static void funct2(final List<?> list) {
list.add(list.get(0)); // !!!!!!!!!!!!!! won't compile !!!!!!!!!
}
}
One might as why do we need <?> when it only restricts what we can do with it (as #Babu_Reddy_H did in the comments). I see the following benefits of the wildcard version:
The caller has to know less about the object he passes in. For example if I have a Map of Lists: Map<String, List<?>> I can pass its values to your function without specifying the type of the list elements. So
If I hand out objects parameterized like this I actively limit what people know about these objects and what they can do with it (as long as they stay away from unsafe casting).
These two make sense when I combine them: List<? extends T>. For example consider a method List<T> merge(List<? extends T>, List<? extends T>), which merges the two input lists to a new result list. Sure you could introduce two more type parameters, but why would you want to? It would be over specifying things.
finally wildcards can have lower bounds, so with lists you can make the add method work, while get doesn't give you anything useful. Of course that triggers the next question: why don't generics have lower bounds?
For a more in depth answer see: When to use generic methods and when to use wild-card? and http://www.angelikalanger.com/GenericsFAQ/FAQSections/TypeArguments.html#FAQ203

Generics makes the collection more type safe.
List<E> : E here is the Type Parameter, which can be used to determine the content type of the list, but there was No way to check what was the content during the runtime.
Generics are checked only during compilation time.
<? extends String> : This was specially build into java, to handle the problem which was with the Type Parameter. "? extends String" means this List can have
objects which IS-A String.
For eg:
Animal class
Dog class extends Animal
Tiger class extends Animal
So using "public void go(ArrayList<Animal> a)" will NOT accept Dog or Tiger as its content but Animal.
"public void go(ArrayList<? extends Animal> a)" is whats needed to make the ArrayList take in Dog and Tiger type.
Check for references in Head First Java.

List<E> as a parameter type says that the parameter must be a list of items with any object type. Moreover, you can bind the E parameter to declare references to list items inside the function body or as other parameter types.
The List<?> as a parameter type has the same semantics, except that there is no way to declare references to the items in the list other than to use Object. Other posts give additional subtle differences.

The first is a function that accepts a parameter that must be a list of items of E type.
the second example type is not defined
List<?> list
so you can pass list of any type of objects.

I usually explain the difference between <E> and <?> by a comparison with logical quantifications, that is, universal quantification and existential quantification.
corresponds to "forall E, ..."
corresponds to "there exists something(denoted by ) such that ...."
Therefore, the following generic method declaration means that, for all class type E, we define funct1
public static <E> void funct1 (List<E>; list1) {
}
The following generic method declaration means that, for some existing class denoted by <?>, we define funct2.
public static void funct2(List<?> list) {
}

(Since your edit) Those two function signatures have the same effect to outside code -- they both take any List as argument. A wildcard is equivalent to a type parameter that is used only once.

In addition to those differences mentioned before, there is also an additional difference: You can explicitly set the type arguments for the call of the generic method:
List<Apple> apples = ...
ClassName.<Banana>funct2(apples); // for some reason the compiler seems to be ok
// with type parameters, even though the method has none
ClassName.<Banana>funct1(apples); // compiler error: incompatible types: List<Apple>
// cannot be converted to List<Banana>
(ClassName is the name of the class containing the methods.)

In this context, both wild card (?) and type parameter (E) will do the same for you. There are certain edges based on the use cases.
Let's say if you want to have a method which may have more than one params like:
public void function1(ArrayList<?> a, ArrayList<?> b){
// some process
}
public <T> void function2(ArrayList<T> a, ArrayList<T> b){
// some process
}
in function1 a can be AL of String and b can be AL of the Integer so it is not possible to control the type of both the params but this is easy for the function2.
We should use the Type Params (function 2) if we want to use the type later in the method or class
There are some features in WildCard and Type param:
WildCard(?)
It support the upper and lower bound in the type while the Type param (E) supports only upper bound.
Type Param(E)
SomeTime we do not need to pass the actual type ex:
ArrayList<Integer> ai = new ArrayList<Integer>();
ArrayList<Double> ad = new ArrayList<Double>();
function2(ai, ad);
//It will compile and the T will be Number.
In this case, the compiler infers the type argument for us based on the type of actual arguments

Benefits of using ? instead of T for type parameter in Java generics? [duplicate]

I am reading about generic methods from OracleDocGenericMethod. I am pretty confused about the comparison when it says when to use wild-card and when to use generic methods.
Quoting from the document.
interface Collection<E> {
public boolean containsAll(Collection<?> c);
public boolean addAll(Collection<? extends E> c);
}
We could have used generic methods here instead:
interface Collection<E> {
public <T> boolean containsAll(Collection<T> c);
public <T extends E> boolean addAll(Collection<T> c);
// Hey, type variables can have bounds too!
}
[…]
This tells us that the type argument is being used for polymorphism;
its only effect is to allow a variety of actual argument types to be
used at different invocation sites. If that is the case, one should
use wildcards. Wildcards are designed to support flexible subtyping,
which is what we're trying to express here.
Don't we think wild card like (Collection<? extends E> c); is also supporting kind of
polymorphism? Then why generic method usage is considered not good in this?
Continuing ahead, it states,
Generic methods allow type parameters to be used to express
dependencies among the types of one or more arguments to a method
and/or its return type. If there isn't such a dependency, a generic
method should not be used.
What does this mean?
They have presented the example
class Collections {
public static <T> void copy(List<T> dest, List<? extends T> src) {
...
}
[…]
We could have written the signature for this method another way,
without using wildcards at all:
class Collections {
public static <T, S extends T> void copy(List<T> dest, List<S> src) {
...
}
The document discourages the second declaration and promotes usage of first syntax? What's the difference between the first and second declaration? Both seems to be doing the same thing?
Can someone put light on this area.

There are certain places, where wildcards, and type parameters do the same thing. But there are also certain places, where you have to use type parameters.
If you want to enforce some relationship on the different types of method arguments, you can't do that with wildcards, you have to use type parameters.
Taking your method as example, suppose you want to ensure that the src and dest list passed to copy() method should be of same parameterized type, you can do it with type parameters like so:
public static <T extends Number> void copy(List<T> dest, List<T> src)
Here, you are ensured that both dest and src have same parameterized type for List. So, it's safe to copy elements from src to dest.
But, if you go on to change the method to use wildcard:
public static void copy(List<? extends Number> dest, List<? extends Number> src)
it won't work as expected. In 2nd case, you can pass List<Integer> and List<Float> as dest and src. So, moving elements from src to dest wouldn't be type safe anymore.
If you don't need such kind of relation, then you are free not to use type parameters at all.
Some other difference between using wildcards and type parameters are:
If you have only one parameterized type argument, then you can use wildcard, although type parameter will also work.
Type parameters support multiple bounds, wildcards don't.
Wildcards support both upper and lower bounds, type parameters just support upper bounds. So, if you want to define a method that takes a List of type Integer or it's super class, you can do:
public void print(List<? super Integer> list) // OK
but you can't use type parameter:
public <T super Integer> void print(List<T> list) // Won't compile
References:
Angelika Langer's Java Generics FAQs

Consider following example from The Java Programming by James Gosling 4th edition below where we want to merge 2 SinglyLinkQueue:
public static <T1, T2 extends T1> void merge(SinglyLinkQueue<T1> d, SinglyLinkQueue<T2> s){
// merge s element into d
}
public static <T> void merge(SinglyLinkQueue<T> d, SinglyLinkQueue<? extends T> s){
// merge s element into d
}
Both of the above methods have the same functionality. So which is preferable? Answer is 2nd one. In the author's own words :
"The general rule is to use wildcards when you can because code with wildcards
is generally more readable than code with multiple type parameters. When deciding if you need a type
variable, ask yourself if that type variable is used to relate two or more parameters, or to relate a parameter
type with the return type. If the answer is no, then a wildcard should suffice."
Note: In book only second method is given and type parameter name is S instead of 'T'. First method is not there in the book.

In your first question: It means that if there is a relation between the parameter's type and the method's return type then use a generic.
For example:
public <T> T giveMeMaximum(Collection<T> items);
public <T> Collection<T> applyFilter(Collection<T> items);
Here you are extracting some of the T following a certain criteria. If T is Long your methods will return Long and Collection<Long>; the actual return type is dependent on the parameter type, thus it is useful, and advised, to use generic types.
When this is not the case you can use wild card types:
public int count(Collection<?> items);
public boolean containsDuplicate(Collection<?> items);
In this two example whatever the type of the items in the collections the return types will be int and boolean.
In your examples:
interface Collection<E> {
public boolean containsAll(Collection<?> c);
public boolean addAll(Collection<? extends E> c);
}
those two functions will return a boolean whatever is the types of the items in the collections. In the second case it is limited to instances of a subclass of E.
Second question:
class Collections {
public static <T> void copy(List<T> dest, List<? extends T> src) {
...
}
This first code allow you to pass an heterogeneous List<? extends T> src as a parameter. This list can contain multiple elements of different classes as long as they all extends the base class T.
if you had:
interface Fruit{}
and
class Apple implements Fruit{}
class Pear implements Fruit{}
class Tomato implements Fruit{}
you could do
List<? extends Fruit> basket = new ArrayList<? extends Fruit>();
basket.add(new Apple());
basket.add(new Pear());
basket.add(new Tomato());
List<Fruit> fridge = new ArrayList<Fruit>();
Collections.copy(fridge, basket);// works
On the other hand
class Collections {
public static <T, S extends T> void copy(List<T> dest, List<S> src) {
...
}
constrain List<S> src to be of one particular class S that is a subclass of T. The list can only contain elements of one class (in this instance S) and no other class, even if they implement T too. You wouldn't be able to use my previous example but you could do:
List<Apple> basket = new ArrayList<Apple>();
basket.add(new Apple());
basket.add(new Apple());
basket.add(new Apple());
List<Fruit> fridge = new ArrayList<Fruit>();
Collections.copy(fridge, basket); /* works since the basket is defined as a List of apples and not a list of some fruits. */

Wildcard method is also generic - you could call it with some range of types.
The <T> syntax defines a type variable name. If a type variable has any use (e.g. in method implementation or as a constraint for other type), then it makes sense to name it, otherwise you could use ?, as anonymous variable. So, looks like just a short-cut.
Moreover, the ? syntax is not avoidable when you declare a field:
class NumberContainer
{
Set<? extends Number> numbers;
}

I will try and answer your question, one by one.
Don't we think wild card like (Collection<? extends E> c); is also
supporting kind of polymorphism?
No. The reason is that the bounded wildcard has no defined parameter type. It is an unknown. All it "knows" is that the "containment" is of a type E (whatever defined). So, it cannot verify and justify whether the value provided matches the bounded type.
So, it's no sensible to have polymorphic behaviours on wildcards.
The document discourages the second declaration and promotes usage of
first syntax? What's the difference between the first and second
declaration? Both seems to be doing the same thing?
The first option is better in this case as T is always bounded, and source will definitely have values (of unknowns) that subclasses T.
So, suppose that you want to copy all list of numbers, the first option will be
Collections.copy(List<Number> dest, List<? extends Number> src);
src, essentially, can accept List<Double>, List<Float>, etc. as there is an upper bound to the parameterized type found in dest.
The 2nd option will force you to bind S for every type you want to copy, like so
//For double
Collections.copy(List<Number> dest, List<Double> src); //Double extends Number.
//For int
Collections.copy(List<Number> dest, List<Integer> src); //Integer extends Number.
As S is a parameterized type that needs binding.
I hope this helps.

One other difference which is not listed here.
static <T> void fromArrayToCollection(T[] a, Collection<T> c) {
for (T o : a) {
c.add(o); // correct
}
}
But the following will result in compile time error.
static <T> void fromArrayToCollection(T[] a, Collection<?> c) {
for (T o : a) {
c.add(o); // compile time error
}
}

? means unknown
The general rule applies:
You can read from it, but not write
given simple pojo Car
class Car {
void display(){
}
}
This will compile
private static <T extends Car> void addExtractedAgain1(List<T> cars) {
T t = cars.get(1);
t.display();
cars.add(t);
}
This method won't compile
private static void addExtractedAgain2(List<? extends Car> cars) {
Car car = cars.get(1);
car.display();
cars.add(car); // will not compile
}
Another example
List<?> hi = Arrays.asList("Hi", new Exception(), 0);
hi.forEach(o -> {
o.toString() // it's ok to call Object methods and methods that don't need the contained type
});
hi.add(...) // nothing can be add here won't compile, we need to tell compiler what the data type is but we do not know

As far as I understand, there is only one use case when wildcard is strictly needed (i.e. can express something that you can not express using explicit type parameters). This is when you need to specify a lower bound.
Apart from that however wildcards serve to write more concise code, as described by the following statements in the document you mention:
Generic methods allow type parameters to be used to express
dependencies among the types of one or more arguments to a method
and/or its return type. If there isn't such a dependency, a generic
method should not be used.
[...]
Using wildcards is clearer and more concise than declaring explicit
type parameters, and should therefore be preferred whenever possible.
[...]
Wildcards also have the advantage that they can be used outside of
method signatures, as the types of fields, local variables and arrays.

Mainly -> Wildcards enforce generics at the parameter/argument level of a Non-Generic method.
Note. It can also be performed in genericMethod by default, but here instead of ? we can use T itself.
package generics;
public class DemoWildCard {
public static void main(String[] args) {
DemoWildCard obj = new DemoWildCard();
obj.display(new Person<Integer>());
obj.display(new Person<String>());
}
void display(Person<?> person) {
//allows person of Integer,String or anything
//This cannnot be done if we use T, because in that case we have to make this method itself generic
System.out.println(person);
}
}
class Person<T>{
}
SO wildcard has its specific usecases like this.

Generic extend can't add the same type

I'm new to these generic types. In the below code, I created a method that accepts a List of items that extends "String".
My Question? - When the list can be assigned with a new list that is , why can't a string "test" can be added...It gives me a compilation error.
public class Child {
public void takeList(List<? extends String> list){
list = new ArrayList<String>();
list.add("test");
}
}

Because it's not the runtime type that's relevant here. list is still of type List<? extends String>, you've just happened to assign it to a new ArrayList<String>(). Consider this:
list = rand() ? new ArrayList<String>() : new ArrayList<NotString>();
The compiler could not possibly tell if list.add("test") will be valid -- it only makes decisions based on the compile-time type of list.
Note that in reality nothing extends String, it's a final class.

There's a subtle difference. It takes a list that contains one type of thing (a thing that extends string). This list may be a subclass of String and therefore not be a String iyswim. See http://docs.oracle.com/javase/tutorial/java/generics/upperBounded.html Upper bounded wildcards.
If it was
public void takeList(List<? *super* String> list){
Then you could add strings to it, because the list is guaranteed to be able to accept Strings.

When you have a variable with a wildcard, and a method that takes the generic type parameter, Java cannot ensure type safety. It must disallow this call.
Consider a List<? extends Animal> for example. You may have assigned it List<Dog>, but the variable could be assigned a List<Squid>. You shouldn't be allowed to add a Dog to such a list.
To allow the add method to be called, remove the wildcard.
public void takeList(List<String> list){
Besides, String is final, so there really is no point to saying ? extends String.

Why is there an extra <E> in this generic method?

I learned java generics some time ago, but now I'm learning collections and found some code that I don't understand. Here is the code:
static <E> List<E> nCopies(int n, E value)
It is from class java.util.Collections.
My question is why there is:
<E> List<E>
and not only
List<E>
Obviously I am missing something, can someone clarify this for me?

You use the <E> to typify the method you are defining.
The most common example of generics is to have a typified class like this:
public class SomeClass<E> {
...
}
Then, when you are creating a new object of that class you define the type directly like this:
new SomeClass<String>();
That way any method in that class that refers to <E>, will treat <E> as a String, for that instance.
Now consider a static method (which is not bound to any particular instance of a class), in order to typify that method you have use another kind of typification which applies to methods, like this:
static <E> List<E> nCopies(int n, E value)
You use the <E> before the return type to say "this particular method will consider some E when it executes". What <E> will be is decided when you invoke the method:
nCopies(3, "a");
In this example <E> will be a String, so the return type will be a List<String>.
Finally, you can even mix them both:
public class SomeClass<E> {
public <F> void doSomething(E e, F f) {
...
}
}
In this case, if you have an instance of SomeClass, the E in the doSomething method will always be String (for that instance), but the F can be anything you want it to be.

In <E> List<E>, the first <E> denotes that E is a type parameter. If you hadn't specified it, then Java would think the E in E value referred to an actual class named E, and ask you to import it. See generic methods.

The <E> is required to tell the compiler that you intend to use E as a type parameter, the same way you do when you make a generic class (e.g. public interface List<E>).
Since there is no rule (only conventions) on interface or class names being more than one character, and no rule (only conventions) that type parameter names have to be one character, the compiler would not know you intended it to be a type parameter rather than a concrete class name.
Edit
A lot of people have been saying this is directly related to static methods. That is not true. You can have an instance method that is generic on its own type parameters as well (though typically, the type parameters will be related to the class type parameters).
Here's an example of where you could have this:
public class MyList<E> {
public <N super E> MyList<N> createCopy() {
//...
}
}
This method would allow you to create a copy of the list but not restrain you to using the same type as the list you have, but rather allowing you to use a supertype. For example:
MyList<Integer> integers = createList(1, 2, 5);
MyList<Number> numbers = integers.createCopy();

List<E> is the return type for the method whereas <E> is the type being passed in (This is inferred by the compiler from what is being passed as E value).
static <E> List<E> someMethod(E myObject)
{
E objectOfMyType = myObject;
List<E> myList = new ArrayList<E>();
...
return myList;
}
This would be called as:
MyObject o = new MyObject();
List<MyObject> myList = SomeClass.someMethod(o);
IMHO the syntax for methods is kinda goofy, but there you have it. The relavent Oracle tutorial is here:
http://download.oracle.com/javase/tutorial/extra/generics/methods.html

in simple words: to indicate that E is not a class. List<E> would be a valid return type if E was a class (or interface) - despite not recommended, a class can be named with a single letter (as a type variable also could be named with more letters, even using an existing class name, to confuse anyone: static <Integer> List<Integer> method() {...}).