I am studying "Java in a Nutshell" 5th Edition. In p169, it says "Remember,
though, that type variables exist only at compile time, so you can’t use a type variable
with the runtime operators instanceof and new.". If I understand it correctly (Am I??), new and instanceof should not be used with type variables. But after I did some tests, I became more confused.
List< ?> l = new ArrayList< ?>(); // not compiled and I may know why
List< E> l = new ArrayList< E>(); // compiled!! WHY??
Could anyone tell me why (2) is allowed??
Really thanks for any information.
The statement "you can't use type variable with new" means you can't do this:
E e = new E(); // can't do this
But you can pass a type variable through to another class for it to use:
class MyClass<E> {
List<E> = new ArrayList<E>(); // OK
}
You're not instantiating E, you're instantiating an ArrayList with the type E.
Regarding using ?, that's a different issue. The ? means "unknown" type, while a variable can be of an unknown type, you can't instantiate a generic class by specifying unknown type - you must pass a known type through to the generic class.
You can however code this:
List<?> list = new ArrayList<E>();
Unless you've defined it, E isn't a class.
E is just a way to make reference to a class during compilation. That way, proper methods can be called with it.
List<String> list1 = new ArrayList<String>(); // Diamond types for Java 7+
List<Integer> list2 = new ArrayList<Integer>();
// Populate lists
list1.get(0).charAt(0); // Perfectly fine, String#charAt(int) exists
list2.get(0).charAt(0); // Error! Integer doesn't have a charAt(int) method!
So, the type E inside the ArrayList class is used to ensure that you can perform operations on the elements if you know the type, as if it were that type. This avoids running into issues when calling on methods that only exist in the specific class, as I showed with the charAt(int) method.
Adding onto this, it can be pointed out why you can't make the call with the ?.
List<?> list1 = new ArrayList<String>(); // Fine
List<?> list2 = new ArrayList<?>(); // What type is this list composed of?
It doesn't make sense to make a list of an unknown type. If you want a list of a generic type, you can always pass the Object into the generics, but there was no reason to add support for creating instances with an unknown type.
Bohemian answered a bit of the "what," but I'd like to take a crack at some of the "why."
Let's start with what you can't do. It all comes down to erasure, which basically means that when your code has a List<String>, the bytecode information only has a List. Anything between the angle brackets is missing from the bytecode (and thus from the runtime).
So, why can't you do foo instanceof E? Well, because that's a runtime check, and E isn't known at runtime.
And why can't you do new E()? Again, because that requires the runtime to know what E is (how else can it call the right constructor? Or even ensure that a no-arg constructor exists?), but that information isn't available to it.
In all these cases, the thing to keep in mind is that the language isn't preventing you from doing something just for the sake of preventing you. It's preventing you from doing it because it'd be impossible to execute at runtime.
So, why can you type new ArrayList<String>()? Because the only information that the runtime needs for that is the fact that an ArrayList is being instantiated, and it has that information. The fact that it's an ArrayList of Strings doesn't matter to the runtime, because erasure means that the ArrayList doesn't know or care about what it's a list of, at runtime.
Related
I believe that the type ? in generics is a specific unknown type. Which means, declaring let's say a list of that type would prevent us from adding any type of object into it.
List<?> unknownList;
unknownList.add(new Object()); // This is an error.
The compiler gives an error as expected.
But when the unknown type is a second level generics, the compiler doesn't seem to care.
class First<T> {}
List<First<?>> firstUnknownList;
// All these three work fine for some reason.
firstUnknownList.add(new First<>());
firstUnknownList.add(new First<Integer>());
firstUnknownList.add(new First<String>());
I thought probably the compiler doesn't care about generic parameter in the second level at all, but it's not the case,
List<First<Integer>> firstIntegerList;
firstIntegerList.add(new First<String>()); // This gives a compiler error as expected.
So, why does the compiler allow us adding any kind of element when only an unknown element (and hence nothing) is acceptable in the second example?
Note: Compiler Java 1.8
You can add anything to a List<T> that you can store in a reference of type T:
T item = ...
List<T> list = new ArrayList<>();
list.add(item);
First<?> is a supertype of First<T>; so you can store a reference to a First<T> in a variable of type First<?>:
First<?> first = new First<String>();
So, substituting T for First<?> above:
First<?> item = new First<String>();
List<First<?>> list = new ArrayList<>();
list.add(item);
All that is happening in OP's example is that the temporary variable item is omitted:
firstUnknownList.add(new First<String>());
However, if you do this with the firstIntegerList example:
First<Integer> item = new First<String>(); // Compiler error.
List<First<Integer>> list = new ArrayList<>();
list.add(item);
it is clear why that's not allowed: you can't make the assignment of item.
It's also possible to see that you can't do anything unsafe with the contents of that list.
If you add a couple of methods to the interface:
interface First<T> {
T producer();
void consumer(T in);
}
Now, consider what you can do with the elements that you added to the list:
for (First<?> first : firstUnknownList) {
// OK at compile time; OK at runtime unless the method throws an exception.
Object obj = first.producer();
// OK at compile time; may fail at runtime if null is not an acceptable parameter.
first.consumer(null);
// Compiler error - you can't have a reference to a ?.
first.consumer(/* some maybe non-null value */);
}
so there isn't actually anything that you can really do with elements of that list that would violate type safety (provided you don't do anything willful to violate it, like using raw types). You can demonstrate that generic producer/consumer methods are similarly safe or forbidden by the compiler.
So there is no reason not to allow you to do this.
It's all about subtype/supertype-relationships.
List<?> is a list that contains elements of an unknown (but specific) type. You never know which type exactly is contained in this list. So you may not add objects to it, because they may be of the wrong type:
List<Integer> ints = new ArrayList<Integer>();
List<?> unknowns = ints;
// It this worked, the list would contain a String....
unknowns.add("String");
// ... and this would crash with some ClassCastException
Integer i = ints.get(0);
It may also be clear that you can do
List<Number> numbers = null;
Integer integer = null;
numbers.add(integer);
This works because Number is a true supertype of Integer. It does not violate the type safety to add an object of a more specific type to a list.
The key point regarding the second example is:
First<?> is a supertype of every First<T>
You can always to something like
First<Integer> fInt = null;
First<Integer> fString = null;
First<?> f = null;
f = fInt; // works
f = fString; // works
So the reason of why you can add a First<String> to a List<First<?>> is the same as why you can add an Integer to a List<Number>: The element that you want to add is of a true subtype of the elements that are expected in the list.
I'll change First interface to Box interface
Box<?> uknownBox gray box with something in it
Box<Apple> appleBox box with apple
List<Box<Apple>> appleBoxList many boxes with apples
List<Box<?>> uknownBoxList many unknown gray boxes
appleBoxList.add(new Box<Orange>()) - can't add box with oranges to list of apple boxes
unknownBoxList.add(new Box<?>()) - we don't know what's in that gray boxes, adding one more unknown gray box changes nothing
unknownBoxList.add(new Box<Orange>()) - same rules when you add specific boxes
unknownBoxList.add(new Box<Apple>()) - since you are not allowed to 'open' them
unknownBoxList = appleBoxList this does not compile to prevent adding gray (possibly not apple) boxes to apple box list. because of that previous operation is legal.
I believe that the type ? in generics is a specific unknown type.
This is slightly inaccurate. Yes, a wildcard type stands for an unknown type, but may stand for different types at different times:
List<?> list = new ArrayList<String>();
list = new ArrayList<Integer>();
The only invariant is that an expression whose type contains a wildcard will always yield a value whose type conforms to that wildcard. Since every value has a type that is not just a wildcard, one could say that wildcard stands for a (more) "specific" type at any time.
Can someone explained, as detailed as possible, the differences between the following types?
List
List<Object>
List<?>
Let me make this more specific. When would I want to use
// 1
public void CanYouGiveMeAnAnswer(List l) { }
// 2
public void CanYouGiveMeAnAnswer(List<Object> l) { }
// 3
public void CanYouGiveMeAnAnswer(List<?> l) { }
As the other posts have noted, you are asking about a Java feature called generics. In C++, this is called templates. This feature in Java is usually easier to work with than the that found in C++.
Let me answer your questions functionally (if that's not a naughty word for OO discussions).
Before generics, there were concrete classes like Vector.
Vector V = new Vector();
Vectors hold any object you give them.
V.add("This is an element");
V.add(new Integer(2));
v.add(new Hashtable());
They do this by casting all values given to it into an Object (the root of all Java classes). When you attempt to retrieve the values stored in your Vector, you need to cast the value back into the original class (if you want to do anything meaningful with it).
String s = (String) v.get(0);
Integer i = (Integer) v.get(1);
Hashtable h = (Hashtable) v.get(2);
Casting gets old fast. More than that, the compiler complains to you about unchecked casts. The most urgent problem with casting like this is that consumers of your Vector have to know the classes of its values at compile time in order to cast correctly. In cases where the producer of the Vector and the consumer of the same are completely isolated from each other (think RPC messages), this can be a fatal issue.
Enter generics. Generics attempt to create strongly typed classes to do generic operations.
ArrayList<String> aList = new ArrayList<String>();
aList.add("One");
String element = aList.get(0); // no cast needed
System.out.println("Got one: " + element);
The Design Patterns book encourages the reader to think in terms of contracts, not concrete types. There is wisdom (and code re-use) in divorcing variables from their implementing class.
With this in mind, you might think that all implementations List objects should do the same set of things: add(), get(), size(), etc. With a little reflection, you can imagine many implementations of List operations that obey the List contract in various ways (e.g. ArrayList). However, the type of data these objects deal with is orthogonal to the actions performed on them.
Put it all together and you'll see the following kinds of code frequently:
List<String> L = new ArrayList<String>();
You should read that as "L is a kind of List that deals with String objects". When you start dealing with Factory classes, it is critical to deal with contracts rather than specific implementations. Factories produce objects of various types at runtime.
Using generics is pretty easy (most of the time).
One day you may decide you want to implement your own generic class. Perhaps you want to write a new database abstraction interface that elides the differencesbetween various data stores. When you define that generic class, you will use <t> as a placeholder for the kind of object that will be manipulated by the methods.
If you are still confused, use the generic classes for List until you are comfortable. Later, you can dive into the implementation with a bit more confidence. Or you can look at the source code for the various List classes that ship with the JRE. Open source is great that way.
Have a look at the Oracle/Sun docs about generics.
Cheers.
In my own simple terms:
List
Would declare an ordinary collection, can hold any type, and will always return Object.
List<Object>
Will create a list that can hold any type of object, but can only get assigned a another List<Object>
For instance this doesn't work;
List<Object> l = new ArrayList<String>();
Of course you can add anything but only can pull Object.
List<Object> l = new ArrayList<Object>();
l.add( new Employee() );
l.add( new String() );
Object o = l.get( 0 );
Object o2 = l.get( 1 );
Finally
List<?>
Will let you assign any type, including
List <?> l = new ArrayList();
List <?> l2 = new ArrayList<String>();
This would be called collection of unknown and since the common denominator of unknown is Object you will be able to fetch Objects ( a coincidence )
The importance of unknown comes when its used with subclassing:
List<? extends Collection> l = new ArrayList<TreeSet>(); // compiles
List<? extends Collection> l = new ArrayList<String>(); // doesn't,
// because String is not part of *Collection* inheritance tree.
I hope using Collection as the type doesn't create confusion, that was the only tree that came to my mind.
The difference here, is that l is a collection of unknow that belongs to the Collection hierarchy.
To add to the already good answers here:
Method arguments:
List<? extends Foo>
good choice if you don't intend to alter the list, and only care that everything in the list is assignable to type 'Foo'. This way, the caller can pass in a List<FooSubclass> and your method works. Usually the best choice.
List<Foo>
good choice if you intend to add Foo objects to the list in your method. The caller may not pass in a List<FooSubclass>, as you intend to add a Foo to the List.
List<? super Foo>
good choice if you intend to add Foo objects to the list, and it's not important what else is in the list (ie, you are ok getting a List<Object> that contains a 'Dog' that has nothing to do with Foo).
Method return values
just like method arguments, but with the benefits reversed.
List<? extends Foo>
Guarantees that everything in the returned List has type 'Foo'. It might be List<FooSubclass> though. Caller cannot add to the List. This is your go-to choice and the most common case by far.
List<Foo>
Just like List<? extends Foo> but also allows the caller to add to the List. Less common.
List<? super Foo>
allows the caller to add Foo objects to the List, but does not guarantee what will be returned from list.get(0)... it could be anything from Foo to Object. The only guarantee is that this won't be a list of 'Dog' or some other choice that would prevent list.add(foo) from being legal. Very rare use case.
I hope that helps. Good luck!
ps. To sum up... two questions...
do you need to add to the List? Do you care what is in the list?
yes yes - use List<Foo>.
yes no - use List<? super Foo>.
no yes - use <? extends Foo> --- most common.
no no - use <?>.
I refer you to the excellent Java Generics tutorial, and the "advanced" Generics tutorial, both available from Sun Microsystems. Another great resource is the Java Generics and Collections book.
I will try to answer this in detail. Before generics we were having only List (a raw list) and it can hold almost anything we can think of.
List rawList = new ArrayList();
rawList.add("String Item");
rawList.add(new Car("VW"));
rawList.add(new Runnable() {
#Override
public void run() {
// do some work.
}
});
The major problem with the raw list is when we want to get any element out of such list it can only guarantee that it would be Object and for that reason we need to use casting as:
Object item = rawList.get(0); // we get object without casting.
String sameItem = (String) rawList.get(0); // we can use casting which may fail at runtime.
So conclusion is a List can store Object (almost everything is Object in Java) and always returns an Object.
Generics
Now lets talk about generics. Consider the following example:
List<String> stringsList = new ArrayList<>();
stringsList.add("Apple");
stringsList.add("Ball");
stringsList.add(new Car("Fiat")); //error
String stringItem = stringsList.get(0);
In the above case we cannot insert anything other than String in stringsList as Java compiler applies strong type checking to generic code and issues errors if the code violates type safety. And we get error when we try to insert a Car instance in it. Also it eliminates cast as you can check when we invoke get method. Check this link for understanding why we should use generics.
List<Object>
If you read about type erasure then you will understand that List<String>, List<Long>, List<Animal> etc. will be having different static types at compile time but will have same dynamic type List at run time.
If we have List<Object> then it can store only Object in it and almost everything is Object in Java. So we can have:
List<Object> objectList = new ArrayList<Object>();
objectList.add("String Item");
objectList.add(new Car("VW"));
objectList.add(new Runnable() {
#Override
public void run() {
}
});
Object item = objectList.get(0); // we get object without casting as list contains Object
String sameItem = (String) objectList.get(0); // we can use casting which may fail at runtime.
It seems List<Object> and List are same but actually they are not. Consider the following case:
List<String> tempStringList = new ArrayList<>();
rawList = tempStringList; // Ok as we can assign any list to raw list.
objectList = tempStringList; // error as List<String> is not subtype of List<Obejct> becuase generics are not convariant.
You can see we can assign any list to raw list and major reason for that is to allow backward compatibility. Also List<String> will be converted to List at run time due to type erasure and assignment will be fine anyways.
But List<Object> means it can only refer to a list of objects and can also store objects only. Even though String is subtype of Object we cannot assign List<String> to List<Object> as generics are not covariant like arrays. They are invariant. Also check this link for more. Also check the difference between List and List<Object> in this question.
List<?>
Now we are left with List<?> which basically means list of unknown type and can refer to any list.
List<?> crazyList = new ArrayList<String>();
List<String> stringsList = new ArrayList<>();
stringsList.add("Apple");
stringsList.add("Ball");
crazyList = stringsList; // fine
The character ? is known as wildcard and List<?> is a list of unbounded wildcard. There are certain points to observe now.
We cannot instantiate this list as the following code will not compile:
List<?> crazyList = new ArrayList<?>(); // any list.
We can say a wildcard parameterized type is more like an interface type as we can use it to refer to an object of compatible type but not for itself.
List<?> crazyList2 = new ArrayList<String>();
We cannot insert any item to it as we don't know what actually the type would be.
crazyList2.add("Apple"); // error as you dont actually know what is that type.
Now question arises When would I want to use List<?>?
You can think of this as a read-only list where you don't care about the type of the items. You can use it to invoke methods like returning the length of the list, printing it etc.
public static void print(List<?> list){
System.out.println(list);
}
You can also check the difference between List, List<?>, List<T>, List<E>, and List<Object> here.
Simplest explanation which is not "RTFM":
List
Will generate lots of compiler warnings, but is mostly equivalent to:
List<Object>
While:
List<?>
basically means its something generic, but you don't know what the generic type is. Its great for getting rid of compiler warnings when you cant modify the return types of other things that just returned List. Its much more useful in the form:
List<? extends SomeOtherThing>
The shortest possible explanation is: The second item is a list that can hold any type, and you can add objects to it:
List<Object>
The first item you list is treated as essentially equivalent to this, except you will get compiler warnings because it is a "raw type".
List
The third is a list that can hold any type, but you cannot add anything to it:
List<?>
Basically, you use the second form (List<Object>) when you truly have a list that can contain any object and you want to be able to add elements to the list. You use the third form (List<?>)when you receive the list as a method return value and you will iterate over the list but never add anything to it Never use the first form (List) in new code compiling under Java 5 or later.
I'd put it this way: While List and List<Object> can contain any type of objects, List<?> contains elements of an unknown type, but once that type is captured, it can only contain elements of that type. Which is why it is the only type safe variant of those three, and therefore generally preferable.
To complement the tutorials mentioned by Rob, here's a wikibook explaining the topic:
http://en.wikibooks.org/wiki/Java_Programming/Generics
Edit:
No restrictions on type of items in list
Items in list must extend Object
Wildcard used by itself, so it matches anything
Would it be naive of me to conclude at this point that there's hardly any/no difference at all?
When would I want to use
public void CanYouGiveMeAnAnswer( List l ){}
When you cant to do all the casting your self.
When would I want to use
public void CanYouGiveMeAnAnswer( List l<Object> ){}
When you want to restrict the type of the List. For instance, this would be an invalid argument.
new ArrayList<String>();
When would I want to use
public void CanYouGiveMeAnAnswer( List l<?> ){}
Mostly never.
List, List<?>, and List<? extends Object> are the same thing. The second is more explicit. For a list of this type, you cannot know what types are legal to put into it, and you don't know anything about the types you can get out of it, except that they will be objects.
List<Object> specifically means that the list contains any sort of object.
Let's say we make a list of Foo:
List<Foo> foos= new ArrayList<Foo>();
It is not legal to put a Bar into foos.
foos.add(new Bar()); // NOT OK!
It is always legal to put anything into a List<Object>.
List<Object> objs = new ArrayList<Object>();
objs.add(new Foo());
objs.add(new Bar());
But you mustn't be allowed to put a Bar into a List<Foo> - that's the whole point. So that means that this:
List<Object> objs = foos; // NOT OK!
is not legal.
But it's ok to say that foos is a list of something but we don't know specifically what it is:
List<?> dontKnows = foos;
But that then means that it must be prohibited to go
dontKnows.add(new Foo()); // NOT OK
dontKnows.add(new Bar()); // NOT OK
because the variable dontKnows does't know what types are legal.
List < Object > is meant to pass input type parameter of an Object. While List < ? > represents Wildcard type. The wildcard < ? > is of Unknown parameter type. The wildcard cannot be used as a type argument for a generic method and cannot be used to create a generic instance of a class. Wildcard can be used to extend a subtype class, List < ? extends Number >. To relax the restriction of an Object type and in this case to relax "Number" Object type.
In terms of List functionality differences, the other responses already answered this question. In terms of applicable rules of Java generics, it's a complicated topic. I wrote an in-depth article about Java generics rules, this is the link: https://medium.com/#royalilin/java-generics-rules-1d05de86e9cb
by using generics, we detect any possible during compilation.
for example,
List<String> list = new ArrayList<String>();
//list.add(new Integer(45)); This will cause compilation error.
list.add("car");
list.add("bus");
list.add("bike");
String vehicle = list.get(0); //compiler-generated cast
when we use raw type instead of generics before Java 1.5, it needs explicit casting.
for example,
List list2 = new ArrayList();
list.add("car");
list.add("bus");
list.add("bike");
String vehicle = (String)list.get(0); //explicit casting is necessary
however with generics, type erasure occurs. that is the type information is lost in runtime.
if, that is so, how does the JVM know what object type it is retrieving during runtime, whether it is a string object or a person object (compiler generated cast above). but this valid with generics, which is can cause runtime errors.
List<Object> test = new ArrayList<Object>();
test.add("hello");
test.add(new Integer(34));
finally, Joshua Bloch mentions on page 115 (item 23, effective java) that
Set<Object> is parameterized type representing a set that can contain objects of any type,
Set<?> is a wild card type representing a set that can contain only objects of some unknown type
and Set is a raw type, which opts out of the generic type system.
I do understand what he means by the above statement. some clarifications will help
The compiler inserts cast operations when retrieving items from generic methods; this is the only way that the JVM knows to treat the result of list.get(0) as a String. This is why heap pollution (inserting the wrong type of object into a generic collection) can result in a ClassCastException at runtime.
Regarding the wildcards:
Set<Object>'s generic type is exactly Object. You can insert and retrieve Object instances from it, but you can't pass a Set<Integer> to a method expecting a Set<Object>, since the method might be planning to add a non-Integer object to the set.
Set<?> has an unspecified generic type. A method can retrieve anything from it as an Object (since everything is an Object) and can call universal methods on it like hashCode or toString, but it can't add anything to the set.
Set, as you mention, is the raw type and shouldn't be used in new code.
I am not very sure, but what I understand by type information is lost at runtime is that there is no way at runtime that a collection is of some specific type. If you add a String to a collection, it will be a String only but the collection does not enforce that all elements should be of type String
Generics are implemented by Java compiler as a front-end conversion
called erasure. Type erasure applies to the use of generics. When
generics are used, they're converted into compile time checks and run
time type casts.
Due to type erasure mechanism this code:
List<String> a = new ArrayList<String>();
a.add("foo");
String x = a.get(0);
gets compiled into this:
List a = new ArrayList();
a.add("foo");
String x = (String) a.get(0);
Notice extra cast inserted into compiled compiled-code after type erasure.
PS: #chrylis has already provided good explanation about your 2nd part of question.
Well, this stackoverflow question here can help you .
Eclipse might be using this method to find out the fields in a class and their generic type if any. Please have a look.
What is the difference between List and List<?>? I know I can't add any element to the List<?>. I have a code:
List<String> myList = new ArrayList<String>();
processList(myList);
processListGeneric(myList);
public static void processList(List myList) {
Iterator it = myList.iterator();
while(it.hasNext())
System.out.println(it.next());
}
public static void processListGeneric(List<?> myList) {
Iterator<?> it = myList.iterator();
while(it.hasNext())
System.out.println(it.next());
}
The name of the two methods cannot be the same, because it causes compile time error. So is there any difference in these two approaches?
Both do the same, but in second case compiler is informed that you really want a list with no type bounds and raises no warnings. If you are working with Java 5 or later you are encouraged to use second approach.
The difference is that you can't add anything to a List<?>, since it's a List of an unknown type.
For example, you are prevented from doing this:
List<Integer> listOfInt = new ArrayList<Integer>();
List<?> list = listOfInt;
list.add("hello?"); // Compile-time error
You can add anything you want to the base type List since the type of the list items is not checked.
List<?> (pronounced "collection of unknown")is a collection whose element type matches anything. It's called a wildcard type for obvious reasons.
Refer to the following code
List<String> myList = new ArrayList<String>();
myList.add("John");
String name = myList.get(0);
System.out.println(name); //print John
List<?> myListNew = myList;
myListNew.add("Sam");//Compile time error
String nameNew = myListNew.get(0);//Compile time error
Object newName = myListNew.get(0);
System.out.println(newName);//Prints John
Since we don't know what the element type of myListNew stands for, we cannot add objects to it. The add() method takes arguments of type E, the element type of the collection. 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.
On the other hand, given a List<?>, we can call get() and make use of the result. The result type is an unknown type, but we always know that it is an object. It is therefore safe to assign the result of get() to a variable of type Object or pass it as a parameter where the type Object is expected.
List is raw type and List< ?> is wildcard type. Take look http://docs.oracle.com/javase/tutorial/extra/generics/wildcards.html
Both of them behave same. Using parametrized notation you are just avoiding any warnings in Java5 and above. You cannot have both the syntax in same java file because due to type erasure compiled unit has two methods with exactly same signature in same class file and thus violating language rules.
Following is what you would be getting from compiler:
Method processList(List) has the same erasure processList(List) as another method in type ...
Just writing List without a type parameter is deprecated, but otherwise the same as writing List<Object>. So the question becomes "What's the difference between List<Object> and List<?> and which one should I use?".
As you already know, you can't add (anything other than null) to a List<?>, so if you need to add to a list, you should use a List<Object> (or a more specific type where applicable, of course). On the other hand, a method that takes a List<Object>, only accepts List<Object>s and not any lists that contain a subclass of Object. That is, it would not accept a List<String>. If the method takes a List<?> however, it accepts any kind of list. So if you don't need to add to the list, you should use List<?> as it is more general.
The second method uses generics (introduced in Java 5).
One important distinction is the <?> represents a single type, not any object like this:
List<? extends Object> myList
So you could say that using the first method (without the wildcard syntax) is more flexible, as you'd be able to add any object to your List. Although, you'll get a (compiler) warning that you declaration should be parameterized.
Using the <?> unbounded wildcard syntax will avoid the warning, but you're telling the compiler it could be a list of any type, instead of actually using generics to enforce type safety. You are strongly encouraged to use generics to assist you in making your application type safe.
If, for example, you know that the list should only ever contain String objects, declare it thus:
List<String> myList
Then you'll avoid unnecessary casting of use of the instanceof operator, etc.
Here's a brief tutorial on generics in Java, for your information:
http://javarevisited.blogspot.co.uk/2011/09/generics-java-example-tutorial.html
Can someone explained, as detailed as possible, the differences between the following types?
List
List<Object>
List<?>
Let me make this more specific. When would I want to use
// 1
public void CanYouGiveMeAnAnswer(List l) { }
// 2
public void CanYouGiveMeAnAnswer(List<Object> l) { }
// 3
public void CanYouGiveMeAnAnswer(List<?> l) { }
As the other posts have noted, you are asking about a Java feature called generics. In C++, this is called templates. This feature in Java is usually easier to work with than the that found in C++.
Let me answer your questions functionally (if that's not a naughty word for OO discussions).
Before generics, there were concrete classes like Vector.
Vector V = new Vector();
Vectors hold any object you give them.
V.add("This is an element");
V.add(new Integer(2));
v.add(new Hashtable());
They do this by casting all values given to it into an Object (the root of all Java classes). When you attempt to retrieve the values stored in your Vector, you need to cast the value back into the original class (if you want to do anything meaningful with it).
String s = (String) v.get(0);
Integer i = (Integer) v.get(1);
Hashtable h = (Hashtable) v.get(2);
Casting gets old fast. More than that, the compiler complains to you about unchecked casts. The most urgent problem with casting like this is that consumers of your Vector have to know the classes of its values at compile time in order to cast correctly. In cases where the producer of the Vector and the consumer of the same are completely isolated from each other (think RPC messages), this can be a fatal issue.
Enter generics. Generics attempt to create strongly typed classes to do generic operations.
ArrayList<String> aList = new ArrayList<String>();
aList.add("One");
String element = aList.get(0); // no cast needed
System.out.println("Got one: " + element);
The Design Patterns book encourages the reader to think in terms of contracts, not concrete types. There is wisdom (and code re-use) in divorcing variables from their implementing class.
With this in mind, you might think that all implementations List objects should do the same set of things: add(), get(), size(), etc. With a little reflection, you can imagine many implementations of List operations that obey the List contract in various ways (e.g. ArrayList). However, the type of data these objects deal with is orthogonal to the actions performed on them.
Put it all together and you'll see the following kinds of code frequently:
List<String> L = new ArrayList<String>();
You should read that as "L is a kind of List that deals with String objects". When you start dealing with Factory classes, it is critical to deal with contracts rather than specific implementations. Factories produce objects of various types at runtime.
Using generics is pretty easy (most of the time).
One day you may decide you want to implement your own generic class. Perhaps you want to write a new database abstraction interface that elides the differencesbetween various data stores. When you define that generic class, you will use <t> as a placeholder for the kind of object that will be manipulated by the methods.
If you are still confused, use the generic classes for List until you are comfortable. Later, you can dive into the implementation with a bit more confidence. Or you can look at the source code for the various List classes that ship with the JRE. Open source is great that way.
Have a look at the Oracle/Sun docs about generics.
Cheers.
In my own simple terms:
List
Would declare an ordinary collection, can hold any type, and will always return Object.
List<Object>
Will create a list that can hold any type of object, but can only get assigned a another List<Object>
For instance this doesn't work;
List<Object> l = new ArrayList<String>();
Of course you can add anything but only can pull Object.
List<Object> l = new ArrayList<Object>();
l.add( new Employee() );
l.add( new String() );
Object o = l.get( 0 );
Object o2 = l.get( 1 );
Finally
List<?>
Will let you assign any type, including
List <?> l = new ArrayList();
List <?> l2 = new ArrayList<String>();
This would be called collection of unknown and since the common denominator of unknown is Object you will be able to fetch Objects ( a coincidence )
The importance of unknown comes when its used with subclassing:
List<? extends Collection> l = new ArrayList<TreeSet>(); // compiles
List<? extends Collection> l = new ArrayList<String>(); // doesn't,
// because String is not part of *Collection* inheritance tree.
I hope using Collection as the type doesn't create confusion, that was the only tree that came to my mind.
The difference here, is that l is a collection of unknow that belongs to the Collection hierarchy.
To add to the already good answers here:
Method arguments:
List<? extends Foo>
good choice if you don't intend to alter the list, and only care that everything in the list is assignable to type 'Foo'. This way, the caller can pass in a List<FooSubclass> and your method works. Usually the best choice.
List<Foo>
good choice if you intend to add Foo objects to the list in your method. The caller may not pass in a List<FooSubclass>, as you intend to add a Foo to the List.
List<? super Foo>
good choice if you intend to add Foo objects to the list, and it's not important what else is in the list (ie, you are ok getting a List<Object> that contains a 'Dog' that has nothing to do with Foo).
Method return values
just like method arguments, but with the benefits reversed.
List<? extends Foo>
Guarantees that everything in the returned List has type 'Foo'. It might be List<FooSubclass> though. Caller cannot add to the List. This is your go-to choice and the most common case by far.
List<Foo>
Just like List<? extends Foo> but also allows the caller to add to the List. Less common.
List<? super Foo>
allows the caller to add Foo objects to the List, but does not guarantee what will be returned from list.get(0)... it could be anything from Foo to Object. The only guarantee is that this won't be a list of 'Dog' or some other choice that would prevent list.add(foo) from being legal. Very rare use case.
I hope that helps. Good luck!
ps. To sum up... two questions...
do you need to add to the List? Do you care what is in the list?
yes yes - use List<Foo>.
yes no - use List<? super Foo>.
no yes - use <? extends Foo> --- most common.
no no - use <?>.
I refer you to the excellent Java Generics tutorial, and the "advanced" Generics tutorial, both available from Sun Microsystems. Another great resource is the Java Generics and Collections book.
I will try to answer this in detail. Before generics we were having only List (a raw list) and it can hold almost anything we can think of.
List rawList = new ArrayList();
rawList.add("String Item");
rawList.add(new Car("VW"));
rawList.add(new Runnable() {
#Override
public void run() {
// do some work.
}
});
The major problem with the raw list is when we want to get any element out of such list it can only guarantee that it would be Object and for that reason we need to use casting as:
Object item = rawList.get(0); // we get object without casting.
String sameItem = (String) rawList.get(0); // we can use casting which may fail at runtime.
So conclusion is a List can store Object (almost everything is Object in Java) and always returns an Object.
Generics
Now lets talk about generics. Consider the following example:
List<String> stringsList = new ArrayList<>();
stringsList.add("Apple");
stringsList.add("Ball");
stringsList.add(new Car("Fiat")); //error
String stringItem = stringsList.get(0);
In the above case we cannot insert anything other than String in stringsList as Java compiler applies strong type checking to generic code and issues errors if the code violates type safety. And we get error when we try to insert a Car instance in it. Also it eliminates cast as you can check when we invoke get method. Check this link for understanding why we should use generics.
List<Object>
If you read about type erasure then you will understand that List<String>, List<Long>, List<Animal> etc. will be having different static types at compile time but will have same dynamic type List at run time.
If we have List<Object> then it can store only Object in it and almost everything is Object in Java. So we can have:
List<Object> objectList = new ArrayList<Object>();
objectList.add("String Item");
objectList.add(new Car("VW"));
objectList.add(new Runnable() {
#Override
public void run() {
}
});
Object item = objectList.get(0); // we get object without casting as list contains Object
String sameItem = (String) objectList.get(0); // we can use casting which may fail at runtime.
It seems List<Object> and List are same but actually they are not. Consider the following case:
List<String> tempStringList = new ArrayList<>();
rawList = tempStringList; // Ok as we can assign any list to raw list.
objectList = tempStringList; // error as List<String> is not subtype of List<Obejct> becuase generics are not convariant.
You can see we can assign any list to raw list and major reason for that is to allow backward compatibility. Also List<String> will be converted to List at run time due to type erasure and assignment will be fine anyways.
But List<Object> means it can only refer to a list of objects and can also store objects only. Even though String is subtype of Object we cannot assign List<String> to List<Object> as generics are not covariant like arrays. They are invariant. Also check this link for more. Also check the difference between List and List<Object> in this question.
List<?>
Now we are left with List<?> which basically means list of unknown type and can refer to any list.
List<?> crazyList = new ArrayList<String>();
List<String> stringsList = new ArrayList<>();
stringsList.add("Apple");
stringsList.add("Ball");
crazyList = stringsList; // fine
The character ? is known as wildcard and List<?> is a list of unbounded wildcard. There are certain points to observe now.
We cannot instantiate this list as the following code will not compile:
List<?> crazyList = new ArrayList<?>(); // any list.
We can say a wildcard parameterized type is more like an interface type as we can use it to refer to an object of compatible type but not for itself.
List<?> crazyList2 = new ArrayList<String>();
We cannot insert any item to it as we don't know what actually the type would be.
crazyList2.add("Apple"); // error as you dont actually know what is that type.
Now question arises When would I want to use List<?>?
You can think of this as a read-only list where you don't care about the type of the items. You can use it to invoke methods like returning the length of the list, printing it etc.
public static void print(List<?> list){
System.out.println(list);
}
You can also check the difference between List, List<?>, List<T>, List<E>, and List<Object> here.
Simplest explanation which is not "RTFM":
List
Will generate lots of compiler warnings, but is mostly equivalent to:
List<Object>
While:
List<?>
basically means its something generic, but you don't know what the generic type is. Its great for getting rid of compiler warnings when you cant modify the return types of other things that just returned List. Its much more useful in the form:
List<? extends SomeOtherThing>
The shortest possible explanation is: The second item is a list that can hold any type, and you can add objects to it:
List<Object>
The first item you list is treated as essentially equivalent to this, except you will get compiler warnings because it is a "raw type".
List
The third is a list that can hold any type, but you cannot add anything to it:
List<?>
Basically, you use the second form (List<Object>) when you truly have a list that can contain any object and you want to be able to add elements to the list. You use the third form (List<?>)when you receive the list as a method return value and you will iterate over the list but never add anything to it Never use the first form (List) in new code compiling under Java 5 or later.
I'd put it this way: While List and List<Object> can contain any type of objects, List<?> contains elements of an unknown type, but once that type is captured, it can only contain elements of that type. Which is why it is the only type safe variant of those three, and therefore generally preferable.
To complement the tutorials mentioned by Rob, here's a wikibook explaining the topic:
http://en.wikibooks.org/wiki/Java_Programming/Generics
Edit:
No restrictions on type of items in list
Items in list must extend Object
Wildcard used by itself, so it matches anything
Would it be naive of me to conclude at this point that there's hardly any/no difference at all?
When would I want to use
public void CanYouGiveMeAnAnswer( List l ){}
When you cant to do all the casting your self.
When would I want to use
public void CanYouGiveMeAnAnswer( List l<Object> ){}
When you want to restrict the type of the List. For instance, this would be an invalid argument.
new ArrayList<String>();
When would I want to use
public void CanYouGiveMeAnAnswer( List l<?> ){}
Mostly never.
List, List<?>, and List<? extends Object> are the same thing. The second is more explicit. For a list of this type, you cannot know what types are legal to put into it, and you don't know anything about the types you can get out of it, except that they will be objects.
List<Object> specifically means that the list contains any sort of object.
Let's say we make a list of Foo:
List<Foo> foos= new ArrayList<Foo>();
It is not legal to put a Bar into foos.
foos.add(new Bar()); // NOT OK!
It is always legal to put anything into a List<Object>.
List<Object> objs = new ArrayList<Object>();
objs.add(new Foo());
objs.add(new Bar());
But you mustn't be allowed to put a Bar into a List<Foo> - that's the whole point. So that means that this:
List<Object> objs = foos; // NOT OK!
is not legal.
But it's ok to say that foos is a list of something but we don't know specifically what it is:
List<?> dontKnows = foos;
But that then means that it must be prohibited to go
dontKnows.add(new Foo()); // NOT OK
dontKnows.add(new Bar()); // NOT OK
because the variable dontKnows does't know what types are legal.
List < Object > is meant to pass input type parameter of an Object. While List < ? > represents Wildcard type. The wildcard < ? > is of Unknown parameter type. The wildcard cannot be used as a type argument for a generic method and cannot be used to create a generic instance of a class. Wildcard can be used to extend a subtype class, List < ? extends Number >. To relax the restriction of an Object type and in this case to relax "Number" Object type.
In terms of List functionality differences, the other responses already answered this question. In terms of applicable rules of Java generics, it's a complicated topic. I wrote an in-depth article about Java generics rules, this is the link: https://medium.com/#royalilin/java-generics-rules-1d05de86e9cb