How can I create a list with a specified type argument?
For example:
LinkedList<Integer> list = createList(LinkedList.class, Integer.class);
I've tried creating a method for it, but the method doesn't include the type argument when creating the new instance.
public static <T, L extends List<T>> L createList(Class<L> listClazz, Class<T> valueClazz) throws Exception
{
return listClazz.getConstructor().newInstance();
//Instead of
// new L<T>();
//does
// new L();
}
I hope my question is clear enough, thank you for any help.
You can't do that. Generics have no business at runtime, so you can't create parameterized instance at runtime using reflection. At runtime, LinkedList<Intege> is just a LinkedList. The type information - Integer is erased at compile time through "type erasure". The point is, why would you like to do that?
//Instead of
// new L<T>();
//does
// new L();
The two "do" exactly the same thing. How do you know it "does" one and not the other? Type parameters are compile-time illusions for type-checking. They are not relevant in what the code "does". To create an instance of LinkedList<T>, you do not need to know what T is, because it doesn't really exist after type erasure.
I think the issue you are having has nothing to do with the code inside the function, but with the signature of the method. By the very fact that the method is declared as it is, and you are passing an expression of type Class<LinkedList> (the type of LinkedList.class) to it, L must be LinkedList, and since the method returns L, the compiler must consider its return to by type LinkedList. None of this has anything to do with the code inside the method.
If you want the method's return type to be considered as LinkedList<Integer>, then just pass in an expression of type Class<LinkedList<Integer>>. Simple, right? How do you get an expression of type Class<LinkedList<Integer>>? For one, you could do (Class<LinkedList<Integer>>)(Class<?>)LinkedList.class.
If that seems kind of bogus, it's because Class is fundamentally a runtime thing -- it allows runtime creation of an object from that class. Using it with Generics, a compile-time type-checking mechanism, is fundamentally flawed. There is just one class object in existence for any class, e.g. LinkedList. Is it LinkedList or LinkedList<Integer> or LinkedList<String>? Well, it could be all of these, in the sense the the newInstance() method could create any of them (there is no difference at runtime).
Related
I have seen in multiple different places people who instantiate a list or ArrayList like:
List<?> l = new ArrayList<>();
What type is ?? Does this mean that it can hold any types in it? If so, why would this be used instead of just and ArrayList?
Does this mean that it can hold any types in it?
No. It means that your l variable could be referring to a list parameterized with any type. So it's actually a restriction: you will not be allowed to add any object to l because you have no idea which items it accepts. To give a concrete example, l could be a List<String> or it could be a List<ExecutorService>.
As correctly pointed by Marko, its an unknown restriction on the List type.
The Java docs says that:
The unbounded wildcard type is specified using the wildcard character
(?), for example, List<?>. This is called a list of unknown type.
There are two scenarios where an unbounded wildcard is a useful
approach:
If you are writing a method that can be implemented using functionality provided in the Object class.
When the code is using methods in the generic class that don't depend on the type parameter. For example, List.size or List.clear.
In fact, Class<?> is so often used because most of the methods in
Class do not depend on T.
Let me make this a long bed time story; read it to fall asleep:)
Let's begin with this point -- To invoke a generic method, its type arguments must be supplied. (Unless the method is invoked in a "raw" manner, i.e. in the erased form, which is another topic:)
For example, to invoke Collections.<T>emptyList(), T must be supplied. It can be supplied explicitly by the programmer --
List<String> list = Collections.<String>emptyList(); // T=String
But that is tedious, and kind of dumb. Obviously in this context, T can only be String. It's stupid if the programmer has to repeat the obvious.
That's where type inference is helpful. We can omit the type argument, and the compiler can infer what the programmer intends it to be
List<String> list = Collections.emptyList(); // T=String is implied
Remember, <String> is still supplied, by the programmer, implicitly.
Supposedly, the programmer is the all-knowing dictator of all type arguments, and, the compiler and the programmer have a common understanding on when type arguments can be omitted and inferable from context. When the programmer omits a type argument, he knows the compiler can infer it exactly as he intended, based on a rigorous algorithm (which he masters:)
It is not the compiler's discretion to pick and choose type arguments, rather, the programmer does, and conveys it to the compiler.
Realistically, type inference is so complex, few no programmer has any idea what's going on in a lot of cases:) The programmer is more like a dictator making vague commands, and the compiler tries its best to make sense out of it. We mostly write code on intuition, not paying attention to details, and we sort of believe that the code does what we want if the compiler approves it.
In any case, all type arguments are fixed precisely and predictably at compile time. Any omitted type argument is equivalent to an explicitly specified one.
Some type arguments are "undenotable", e.g. a type variable introduced by capture conversion. They can not be explicitly specified, they can only be inferred. (Nevertheless the programmer is supposed to know what they are, even though they cannot be named)
In the previous example, T can only be inferred as String, there's no other choices. But in a lot of cases, there are more candidates for T, and the type inference algorithm must have a strategy to resolve it to one of the candidates. For example, consider this lonely statement
Collections.emptyList();
T could be any type; T is resolved to Object, because, well, there's no good reason to resolve it to anything else, like Integer or String etc. Object is more special because it's the supertype of all.
Now, let's get to constructors. Formally speaking, constructors are not methods. But they are very much alike in a lot of aspects. Particularly, type inference on constructors is almost the same as on methods. Invoking a constructor of a class CLASS takes the form of new CLASS(args).
Just like methods, a constructor can be generic, with its own type parameters. For example,
class Bar
{
<T>Bar(T x){ .. }
and type inference works on generic constructors too
new Bar("abc"); // inferred: T=String
To explicitly supply type arguments for a constructor,
new <String>Bar("abc");
It's pretty rare though that a constructor is generic.
A generic constructor is different from a generic CLASS! Consider this
class Foo<T>
{
Foo(T x){ .. }
The class is generic, the constructor is not. To invoke the constructor for class Foo<String>, we do
new Foo<String>(""); // CLASS = Foo<String>
Method type inference we've been talking about so far does not apply here, because the constructor is not even generic. In Java 5/6, there is no type inference on CLASS, therefore <String> must be explicitly specified. It's stupid, because <String> is obvious in this context. There were workarounds (i.e. using static factory methods), but people were of course very upset and demanded a solution.
In Java 7, this problem is solved by "diamond inference" -
new Foo<>(""); // inferred: T=String
"diamond" refers to the curious <> operator. It is required; we cannot simply write
new Foo("");
because that already had a different meaning - invoking the constructor of "raw" Foo.
With diamond inference, we can do things we couldn't in Java 5/6
List<Object> list = new ArrayList<>(); // Java 7. inferred: E=Object
// equivalent to
List<Object> list = new ArrayList<Object>(); // <Object> is required in Java 5/6
Remember, T=Object is still supplied, through diamond inference.
Finally, we come back to your original question
List<?> list = new ArrayList<>();
Here, E=Object is inferred (what else?). The code is equivalent to
List<?> list = new ArrayList<Object>();
Yep, the list object is indeed an ArrayList<Object>, not ArrayList<SomethingElse>.
Also note that the following would be illegal and nonsensical
List<?> list = new ArrayList<?>();
^^^
CLASS in new CLASS(args) must be a concrete type. We can only instantiate an ArrayList of a specific element type.
The declared type List<?> of variable list is too general though. For a local variable, it is the best practice IMO to declare it in its more specific type
ArrayList<Object> list = new ArrayList<>();
Don't use <?> here - it just causes confusion to everybody.
On a related note, a lot of people would argue for "program against interface"
List<Object> list = new ArrayList<>();
^^^^
That is wrong IMO. Who are we providing abstraction for in a local block? Use the most specific type in implementation for max clarity;
use abstract types in interfaces.
zzzzzzzzzz
Methods that are generic using the T parameter can for sure be handy. However, I am curious what the use of a generic method would be if you pass an argument such as Class<T> clazz to the method. I've come up with a case that maybe could be an possible use. Perhaps you only want to run a part of the method based on the type of class. For example:
/** load(File, Collection<T>, Class<T>)
* Creates an object T from an xml. It also prints the contents of the collection if T is a House object.
* #return T
* Throws Exception
*/
private static <T> T void load(File xml, Collection<T> t, Class<T> clazz) throws Exception{
T type = (T) Jaxb.unmarshalFile(xml.getAbsolutePath(), clazz); // This method accepts a class argument. Is there an alternative to passing the class here without "clazz"? How can I put "T" in replace of "clazz" here?
if (clazz == House.class) {
System.out.println(t.toString());
} else {
t.clear();
}
return T;
}
Is this an accepted practice? When is the Class<T> clazz argument useful with generic methods?
Is this an accepted practice?
Well, to me.. no not really. To me, it seems somewhat pointless when you can simply define some boundaries on the type of T. For example:
private static <T extends House> void load(Collection<T> t)
This will guarantee that either the object is of type House or of a subclass of House, but then again if you only want an instance of type House or it's subclasses, it should really just be:
private static void load(Collection<House> houses)
The idea of generics is to make a method or a class more malleable and extensible, so to me it seems counter-intuitive to start comparing class types in the method body, when the very notion of generics is to abstract away from such details.
I'd only pass class objects if the generic type could not be derived otherwise. In your case, the compiler should be able to infer T from the collection. To treat specific objects differently, I'd use polymorphism - e.g. House#something() and Other#something(), and just call anyObject.something().
I think it is acceptable but if it can be avoided then you should. Typically, if you can have different methods which accepts different type, then do it instead of one method which uses if clauses to do something different depending on the type of the parameter. You could also delegates to the class the operation you want to make specific for a given type.
In your case, you could simply test the type of each element of the collection using instanceof, to do what you need for the specific type. But it won't work if the list is empty.
A typical use is if you need to get the type to create it and you can find it from another way. For instance, Spring uses it to load a bean from its name:
<T> T getBean(Class<T> requiredType)
In that case, it cannot be avoided (without having to cast).
If the returned value or other parameters types are dependent or need to be equal, generics will add compile time checks, so that there's no need to cast to T.
Examples
<T> T createNewInstanceOfType(Class<T> type);
<T> void addValueToCollection(Collection<T> collection,T value);
<T> List<Class<? extends T>> findSubClassesInClasspath(Class<T> superType);
Raw types
It is still possible to defer a casting error until runtime (ClassCastException) with some casts, e.g. with implicit casts from non-generic (raw) types to generic ones:
List nonGenericList = new ArrayList();
nonGenericList.add(new Integer(42));
List<String> wreckedList = nonGenericList;
The compiler will generate a bunch of warnings, unless you suppress them with annotations or compiler settings.
Compiler Settings (Eclipse):
For example, the usage of raw types generates a warning per default, one can treat warnings as errors and even as fatal errors:
You would pass a Class<T> argument in generics if, and only if, you would pass a Class argument before generics. In other words, only if the Class object is used in some way. Generics serves as a compile-time type checking tool. However, what arguments you pass should be determined by the runtime logic of the program, and should be irrelevant of generics.
I haven't seen passing a Class object in order to check the runtime type of an object as a common use case for generics. If you're doing that, there's a good chance that there's a better way to set up your class structure.
What I have seen is if you need to create a new instance of the class in question, or otherwise use reflection. In that case you do have to pass the Class object, because Java cannot derive it at runtime thanks to type erasure.
In your case actually having the Generic parameter is not strictly needed.
Since the output of the function you are describing does not depend on the type of the input you might as well use wild cards.
private static void stuff(Collection<?> t){
Object next = t.iterator().next(); //this is ugly and inefficient though
if(next instanceof House){
System.out.print(next.toString());
}else{
t.clear();
}
}
The only time you should use generic parameter is when the type of the result of a function will be dependent of the type of the parameters.
You will need to pass the Class corresponding to the type when your code will need it; most of the time this happens when:
- You need to cast/type check objects to T
- There is serialization/deserialization involved.
- You cannot access any instance of T in your function and you cannot call the getClass() method when you need it.
Passing a Class on every generic function will result in you passing an unnecessary parameter most of the time, which is regarded as bad practice.
I answered a similar discussion in the past:
When to use generic methods and when to use wild-card?
How can call public <T> T doit(Class<T> clazz); using MyClass<String>.class as clazz where I can not instantiate or extend MyClass.
EDIT: 'David Winslow' and 'bmargulies' responses are correct (MyClass<String>) doit(MyClass.class); works for the original question BUT surprisingly when the method returns say MyClass<T> instead of T casting will not compile any more.
Edit: I have replaced List with MyClass and added the condition to my original question.
Use List.class. Because of type erasure type parameters to Java classes are entirely a compile-time construct - even if List<String>.class was valid syntax, it would be the exact same class as List<Date>.class, etc. Since reflection is by nature a runtime thing, it doesn't deal well with type parameters (as implemented in Java).
If you want to use the Class object to (for example) instantiate a new List instance, you can cast the result of that operation to have the appropriate type parameter.
List<String> list = (List<String>)(ArrayList.class.newInstance());
I've seen similar questions asked several times, for example
Acquiring generic class type
There are legitimate reasons to construct static generic types. In op' case, he would probably like to
MyClass<String> result = doit(MyClass<String>.class);
Without language syntax support, casting is the correct way to go. If this is needed quite often, the casting should be put in a method, as
public class MyClass<T>
{
#SuppressWarnings("unchecked")
// may need a better method name
static public <T2> Class<MyClass<T2>> of(Class<T2> tClass)
{
return (Class<MyClass<T2>>)(Class<?>)(MyClass.class);
}
}
MyClass<String> result = doit(MyClass.of(String.class)); // no warning
We can supress the warning on that method alone, after making sure the cast is safe. Any call site will not see the warning.
This is all compile time casting game. At runtime all the type parameters are erased, and really only the naked class object is passed around. The of method will most likely be optimized off, so to JVM the last line is nothing but
MyClass result = doit(MyClass.class)
There are also times when at runtime we need a complete MyClass<String> type. A ParameterizedType object needs to be obtained to represent MyClass<String>.
When the two requirements are combined together, that is, we need a compile time expression regarding MyClass and String that will evaluate at runtime to a ParameterizedType
ParameterizedType type_MyClass_String = ???? MyClass ?? String ???
There is a technique involving an anonymous subclass of MyClass<String>
ParameterizedType type_MyClass_String = superTypeOf( new MyClass<String>(){} );
which I find quite disturbing.
See http://jackson.codehaus.org/1.7.0/javadoc/org/codehaus/jackson/type/TypeReference.html and the references that it references for a comprehensive discussion of the issues around generics.
the bottom line is that, if you really want to work with generic types in this way, you have to stop using Class and start using Type and its subclasses.
Contrary to your comment on another answer, you can write List<List<String>> obj = (List<List<String>>) doit(List.class);, you just can't avoid a warning when you write it.
Since after your update your question does not appear to be an exact duplicate:
You would need to call getClass() on an instance of MyClass. Better have a dummy static final instance somewhere:
public static final MyClass INSTANCE = new MyClass();
...
return (Class<MyClass<String>>) instance.getClass();
T corresponds to List, so any reference to String as the generic paramter of List is irrelevant.
How to do MyClass<String>.class in
Java?
You can't.
Generics in Java use type erasure; the type of the parametrized argument is enforced during compilation, but it is lost after compilation. The resulting byte code for an instance of a generic class does not contain any run-time meta-data on its arguments whatsoever.
As it is now, it is just not possible, a major language design blunder IMO.
I am still experimenting with how Java handles generics. I stumbled upon the fact/issue/thing that if you have a generic interface like A<T>, you cannot really check afterwards if some object is actually implementing A<B> or A<C>.
I wondered if that could cause actual problems.
Now I have tried this code:
static interface A<T> { void foo(T obj); }
static class B implements A<B> {
public void foo(B obj) { obj.bar(); }
void bar() {}
}
static {
assert (new B() instanceof A<?>);
((A<?>) new B()).foo(new Object());
}
This gives me this error (for the foo-call):
The method foo(capture#1-of ?) in the type Main.A<capture#1-of ?> is not applicable for the arguments (Object)
I wonder why that is. Eclipse tells me that the signature of foo after the cast to A<?> is foo(? obj) which I thought is the same as foo(Object obj).
The assert succeeds.
What I tried to figure out is at what point exactly does it cast the object when I call the foo function.
Also, how can I call foo from A<?>? This is the thing I actually need to be able to do. Or is that impossible with any other parameter than null?
A more real-world example where I actually wonder about this: I use the Comparable<T> interface a lot. That case is actually even more complicated; I might open another question about that if this here doesn't answer it.
I wonder why that is. Eclipse tells me that the signature of foo after the cast to A is foo(? obj) which I thought is the same as foo(Object obj).
Nope, absolutely not. Imagine A<T> is List<T> with foo(T) being add(T) and so A<?> is List<?>. Should you be able to do this?
List<String> strList = new ArrayList<String>();
List<?> wcList = strList;
wcList.add(Integer.valueOf(6)); //possible if add(?) is same as add(Object)
//...
String str = strList.get(0);
Of course not, since you'd get a ClassCastException in the final line.
What foo(?) really means is that the method applies to some unknown but specific type. You can't typically invoke these methods unless you pass null as the parameter, which is acceptable to assign to any reference type.
If you have "foo(? obj)" then the ? could be any type. If it is say String then you can't pass, say, an Integer to it. All you can pass is null.
Casting and use of instanceof should normally be avoided unless unavoidable (such as implementing equals), particularly with generics.
The compiler is correct because it does a compile-cast test on compile-time.
((A<?>) new B()).foo(new Object());
is errornous because the compiler expected
((A<?>) new B()).foo(A object)....
meaning that it wanted anything of type A or its children. Object is the parent of A and it doesn't match the compile-test Parameter type for compilation.
When you check if an instance of B is an instance of A<?> you just do the same thing as new B() instanceof A. You don't really check how T is set, it's just "something".
Later in the code with the cast, you tell that you'll use B as a A<?> so the variable will have the characteristic of a A but the generic type is still "something". This "something" exists and is probably a specified class but your don't care of the exact type.
That's why when you use the foo() method which take a T in parameter, you can't pass a "something" in parameter, because you don't know what it is, it could be an Object but it could be anything else.
Because of this the compiler tells you foo(capture#1-of ?) isn't applicable for the argument Object. The needed parameter is "something" but not necessarily an Object.
Then, when would you need this feature ?
For example if you work with a map, if you don't really care of the type of the key (if you only work with the values() method for example), you could do something like this :
Map<?, V> m
This way you won't be able to use features related to the key of the map (but you don't care about that), but you'll be able to use a map with any kind of key.
No, foo(? obj) is not actually the same as foo(Object obj). The difference is, when the parameter type is Object, it's explicitly stating that any type of object is legal. With a parameter type of ?, the method states that it does not know what type of object is legal... therefore nothing is legal except for null.
The reasoning for this becomes apparent when you consider a List rather than an arbitrary interface. Look at this method:
public void foo(List<?> list) {
list.add(...); // what can we add here?
}
The ? indicates that any type of List is acceptable... the List passed in could be a List<String>, a List<Integer> or a List<Map<Foo, Bar>>. There's no way of knowing. Note that this is only a problem with methods that consume generic parameters, such as add or your foo method. For methods that produce (return) an object of the generic type, it's fine to use such a method and assign the result as an Object. This, unlike the consumer method, cannot corrupt the internal state of the generic object.
Also notice that when you call ((A<?>) new B()).foo(new Object()), you're trying to do something illegal... if something (such as the Object) that is not a B were to be passed to B's foo method, it would explode at runtime. The compiler is correctly preventing you from doing this.
You may also want to check out my answer to another question here which explains some things about bounded wildcard types and such.
This is a simple explanation:
A<?> means A of unknown parametrized type. Given
A<?> ref = new B()
ref could point to any A: A<Object, A<String>, anything. So you can not call A.foo(new Object()) because in case of A<?> reference there is no way to know which parameter ref.foo() accepts. Actually the only thing valid is ref.foo(null).
Many have clarified that point about ? but noone has really answered the main question yet, so here it is:
It is possible to call A#foo like this:
((A) new B()).foo(new Object());
The cast then is done inside of foo.
I currently posses a List of Objects(Using Java 1.3), and let's say that I wanted to cast one of the Objects returned from list.get(i) to a type of which I only know the name of the Class as a String. Essentially, how do I
Object o = (classname)list.get(i); where className is a String variable of a className.
I thought that I could use ( Class.forName(className) )list.get(i), but I received a syntax error claiming that I forgot a semicolon.
Unfortunately, since I am using Java 1.3, I do not have access to the Class.cast(Object) method.
What is the name of the class used when casting to another type in Java 1.3? Is there some method that can give me the correct type I need with a String parameter of the class name?
what is the point of casting when all you do is assign the result to object?
All you would achieve is an exception if it did not implement the interface/extend or was the class or do nothing if it did.
For that a simple:
public static boolean IsInstance(object x, String className)
{
Class cls = Class.forName(className);
return cls.isInstance(x);
}
is sufficient (and cleaner)
If you were to the use reflection to get at the fields/methods of the class that's just fine
No, and you can't do this across most languages.
The reason is that the type to cast to has to be known at compile time, not at runtime (which is what you are trying to do).
If you think about it, it makes sense, because given that the variable could be any type name, how are you supposed to access the various members? You can't, not unless they are defined in a base type/interface that all instances implement, in which case you should just use that.
One scenario where the need for this arises is when enforcing type safety with a legacy system. For example, suppose you have a persistence system like Hibernate that provides a raw List of results from a "finder" method. Casting this raw List to a parameterized type will result in an unchecked warning, and if the List contains an object of the wrong type, a ClassCastException can be raised at an unspecified time in some distantly related code. It may be best to validate the contents of the list up front, using a mechanism like the OP suggests.
Here's the Java 1.3 version (without generics):
private static void checkType(Collection objs, String className)
throws ClassNotFoundException
{
Class clz = Class.forName(className);
Iterator i = objs.iterator();
while (i.hasNext()) {
Object obj = i.next();
if (!clz.isInstance(obj)) {
throw new ClassCastException();
}
}
}
In Java 5 and later, with generics, you can do something similar with the Class.cast() method to verify the contents of a collection, justifying the use of a SuppressWarnings annotation. In our review process, suppressing a warning without some "proof" that it is safe is filed as a bug.
I assume that you really wanted to write the following, instead of using Object on the left side. Since otherwise, it's really just about checking whether the object in the list is of the right type.
ClassName o = (classname)list.get(i);
Well, Java is statically typed. It's not possible that you give it a string and it gives you the corresponding static type, so that you can go without casting. Even with generics and Class<T>.cast, the cast destination type is not given by a string, but by the generic type-argument T, which is known at compile-time. You have to manually cast to the right type, or keep using the most common type (may be Object in your case).
If you do Class.forName(className), it gives you back an object of the type Class which contains information about the type at runtime, so that it allows you to do
Class.forName("my.stuff.MyClass").newInstance()
But the cast wants a type - not an object of some type. That is why the compiler told you there is something wrong with that code.
The static type of the reference returned by that is of Object. This is important: The dynamic type of an object that is referenced, and the static type of the reference that points to that object. The dynamic type of the object is what can be "controlled" by a string (by using Class.forName), but the static type of the reference that you have to do with at compile time, and that is (just to give an example) used to select functions that overload each other, can not be determined by a string.
The question was answered already, but I'd like to add that it seems a bit dubious that you should have a List in which you keep several different kinds of objects (in this case, any objects), yet you'd apparently like to invoke operations on them that are specific to each different type...
What's the point of this collection? Don't the instances you keep in it have anything in common - any common supertype that you could cast them into?