How remove the:
Type safety: The expression of type
List[] needs unchecked conversion to conform to List<Object>[]
compiler warning in the following expression:
List<Object>[] arrayOfList = new List[10];
Afaik the only way is to use #SuppressWarnings("unchecked"). At least if you want to avoid the raw type warning that occurs in Matthew's answer.
But you should rethink your design. An Array of Lists? Is that really necessary? Why not a List of Lists? And if it gets too complicated (a List of List of Map of List to...), use custom data types. This really makes the code much more readable.
And as an unrelated side note: You should write List<Object>[] arrayOfList. The brackets are part of the type, not the variable.
You cannot do any variation of this without a compiler warning. Generics and arrays do not play nice. Though you can suppress it with
#SuppressWarnings("unchecked")
final List<Object> arrayOfList[] = new List[10];
List arrayOfList[] = new List[10];
or
List[] arrayOfList = new List[10];
You can't have generic arrays in Java, so there is no reason to have a reference to one. It would not be type-safe, hence the warning. Of course, you're using <Object>, which means your lists are intended to contain any object. But the inability to have generic arrays still applies.
Also, note that the second version above (with [] attached to the type) is usually considered better style in Java. They are semantically equivalent.
There are a number of ways:
You can add `#SuppressWarnings("unchecked") before the assignment to tell the compiler to ignore the warning
Use raw types List[] arrayOfList = new List[10]; notice that in Java you usually put the [] after the type not the variable when declaring it - though using raw types is discouraged by Sun since they might be removed in a future version.
Don't use arrays, it's usually a bad idea to mix collections with arrays: List<List<Object>> listOfList = new ArrayList<List<Object>>;
Unfortunately, due to the fact that generics are implemented in Java using type erasure, you cannot create an array of a type with type parameters:
// This will give you a compiler error
List<Object>[] arrayOfList = new ArrayList<Object>[10];
See this in Angelika Langer's Java Generics FAQ for a detailed explanation.
You could remove the generics and use raw types, as Matthew Flaschen shows, or use a collection class instead of an array:
List<List<Object>> data = new ArrayList<List<Object>>();
Related
I want to create a two-dimensional array (yes I know that this is actually an array of arrays) holding Optionals. The normal approach for generic array creation does not work though as it fails with a ClassCastException. Here is my code:
#SuppressWarnings("unchecked")
Optional<Integer>[][] arr = (Optional<Integer>[][]) new Object[5][5];
Is there a way to create such an array, if yes what would be the approach for that?
In Java "it is illegal to create an array of a generic type, a parameterized type, or a type parameter". "Why is it illegal to create a generic array? Because it isn’t typesafe. If it were legal, casts generated by the compiler in an otherwise correct program could fail at runtime with a ClassCastException. This would violate the fundamental guarantee provided by the generic type system." [Joshua Bloch - Effective Java]
So what solutions are to be able to create multidimensional arrays?
The recommended one would be to use a container:
List<List<Optional<Integer>>> arr = new ArrayList<>();
for (int i = 0; i < 5; i++) {
arr.add(new ArrayList<Optional<Integer>>());
}
Generics aside, you can't cast an Object[][] to a raw-typed Optional[][]. You'll get a ClassCastException at runtime. The array has to be created as an Optional[][], not as Object[][]. But generics are usually preferred raw types.
It's not that you can never create arrays of generics. You have to do so indirectly. Typically the way to do it is to create arrays of unbounded-wildcard generics, and then do an unchecked cast -- as you've done -- to the right type:
#SuppressWarnings("unchecked")
Optional<Integer>[][] arr = (Optional<Integer>[][]) new Optional<?>[5][5];
The above applies to the creation of any arrays of some specific generic type. In this case, you might consider using OptionalInt instead of Optional<Integer>. This bypasses any concerns about arrays of generics.
(Overall I'm somewhat suspicious of the notion of creating arrays or collections of Optionals of any flavor. It just seems like an odd thing to do. There are often better alternatives. But it might be justified in some cases. Anyway, whether an array of Optionals is appropriate for whatever problem you're trying to solve is a separate question.)
I came across the following code, a simple example of adding elements to List
List list = new ArrayList<Integer>();
ListIterator<Integer> litr = null;
list.add("A");
list.add("1");
list.add(5);
litr = list.listIterator();
while(litr.hasNext()){
System.out.println("UIterating " + litr.next());
}
I expected it to throw an ClassCastException, but rather it wrote this to the console
A
1
5
which looks weird. When i tried:
List<Integer> list = new ArrayList<Integer>();
I got a compile time error.
I would be grateful if someone could explain how the String objects are added to the ArrayList
You assigned the new ArrayList to an untyped List. Generic type restrictions don't apply to an untyped List, it will let you put whatever you want in it. The compiler does not keep track that your untyped List refers to something that was declared with a generic type.
In any case this wouldn't produce a ClassCastException, generics only affect compilation. At runtime
The case where you put the type on the list variable:
List<Integer> list = new ArrayList<Integer>();
is preferred, it should generate a compiler error telling you you're putting the wrong type in the collection.
There's a description of how legacy, non-generic code and generic code interoperate in this article:
In proper generic code, Collection would always be accompanied by a type parameter. When a generic type like Collection is used without a type parameter, it's called a raw type.
Most people's first instinct is that Collection really means Collection<Object>. However, as we saw earlier, it isn't safe to pass a Collection<Part> in a place where a Collection<Object> is required. It's more accurate to say that the type Collection denotes a collection of some unknown type, just like Collection<?>.
But wait, that can't be right either! Consider the call to getParts(), which returns a Collection. This is then assigned to k, which is a Collection<Part>. If the result of the call is a Collection<?>, the assignment would be an error.
In reality, the assignment is legal, but it generates an unchecked warning. The warning is needed, because the fact is that the compiler can't guarantee its correctness. We have no way of checking the legacy code in getAssembly() to ensure that indeed the collection being returned is a collection of Parts. The type used in the code is Collection, and one could legally insert all kinds of objects into such a collection.
So, shouldn't this be an error? Theoretically speaking, yes; but practically speaking, if generic code is going to call legacy code, this has to be allowed. It's up to you, the programmer, to satisfy yourself that in this case, the assignment is safe because the contract of getAssembly() says it returns a collection of Parts, even though the type signature doesn't show this.
This is possible because of how generics are implemented in Java - using type erasure, and because Java supports raw types for backward compatibility with old versions of Java (1.4 and older).
Generics only exist in your source code. The compiler uses them to check the types at compile-time, but then throws away the generics. At runtime, a List<Integer> is just a List of objects, and it doesn't know that it's a list that should contain only Integer objects.
Java supports the use of raw types such as List instead of List<Integer> for backward compatibility with old versions. When you use a raw type, as you are doing in your code above, you get a compiler warning. You should not use raw types in new code - only ever use them when you need to deal with old code that you can't change.
The combination of raw types and type erasure allows you to put types of objects in lists that you shouldn't be putting in there.
Because the List at runtime doesn't know anything about the type that its elements are supposed to have, it doesn't check anything so you will not get a ClassCastException.
What's the difference between using a typed vs. non-typedArrayList in Java?
For example, Using an ArrayList of CustomObject:
Typed:
ArrayList<CustomObject> typedArray = new ArrayList<>();
typedArray.add(new CustomObject);
or non-typed:
ArrayList<> nonTypedArray = new ArrayList<>();
nonTypedArray.add(new CustomObject);
Is there any situation where the latter is preferred? Is there any difference when the ArrayList is holding different datatypes, e.g. an ArrayList of String, Int, etc.?
In the Second Approach, it is not mandatory to add only CustomObject whereas it is in 1st Approach, otherwise, you will get Compilation Error.
ArrayList<CustomObject> typedArray = new ArrayList<>();
typedArray.add(new CustomObject());
This approach is generally preferable as there are no chances of having Class Cast Exception but in second approach there are high chances of that !!
JavaDocs explains it beautifully : Why to prefer Generics
Stronger type checks at compile time.
Elimination of casts.
Enabling programmers to implement generic algorithms.
It's never preferable to use the latter option. I don't think that is even possible. I think you meant:
ArrayList nonTypedArray = new ArrayList();
This syntax is left over from Java 1.4 and earlier. It still compiles for the purposes of backwards compatibility.
Generics was introduced in Java 1.5 which allowed you to specify the types between angled brackets.
It is always preferable to use generics because it is more type-safe.
That is, if you specify
ArrayList<String> typedArray = new ArrayList<String>();
Then you cannot accidentally add an integer to this array list; if you tried to add an integer, the program would not compile.
Of course, Generics ensures type safety at compile time. At runtime ArrayList<String> typedArray = new ArrayList<String>(); becomes ArrayList typedArray = new ArrayList();. This is to maintain backwards compatibility.
What's the difference between using a typed vs. non-typed ArrayList in
Java?
A typed/generic ArrayList is a collection of objects in which the "type" of the object is defined in angled brackets. Generics were introduced in Java 5 to create type-safe collections.
Before Generics the collection was called untyped/raw type collection because there was no way to specify the compiler the type of the collection being created.
The difference between both is to detect type-safe operations at compile time.
In both of your cases, you are adding object(s) of type 'CustomObject' to the ArrayList. There will be no issue while adding elements in the list, as both lists will consider them as typed objects.
Typed:
ArrayList<CustomObject> typedArray = new ArrayList<CustomObject>();
typedArray.add(new CustomObject);
Untyped:
ArrayList<> nonTypedArray = new ArrayList<>();
nonTypedArray.add(new CustomObject);
Is there any situation where the latter is preferred?
I don't think so. As generics are recommended to be used while creating a list to ensure type-safe operations.
Is there any difference when the ArrayList is holding different
datatypes, e.g. an ArrayList of String, Int, etc.?
Surely, there is a reasonable difference. For an untyped list, you will need to add type-cast while fetching elements from a list. As there is a possibility of the compiler throwing a ClassCastException at runtime due to different types of elements.
In runtime, there is absolutely no difference, however in compilation time, using type parameters can save you from a plethora of errors, so it is always preferable to use generics properly.
The only case where raw types are used reasonably is in legacy applications, but even in this case, you try to use typed parameters if you can.
The use of type simplifies your coding removing the need of casting and also stores your data efficiently
https://docs.oracle.com/javase/tutorial/java/generics/why.html
Yeah, I know this is an old post. But I wanted to share an instance where an untyped ArrayList is useful: when you're writing a function that supposed to act on arbitrary element types. For example, suppose you want to make a generic shuffle function that knows how to shuffle an array. Like so:
ArrayList<Die> diceRolls = getGetPossibleDiceRolls();
ArrayList<Card> cardDeck = getPossibleCards();
ArrayList<GirlToDate> blackbook = getBlackbook();
shuffle(diceRolls);
shuffle(cardDeck);
shuffle(blackbook);
.
.
void shuffle(ArrayList array) {
int size = array.size();
for (int i=0; i<size; ++i) {
int r = random.nextInt(size - i) + i;
// Swap
Object t = array.get(i);
array.set(i, array.get(r));
array.set(r, t);
}
}
Some might argue "yeah, but the proper way to do this is to create an interface or subclass of something like a Shuffleable type..." But really?
In Java 1.7 and upwards you should normally use the constructor like this:
ArrayList<MyObject> list = new ArrayList<>();
or else for a more general List object:
List<MyObject> list = new ArrayList<>();
Observe that you only specify the type <MyObject> once, not twice. This makes your code easier to maintain. The <> causes the constructor to return an ArrayList which is already typed to match the field/variable to which it is being assigned - so that no cast will be required in the calling code.
Do not use new ArrayList() as the constructor. This returns an untyped ArrayList which then has to be cast to a type to match the field/variable to which it is being assigned. This means unnecessary type checking and casting and so generally reduces performance.
recently I read a piece of code which seems weird to me. As we know, we need to initialize the generic type in collections when we need to use them. Also, we know Collections can contain Collections as their elements.
The code:
public class Solution {
public static void main(String args[]) {
ArrayList res = returnlist();
System.out.print(res.get(0));
}
public static ArrayList<ArrayList<Integer>> returnlist() {
ArrayList result = new ArrayList();
ArrayList<Integer> content = new ArrayList<Integer>();
content.add(1);
result.add(content);
return result;
}}
My question is
why can we use ArrayList result = new ArrayList(); to create an object, since we have not gave the collection the actual type of element.
why can we use result.add(content); to add a collection to a collection with collection "result" is just a plain collection. We have not defined it as a ArrayList of ArrayList
Java generic collections are not stored with a type to ensure backwards compatibility with pre J2SE 5.0. Type information is removed when added to a generic collection. This is called Type Erasure.
This means that a generic collection can be assigned to a non generic reference and objects in a generic typed collection can be placed in an non generic, nontyped collection.
All Java generics really does is make sure you can't add the wrong type to a generic list and saves you from doing an explicit cast on retrieval; even though it is still done implicitly.
Further to this
the Java section of this answer goes a little deeper into what I just said
this article also covers basically what you were asking in a more complete way
other things to watch out for with Type Erasure
Just adding up to provide summarized answer
Old way :
(A) ArrayList result = new ArrayList();
will create an Arraylist to hold "Object"
New Way :
ArrayList<Integer> content = new ArrayList<Integer>();
this represents an Arraylist which will hold "Integer" objects. This was introduced for compile-time type check purposes.
why ?
Consider the first case. Its input type is Object. We know that Object is the super class of all classes. We can pass in an Integer object, String object and so on. When fetching the data the developer has to perform proper type casting. Say if the developer initially thinks the code will accept Integer objects so he adds the following typecast while fetching the data
Integer integer=(Integer) content.get(0);
This is supposed to work. But if mistakenly he passes a String it will result in run-time error.
How it can be avoided ?
By introducing compile time checks
How it works ?
when we specify parameterized type only Integer objects can be added to the ArrayList collection. Else it will show error.
content.add(3); // will work
content.add("HARSHAD"); // error shown
If parameterized generic types are for type checking purposes how correct data can be retrieved from the list ?
The compiler implicitly performs type conversion. See the sample code
List<Integer> list=new ArrayList<Integer>();
list.add(1);
list.add(2);
Integer integer=list.get(0);
System.out.println(integer);
What the compiler actually does when you perform compilation ?
//perform type erasure
(B) List list=new ArrayList();
list.add(1);
list.add(2);
// the arraylist inturn accepts Object when you run the code
//add casting
Integer integer=(Integer)list.get(0);
Conclusion
If you see the codes (A) and (B) both are the same. Only difference is that in the second case the compiler implicitly does the same operation.
Finally to answer your question ...
ArrayList result = new ArrayList();
is allowed for backward compatibility purposes. Although this is not recommended.
Official link from Oracle docs explaining the same concept.
Generics were added to Java only in Java 5. Before that, when you use a collection, it always meant collection of objects. The old syntax is left as is for backward compatibility. So ArrayList result = new ArrayList() is actually creating an ArrayList<Object>. Since ArrayList is also an object, you can add content to the variable result.
why can we use ArrayList result = new ArrayList(); to create an object, since we have not give the collection the actual type of element.
Because java wants to it backward compatible. Generics is more of compiler feature for ensure type safety, collections can store any type of object at runtime.
Java compiler will not give you compiler error for this but it must have given you compiler warning that it is unsafe to use generic classes without type.
It may be a remnant from before generics came along to java (Java 4 or 5 I think).
List<Object> myList = new ArrayList<String>(); //(hint: no)
Map<Integer> myMap = new HashMap<int>(); // (hint: also no)
Why are the statements wrong in the above declarations?
Let's look at the first example. Think of the operations you should be able to do on a List<Object>: Add, Remove, Retrieve any Object.
You're trying to fill those requirements with a collection that you can Add, Remove, and Retrieve only strings.
List<Object> myList = new ArrayList<String>();
// This should be valid based on the List<Object> interface but obviously
// MyClass isn't String...so what would the statement do?
myList.add(new MyClass());
The second is simply because Generics in Java don't support primitive types. For more information, you could check out:
java - Why don't Generics support primitive types?
For the first one, because Java generics are not covariant. So you can't even do:
ArrayList<Object> myList = new ArrayList<String>();
For more info, see this article: Java theory and practice: Generics gotchas.
For the second one, you cannot use primitives as the generic type.
1) What would happen if anyone added a Long to the list that is referenced by myList?
2) You can't have a primitive there.
The first is impossible. You can use a wildcard:
List<?> myList = new ArrayList<String>();
But now, your list is completely unusable. Adding, removing, etc doesn't compile anymore:
myList.add(new Object()); // Error
myList.add("error"); // Error
You can't use primitives (int, double, ....) with generics. You have to use the wrapper class:
HashMap<Integer> myHap = new HashMap<Integer>();
for the second example Map has 2 generic parameters a Key and Value but the following still wont work
Map<Integer,Object> myMap = new HashMap<int,Object>();
that because as said before primitive types don't work with java generics (besides generic types always derive from Object and type erasure translates all generic variables to type Object)
For all things Java-Generics, refer to Angelika Langer. In this case, Is List<Object> a supertype of List<String>?
Even though Java does support auto-boxing, it doesn't automatically convert <int> to <Integer>. IMO this is partly to remind users that you can get nulls.