Creating an array to store generic types in Java [duplicate]

Answer 1

Problems with generics are by default issued as a warning by the compiler.

After compilation, because of type erasure, they all become ArrayList[] pl2 = new ArrayList[10], but the compiler warns you that this is not good.

Generics have been added to Java, and to be backwards compatible you can use generic with non-generic interchangeably.

Answer 2

 ArrayList<Integer>[] pl2 = new ArrayList<Integer>[10];

Means you don't need to do casting when you retrive data from the ArrayList example in normal case

 ArrayList[] pl2 = new ArrayList[10];
 pl2.put(new Integer(10));
 Integer i = p12.get(0);    // this is wrong
 Integer i = (Integer)p12.get(0);    // this is true with casting

but

 ArrayList<Integer>[] pl2 = new ArrayList<Integer>[10];   
 pl2.put(new Integer(10));
 Integer i = p12.get(0);    // this is true no need for casting

Answer 3

Let's start with question 2 first and then get back to question 1:

Question 2:

> ArrayList[] pl2 = new ArrayList[10]; ArrayList[] pl3 = new ArrayList[10];

What is the difference as far as the reference declaration of p12 and p13 is concerned?

In pl2 ensures better type safety than p13.

If I write for pl2:

pl2[0]=new ArrayList<String>();

it will give me a compiler error stating "cannot convert from ArrayList<String> to ArrayList<Integer>"

Thus it ensures compile time safety.

However if I write for p13

pl3[0]=new ArrayList<String>();
pl3[1]=new ArrayList<Integer>();

it will not throw any error and the onus will be on the developer to code and check properly while extracting data from p13, to avoid any unsafe type conversion during runtime.

Question 1:

That's just probably the way generics work. During the main array initialization, ArrayList<Integer>[] pl2 = new ArrayList[10], the left hand side, ArrayList<Integer>[] pl2, will ensure type safety only when you initialize the ArrayList object in the index position:

pl2[0]=new ArrayList<Integer>();

The right hand side main array declaration = new ArrayList[10] just ensures that the index position will hold ArrayList type items. Also have a look at type erasure concepts in Type Erasure for more information.

Answer 4

The generic info only matters in compile time, it tells the compiler which type could be put into an array, in runtime, all the generic info will be erased, so what matters is how you declare the generic type.

Quoted from Think in Java:

it’s not precisely correct to say that you cannot create arrays of generic types. True, the compiler won’t let you instantiate an array of a generic type. However, it will let you create a reference to such an array. For example:

List<String>[] ls; 

This passes through the compiler without complaint. And although you cannot create an actual array object that holds generics, you can create an array of the non-generified type and cast it:

//: arrays/ArrayOfGenerics.java 
// It is possible to create arrays of generics. 
import java.util.*; 

public class ArrayOfGenerics { 
    @SuppressWarnings("unchecked") 
    public static void main(String[] args) { 
        List<String>[] ls; 
        List[] la = new List[10]; 
        ls = (List<String>[])la; // "Unchecked" warning 
        ls[0] = new ArrayList<String>(); 
        // Compile-time checking produces an error: 
        //! ls[1] = new ArrayList<Integer>(); 

        // The problem: List<String> is a subtype of Object 
        Object[] objects = ls; // So assignment is OK 
        // Compiles and runs without complaint: 
        objects[1] = new ArrayList<Integer>(); 

        // However, if your needs are straightforward it is 
        // possible to create an array of generics, albeit 
        // with an "unchecked" warning: 
        List<BerylliumSphere>[] spheres = 
           (List<BerylliumSphere>[])new List[10]; 
        for(int i = 0; i < spheres.length; i++) 
           spheres[i] = new ArrayList<BerylliumSphere>(); 
    } 
}

Once you have a reference to a List[], you can see that you get some compile-time checking. The problem is that arrays are covariant, so a List[] is also an Object[], and you can use this to assign an ArrayList into your array, with no error at either compile time or run time.

If you know you’re not going to upcast and your needs are relatively simple, however, it is possible to create an array of generics, which will provide basic compile-time type checking. However, a generic container will virtually always be a better choice than an array of generics.

Answer 5

This does not work because generic classes does not belong to Reifiable Types.

The JLS about Array creation expression states :

It is a compile-time error if the [class type] does not denote a reifiable type (§4.7). Otherwise, the [class type] may name any named reference type, even an abstract class type (§8.1.1.1) or an interface type (§9).

The rules above imply that the element type in an array creation expression cannot be a parameterized type, other than an unbounded wildcard.

The definition of Reifiable Types is :

Because some type information is erased during compilation, not all types are available at run time. Types that are completely available at run time are known as reifiable types.

A type is reifiable if and only if one of the following holds:

It refers to a non-generic class or interface type declaration.

It is a parameterized type in which all type arguments are unbounded wildcards (§4.5.1).

It is a raw type (§4.8).

It is a primitive type (§4.2).

It is an array type (§10.1) whose element type is reifiable.

It is a nested type where, for each type T separated by a ".", T itself is reifiable.

For example, if a generic class X<T> has a generic member class Y<U>, then the type X<?>.Y<?> is reifiable because X<?> is reifiable and Y<?> is reifiable. The type X<?>.Y<Object> is not reifiable because Y<Object> is not reifiable.

Answer 6

It seems like you cannot create an array of arraylists with a generic type, according to an answer to Stack Overflow question Create an array of ArrayList elements.