Why can't I create generic array in Java?

Question

I had a problem creating some arrays.

Here is the code

Predicate<Integer[]>[] endOfLine = {
    curPoint -> curPoint[0]==r2,
    curPoint -> curPoint[0]==c2,
    curPoint -> curPoint[0]==r1,
    curPoint -> curPoint[0]==c1
};
Consumer<Integer[]>[] move = {
    curPoint -> { curPoint[0]++; },
    curPoint -> { curPoint[1]++; },
    curPoint -> { curPoint[0]--; },
    curPoint -> { curPoint[1]--; }            
};

and eclipse (maybe compiler?) said:

Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
    Cannot create a generic array of Predicate<Integer[]>

I Googled, and found that generic type parameter array can't be made in Java.

I read about compile time and runtime, and I thought it may be not my problem.

I saw some code which compile compiled:

List<Integer>[] arrayOfLists = new ArrayList[2];
ArrayList<Integer[]> lists =new ArrayList<>();
ArrayList<Consumer []> [] lists = new ArrayList[5];
        
Predicate<Integer[]>[] p = new Predicate[5];
p[0] = a -> a[0] == 0;

Predicate<Integer[]>[] pp = { a -> a[0] == 0 }; //cant be compile

and I read oracle doc and found this:

"Cannot Create Arrays of Parameterized Types"

I concluded my shortcut syntax, {} is the problem. Shortcut syntax create code with parameterized type. Is this correct?

Ravindra Ranwala · Accepted Answer · 2022-01-07 02:56:19Z

3

You can't create a generic array in Java. Arrays are reifiable types, and carry their type information at runtime whereas generics are non reifiable, and their type information is erased after the compile time due to erasure. This is due to the implementation of the type system in java and even though this causes some rough edges and corner cases, it eases the evolution of code into generics. Since generic type information is erased, you don't have them at runtime and that's why you can't create a generic array in java.

There are two solutions to circumvent the issue you are facing. You can either create an Object array and cast it to the generic type you need.

final T[] arr = (T[]) new Object[n]

or else

final Object[] arr = new Object[]

You can do the cast when you get the items from this array like this

T itm = (T) arr[1];

Both of the above approaches work if you don't return this internal array.

However, if you are returning the internal array, you need to create it reflectively, since you need to maintain the proper reifiable type.

static <T> T[] createArr(Class<T> clz) {
    return (T[]) Array.newInstance(clz, 5);
}

edited Jan 7, 2022 at 2:56

answered Jan 3, 2022 at 23:59

Ravindra Ranwala

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7 Comments

Nyerguds Over a year ago

Java generics are fake, plain and simple. All of it is awful compared to c#.

Ravindra Ranwala Over a year ago

Each approach has it's own strengths and weaknesses. Java implementation of generics differs from that of C++. The original implementors were more concerned about the backward compatibility. So, the Java implementation is much more backward compatible than that of C#.

scottb Over a year ago

@Nyerguds "Java generics are fake, plain and simple." ... the use of strongly judgemental language to describe Java;s type system is not helpful. The implementation of generics in Java as a mainly compile-time only phenomenon provides developers with many of the benefits of generic abstraction while maintaining run-time compatibility with legacy systems ... pure and simple. If you were to implement generics in Java at that time while aiming to achieve the same goals, I doubt you could do better.

Ravindra Ranwala Over a year ago

A generic class is a parameterised type, such as Collection<T> or Predicate<T> and hence all the above rules hold true. If you need to understand more about generics, I would encourage you to read the book: amazon.com/Java-Generics-Collections-Development-Process/dp/…

newacct Over a year ago

"You can either create an Object array and cast it to the generic type you need" It would have to be an array of the erasure of T. If T's erasure is Object, then that would be okay. But if say T is <T extends Comparable<T>>, then (T[]) new Object[n] would fail. You would need (T[]) new Comparable[n] instead, since Comparable is the erasure of T. In the OP's case, they want to create an array of a parameterized type. The erasure of a parameterized type is just the raw type. They should use (Predicate<Integer[]>[])new Predicate[n] and (Consumer<Integer[]>[])new Consumer[n]

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