Why are arrays covariant but generics are invariant?

孤人 提交于 2021-02-11 12:04:35

问题


From Effective Java by Joshua Bloch,

  1. Arrays differ from generic type in two important ways. First arrays are covariant. Generics are invariant.
  2. Covariant simply means if X is subtype of Y then X[] will also be sub type of Y[]. Arrays are covariant As string is subtype of Object So

    String[] is subtype of Object[]

    Invariant simply means irrespective of X being subtype of Y or not ,

     List<X> will not be subType of List<Y>.
    

My question is why the decision to make arrays covariant in Java? There are other SO posts such as Why are Arrays invariant, but Lists covariant?, but they seem to be focussed on Scala and I am not able to follow.


回答1:


Via wikipedia:

Early versions of Java and C# did not include generics (a.k.a. parametric polymorphism).

In such a setting, making arrays invariant rules out useful polymorphic programs. For example, consider writing a function to shuffle an array, or a function that tests two arrays for equality using the Object.equals method on the elements. The implementation does not depend on the exact type of element stored in the array, so it should be possible to write a single function that works on all types of arrays. It is easy to implement functions of type

boolean equalArrays (Object[] a1, Object[] a2);
void shuffleArray(Object[] a);

However, if array types were treated as invariant, it would only be possible to call these functions on an array of exactly the type Object[]. One could not, for example, shuffle an array of strings.

Therefore, both Java and C# treat array types covariantly. For instance, in C# string[] is a subtype of object[], and in Java String[] is a subtype of Object[].

This answers the question "Why are arrays covariant?", or more accurately, "Why were arrays made covariant at the time?"

When generics were introduced, they were purposefully not made covariant for reasons pointed out in this answer by Jon Skeet:

No, a List<Dog> is not a List<Animal>. Consider what you can do with a List<Animal> - you can add any animal to it... including a cat. Now, can you logically add a cat to a litter of puppies? Absolutely not.

// Illegal code - because otherwise life would be Bad
List<Dog> dogs = new List<Dog>();
List<Animal> animals = dogs; // Awooga awooga
animals.add(new Cat());
Dog dog = dogs.get(0); // This should be safe, right?

Suddenly you have a very confused cat.

The original motivation for making arrays covariant described in the wikipedia article didn't apply to generics because wildcards made the expression of covariance (and contravariance) possible, for example:

boolean equalLists(List<?> l1, List<?> l2);
void shuffleList(List<?> l);



回答2:


The reason is that every array knows its element type during runtime, while generic collection doesn't because of type erasure.

For example:

String[] strings = new String[2];
Object[] objects = strings;  // valid, String[] is Object[]
objects[0] = 12; // error, would cause java.lang.ArrayStoreException: java.lang.Integer during runtime

If this was allowed with generic collections:

List<String> strings = new ArrayList<String>();
List<Object> objects = strings;  // let's say it is valid
objects.add(12);  // invalid, Integer should not be put into List<String> but there is no information during runtime to catch this

But this would cause problems later when someone would try to access the list:

String first = strings.get(0); // would cause ClassCastException, trying to assign 12 to String



回答3:


May be this help:-

Generics are not covariant

Arrays in the Java language are covariant -- which means that if Integer extends Number (which it does), then not only is an Integer also a Number, but an Integer[] is also a Number[], and you are free to pass or assign an Integer[] where a Number[] is called for. (More formally, if Number is a supertype of Integer, then Number[] is a supertype of Integer[].) You might think the same is true of generic types as well -- that List<Number> is a supertype of List<Integer>, and that you can pass a List<Integer> where a List<Number> is expected. Unfortunately, it doesn't work that way.

It turns out there's a good reason it doesn't work that way: It would break the type safety generics were supposed to provide. Imagine you could assign a List<Integer> to a List<Number>. Then the following code would allow you to put something that wasn't an Integer into a List<Integer>:

List<Integer> li = new ArrayList<Integer>();
List<Number> ln = li; // illegal
ln.add(new Float(3.1415));

Because ln is a List<Number>, adding a Float to it seems perfectly legal. But if ln were aliased with li, then it would break the type-safety promise implicit in the definition of li -- that it is a list of integers, which is why generic types cannot be covariant.




回答4:


Arrays are covariant for at least two reasons:

  • It is useful for collections that hold information which will never change to be covariant. For a collection of T to be covariant, its backing store must also be covariant. While one could design an immutable T collection which did not use a T[] as its backing store (e.g. using a tree or linked list), such a collection would be unlikely to perform as well as one backed by an array. One might argue that a better way to provide for covariant immutable collections would have been to define a "covariant immutable array" type they could use a backing store, but simply allowing array covariance was probably easier.

  • Arrays will frequently be mutated by code which doesn't know what type of thing is going to be in them, but won't put into the array anything which wasn't read out of that same array. A prime example of this is sorting code. Conceptually it might have been possible for array types to include methods to swap or permute elements (such methods could be equally applicable to any array type), or define an "array manipulator" object which hold a reference to an array and one or more things that had been read from it, and could include methods to store previously-read items into the array from which they had come. If arrays were not covariant, user code would not be able to define such a type, but the runtime could have included some specialized methods.

The fact that arrays are covariant may be viewed as an ugly hack, but in most cases it facilitates the creation of working code.




回答5:


An important feature of parametric types is the ability to write polymorphic algorithms, i.e. algorithms that operate on a data structure regardless of its parameter value, such as Arrays.sort().

With generics, that's done with wildcard types:

<E extends Comparable<E>> void sort(E[]);

To be truly useful, wildcard types require wildcard capture, and that requires the notion of a type parameter. None of that was available at the time arrays were added to Java, and makings arrays of reference type covariant permitted a far simpler way to permit polymorphic algorithms:

void sort(Comparable[]);

However, that simplicity opened a loophole in the static type system:

String[] strings = {"hello"};
Object[] objects = strings;
objects[0] = 1; // throws ArrayStoreException

requiring a runtime check of every write access to an array of reference type.

In a nutshell, the newer approach embodied by generics makes the type system more complex, but also more statically type safe, while the older approach was simpler, and less statically type safe. The designers of the language opted for the simpler approach, having more important things to do than closing a small loophole in the type system that rarely causes problems. Later, when Java was established, and the pressing needs taken care of, they had the resources to do it right for generics (but changing it for arrays would have broken existing Java programs).




回答6:


Generics are invariant: from JSL 4.10:

...Subtyping does not extend through generic types: T <: U does not imply that C<T> <: C<U> ...

and a few lines further, JLS also explains that
Arrays are covariant (first bullet):

4.10.3 Subtyping among Array Types

enter image description here




回答7:


I think they made a wrong decision at the first place that made array covariant. It breaks the type safety as it described here and they got stuck with that because of backward compatibility and after that they tried to not make the same mistake for generic. And that's one of the reasons that Joshua Bloch prefers lists to arra ys in Item 25 of book "Effective Java(second edition)"




回答8:


My take: When code is expecting an array A[] and you give it B[] where B is a subclass of A, there's only two things to worry about: what happens when you read an array element, and what happens if you write it. So it's not hard to write language rules to ensure that type safety is preserved in all cases (the main rule being that an ArrayStoreException could be thrown if you try to stick an A into a B[]). For a generic, though, when you declare a class SomeClass<T>, there can be any number of ways T is used in the body of the class, and I'm guessing it's just way too complicated to work out all the possible combinations to write rules about when things are allowed and when they aren't.




回答9:


We can’t write List<Object> l = new ArrayList<String>(); because Java is trying to protect us from a runtime exception. You might think this would mean that we can’t write Object[] o = new String[0];. That isn’t the case. This code does compile:

Integer[] numbers = { new Integer(42)};
Object[] objects = numbers;
objects[0] = "forty two"; // throws ArrayStoreException

Although the code does compile, it throws an exception at runtime. With arrays, Java knows the type that is allowed in the array. Just because we’ve assigned an Integer[] to an Object[] doesn’t change the fact that Java knows it is really an Integer[].

Due to type erasure, we have no such protection for an ArrayList. At runtime, the ArrayList doesn’t know what is allowed in it. Therefore, Java uses the compiler to prevent this situation from coming up in the first place. OK, so why doesn’t Java add this knowledge to ArrayList? The reason is backward compatibility; that is, Java is big on not breaking existing code.

OCP reference.



来源:https://stackoverflow.com/questions/63096043/generic-list-not-accepting-subtype

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