Covariance, Invariance and Contravariance explained in plain English?

Solution 1:

Some say it is about relationship between types and subtypes, other say it is about type conversion and others say it is used to decide whether a method is overwritten or overloaded.

All of the above.

At heart, these terms describe how the subtype relation is affected by type transformations. That is, if A and B are types, f is a type transformation, and ≤ the subtype relation (i.e. A ≤ B means that A is a subtype of B), we have

• f is covariant if A ≤ B implies that f(A) ≤ f(B)
• f is contravariant if A ≤ B implies that f(B) ≤ f(A)
• f is invariant if neither of the above holds

Let's consider an example. Let f(A) = List<A> where List is declared by

class List<T> { ... } 

Is f covariant, contravariant, or invariant? Covariant would mean that a List<String> is a subtype of List<Object>, contravariant that a List<Object> is a subtype of List<String> and invariant that neither is a subtype of the other, i.e. List<String> and List<Object> are inconvertible types. In Java, the latter is true, we say (somewhat informally) that generics are invariant.

Another example. Let f(A) = A[]. Is f covariant, contravariant, or invariant? That is, is String[] a subtype of Object[], Object[] a subtype of String[], or is neither a subtype of the other? (Answer: In Java, arrays are covariant)

This was still rather abstract. To make it more concrete, let's look at which operations in Java are defined in terms of the subtype relation. The simplest example is assignment. The statement

x = y;

will compile only if typeof(y) ≤ typeof(x). That is, we have just learned that the statements

ArrayList<String> strings = new ArrayList<Object>();
ArrayList<Object> objects = new ArrayList<String>();

will not compile in Java, but

Object[] objects = new String[1];

will.

Another example where the subtype relation matters is a method invocation expression:

result = method(a);

Informally speaking, this statement is evaluated by assigning the value of a to the method's first parameter, then executing the body of the method, and then assigning the methods return value to result. Like the plain assignment in the last example, the "right hand side" must be a subtype of the "left hand side", i.e. this statement can only be valid if typeof(a) ≤ typeof(parameter(method)) and returntype(method) ≤ typeof(result). That is, if method is declared by:

Number[] method(ArrayList<Number> list) { ... }

none of the following expressions will compile:

Integer[] result = method(new ArrayList<Integer>());
Number[] result = method(new ArrayList<Integer>());
Object[] result = method(new ArrayList<Object>());

but

Number[] result = method(new ArrayList<Number>());
Object[] result = method(new ArrayList<Number>());

will.

Another example where subtyping matters is overriding. Consider:

Super sup = new Sub();
Number n = sup.method(1);

where

class Super {
    Number method(Number n) { ... }
}

class Sub extends Super {
    @Override 
    Number method(Number n);
}

Informally, the runtime will rewrite this to:

class Super {
    Number method(Number n) {
        if (this instanceof Sub) {
            return ((Sub) this).method(n);  // *
        } else {
            ... 
        }
    }
}

For the marked line to compile, the method parameter of the overriding method must be a supertype of the method parameter of the overridden method, and the return type a subtype of the overridden method's one. Formally speaking, f(A) = parametertype(method asdeclaredin(A)) must at least be contravariant, and if f(A) = returntype(method asdeclaredin(A)) must at least be covariant.

Note the "at least" above. Those are minimum requirements any reasonable statically type safe object oriented programming language will enforce, but a programming language may elect to be more strict. In the case of Java 1.4, parameter types and method return types must be identical (except for type erasure) when overriding methods, i.e. parametertype(method asdeclaredin(A)) = parametertype(method asdeclaredin(B)) when overriding. Since Java 1.5, covariant return types are permitted when overriding, i.e. the following will compile in Java 1.5, but not in Java 1.4:

class Collection {
    Iterator iterator() { ... }
}

class List extends Collection {
    @Override 
    ListIterator iterator() { ... }
}

I hope I covered everything - or rather, scratched the surface. Still I hope it will help to understand the abstract, but important concept of type variance.

Solution 2:

Variance is about relationships between classes with different generics parameters. Their relationships are the reason why we can cast them.

Co and Contra variance are pretty logical things. Language type system forces us to support real life logic. It's easy to understand by example.

Covariance

For instance you want to buy a flower and you have two flowers shop in your city: rose shop and daisy shop.

If you ask someone "where is the flowers shop?" and someone tells you where is rose shop, would it be okay? yes, because rose is a flower, if you want to buy a flower you can buy a rose. The same applies if someone replied you with the address of the daisy shop. This is example of covariance: you are allowed to cast A<C> to A<B>, where C is a subclass of B, if A produces generic values (returns as a result from the function). Covariance is about producers.

Types:

class Flower {  }
class Rose extends Flower { }
class Daisy extends Flower { }

interface FlowerShop<T extends Flower> {
    T getFlower();
}

class RoseShop implements FlowerShop<Rose> {
    @Override
    public Rose getFlower() {
        return new Rose();
    }
}

class DaisyShop implements FlowerShop<Daisy> {
    @Override
    public Daisy getFlower() {
        return new Daisy();
    }
}

Question is "where is the flower shop?", answer is "rose shop there":

static FlowerShop<? extends Flower> tellMeShopAddress() {
    return new RoseShop();
}

Contravariance

For instance you want to gift flower to your girlfriend. If you girlfriend loves any flower, can you consider her as a person who loves roses, or as a person who loves daisies? yes, because if she loves any flower she would love both rose and daisy. This is an example of the contravariance: you’re allowed to cast A<B> to A<C>, where C is subclass of B, if A consumes generic value. Contravariance is about consumers.

Types:

interface PrettyGirl<TFavouriteFlower extends Flower> {
    void takeGift(TFavouriteFlower flower);
}

class AnyFlowerLover implements PrettyGirl<Flower> {
    @Override
    public void takeGift(Flower flower) {
        System.out.println("I like all flowers!");
    }

}

You're considering your girlfriend who loves any flower as someone who loves roses, and giving her a rose:

PrettyGirl<? super Rose> girlfriend = new AnyFlowerLover();
girlfriend.takeGift(new Rose());

You can find more at the Source.