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Unit 26: Wildcards

After going through this unit, students should:

  • be aware of the meaning of wildcard ? and bounded wildcards
  • know how to use wildcards to write methods that are more flexible in accepting a range of types
  • know that upper-bounded wildcard is covariant and lower-bounded wildcard is contravariant
  • know the PECS principle and how to apply it
  • be aware that the unbounded wildcard allows us to not use raw types in our programs

contains with Array<T>

Now that we have our Array<T> class, let's modify our generic contains method and replace the type of the argument T[] with Array<T>.

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class A {
  // version 0.5 (with generic array)
  public static <T> boolean contains(Array<T> array, T obj) {
    for (int i = 0; i < array.getLength(); i++) {
      T curr = array.get(i);
      if (curr.equals(obj)) {
        return true;
      }
    }
    return false;
  }
}

Similar to the version that takes in T[], using generics allows us to constrain the type of the elements of the array and the object to search for to be the same. This allows the following code to type-check correctly: 

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Array<String> stringArray;
Array<Circle> circleArray;
Circle circle;
 :
A.<String>contains(stringArray, "hello"); // ok
A.<Circle>contains(circleArray, circle); // ok

But trying to search for a circle in an array of strings would lead to a type error: 

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A.<String>contains(stringArray, circle); // error

Consider now having an array of shapes. 

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Array<Shape> shapeArray;
Array<Circle> circleArray;
Shape shape;
Circle circle;
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A.<Shape>contains(shapeArray, shape); // ok
A.<Circle>contains(circleArray, circle); // ok

As expected, we can pass Shape as the argument for T, and search for a Shape in an instance of Array<Shape>. Similarly, we can pass Circle as the argument for T and search for a Circle in an instance of Array<Circle>.

We could also look for a Circle instance from Array<Shape> if we pass Shape as the argument for T. 

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A.<Shape>contains(shapeArray, circle); // ok

Note that we can pass in a Circle instance as a Shape, since Circle <: Shape.

Recall that generics are invariant in Java, i.e, there is no subtyping relationship between Array<Shape> and Array<Circle>. Array<Circle> is not a subtype of Array<Shape>. Otherwise, it would violate the Liskov Substitution Principle, we can put a square into an Array<Shape> instance, but we can't put a square into an Array<Circle> instance.

So, we can't call: 

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A.<Circle>contains(shapeArray, circle); // compilation error

The following would result in compilation errors as well: 

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A.<Shape>contains(circleArray, shape); // compilation error
A.<Circle>contains(circleArray, shape); // compilation error

Thus, with our current implementation, we can't look for a shape (which may be a circle) in an array of circles, even though this is something reasonable that a programmer might want to do. This constraint is due to the invariance of generics -- while we avoided the possibility of run-time errors by avoiding covariance arrays, our methods have become less general.

Let's see how we can fix this with bounded type parameters first. We can introduce another type parameter, say S, to remove the constraints that the type of the array must be the same as the type of the object to search for. I.e., we change from 

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  public static <T> boolean contains(Array<T> array, T obj) { .. }

to: 

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  public static <S,T> boolean contains(Array<T> array, S obj) { .. }

But we don't want to completely decouple T and S, as we want T to be a subtype of S. We can thus make T a bounded type parameter, and write: 

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  public static <S, T extends S> boolean contains(Array<T> array, S obj) { .. }

Now, we can search for a shape in an array of circles. 

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     A.<Shape,Circle>contains(circleArray, shape);

Copying to and from Array<T>

Let's consider another example. Let's add two methods copyFrom and copyTo, to Array<T> so that we can copy to and from one array to another.

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// version 0.4 (with copy)
class Array<T> {
  private T[] array;

  Array(int size) {
  // The only way we can put an object into the array is through
  // the method set() and we only put an object of type T inside.
  // So it is safe to cast `Object[]` to `T[]`.
  @SuppressWarnings("unchecked")
    T[] a = (T[]) new Object[size];
  this.array = a;
  }

  public void set(int index, T item) {
    this.array[index] = item;
  }

  public T get(int index) {
    return this.array[index];
  }

  public void copyFrom(Array<T> src) {
    int len = Math.min(this.array.length, src.array.length);
    for (int i = 0; i < len; i++) {
        this.set(i, src.get(i));
    }
  }

  public void copyTo(Array<T> dest) {
    int len = Math.min(this.array.length, dest.array.length);
    for (int i = 0; i < len; i++) {
        dest.set(i, this.get(i));
    }
  }
}

With this implementation, we can copy, say, an Array<Circle> to another Array<Circle>, an Array<Shape> to another Array<Shape>, but not an Array<Circle> into an Array<Shape>, even though each circle is a shape!

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Array<Circle> circleArray;
Array<Shape> shapeArray;
  :
shapeArray.copyFrom(circleArray); // error
circleArray.copyTo(shapeArray); // error

Upper-Bounded Wildcards

Let's consider the method copyFrom. We should be able to copy from an array of shapes, an array of circles, an array of squares, etc, into an array of shapes. In other words, we should be able to copy from an array of any subtype of shapes into an array of shapes. Is there such a type in Java?

The type that we are looking for is Array<? extends Shape>. This generic type uses the wildcard ?. Just like a wild card in card games, it is a substitute for any type. A wildcard can be bounded. Here, this wildcard is upper-bounded by Shape, i.e., it can be substituted with either Shape or any subtype of Shape.

The upper-bounded wildcard is an example of covariance. The upper-bounded wildcard has the following subtyping relations:

  • If S <: T, then A<? extends S> <: A<? extends T> (covariance)
  • For any type S, A<S> <: A<? extends S>

For instance, we have:

  • Array<Circle> <: Array<? extends Circle>
  • Since Circle <: Shape, Array<? extends Circle> <: Array<? extends Shape>
  • Since subtyping is transitive, we have Array<Circle> <: Array<? extends Shape>

Because Array<Circle> <: Array<? extends Shape>, if we change the type of the parameter to copyFrom to Array<? extends T>, 

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  public void copyFrom(Array<? extends T> src) {
    int len = Math.min(this.array.length, src.array.length);
    for (int i = 0; i < len; i++) {
        this.set(i, src.get(i));
    }
  }

We can now call: 

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shapeArray.copyFrom(circleArray); // ok

without error.

Lower-Bounded Wildcards

Let's now try to allow copying of an Array<Circle> to Array<Shape>. 

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circleArray.copyTo(shapeArray);

by doing the same thing: 

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  public void copyTo(Array<? extends T> dest) {
    int len = Math.min(this.array.length, dest.array.length);
    for (int i = 0; i < len; i++) {
        dest.set(i, this.get(i));
    }
  }

The code above would not compile. We will get the following somewhat cryptic message when we compile with the -Xdiags:verbose flag: 

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Array.java:32: error: method set in class Array<T> cannot be applied to given types;
        dest.set(i, this.get(i));
          ^
  required: int,CAP#1
  found: int,T
  reason: argument mismatch; T cannot be converted to CAP#1
  where T is a type-variable:
    T extends Object declared in class Array
  where CAP#1 is a fresh type-variable:
    CAP#1 extends T from capture of ? extends T
1 error

Let's try not to understand what the error message means first, and think about what could go wrong if the compiler allows: 

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        dest.set(i, this.get(i));

Here, we are trying to put an instance with compile-time type T into an array that contains elements with the compile-time type of T or subtype of T.

The copyTo method of Array<Shape> would allow an Array<Circle> as an argument, and we would end up putting instance with compile-time type Shape into Array<Circle>. If all the shapes are circles, we are fine, but there might be other shapes (rectangles, squares) in this instance of Array<Shape>, and we can't fit them into Array<Circle>! Thus, the line 

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        dest.set(i, this.get(i));

is not type-safe and could lead to ClassCastException during run-time.

Where can we copy our shapes into? We can only copy them safely into an Array<Shape>, Array<Object>, Array<GetAreable>, for instance. In other words, into arrays containing Shape or supertype of Shape.

We need a wildcard lower-bounded by Shape, and Java's syntax for this is ? super Shape. Using this new notation, we can replace the type for dest with:

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  public void copyTo(Array<? super T> dest) {
    int len = Math.min(this.array.length, dest.array.length);
    for (int i = 0; i < len; i++) {
        dest.set(i, this.get(i));
    }
  }

The code would now type-check and compile.

The lower-bounded wildcard is an example of contravariance. We have the following subtyping relations:

  • If S <: T, then A<? super T> <: A<? super S> (contravariance)
  • For any type S, A<S> <: A<? super S>

For instance, we have:

  • Array<Shape> <: Array<? super Shape>
  • Since Circle <: Shape, Array<? super Shape> <: Array<? super Circle>
  • Since subtyping is transitive, we have Array<Shape> <: Array<? super Circle>

The line of code below now compiles: 

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circleArray.copyTo(shapeArray);

Our new Array<T> is now 

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// version 0.5 (with flexible copy using wildcards)
class Array<T> {
  private T[] array;

  Array(int size) {
  // The only way we can put an object into the array is through
  // the method set() and we only put an object of type T inside.
  // So it is safe to cast `Object[]` to `T[]`.
  @SuppressWarnings("unchecked")
    T[] a = (T[]) new Object[size];
  this.array = a;
  }

  public void set(int index, T item) {
    this.array[index] = item;
  }

  public T get(int index) {
    return this.array[index];
  }

  public void copyFrom(Array<? extends T> src) {
    int len = Math.min(this.array.length, src.array.length);
    for (int i = 0; i < len; i++) {
        this.set(i, src.get(i));
    }
  }

  public void copyTo(Array<? super T> dest) {
    int len = Math.min(this.array.length, dest.array.length);
    for (int i = 0; i < len; i++) {
        dest.set(i, this.get(i));
    }
  }
}

PECS

Now we will introduce the rule that governs when we should use the upper-bounded wildcard ? extends T and a lower-bounded wildcard ? super T. It depends on the role of the variable. If the variable is a producer that returns a variable of type T, it should be declared with the wildcard ? extends T. Otherwise, if it is a consumer that accepts a variable of type T, it should be declared with the wildcard ? super T.

As an example, the variable src in copyFrom above acts as a producer. It produces a variable of type T. The type parameter for src must be either T or a subtype of T to ensure type safety. So the type for src is Array<? extends T>.

On the other hand, the variable dest in copyTo above acts as a consumer. It consumes a variable of type T. The type parameter of dest must be either T or supertype of T for it to be type-safe. As such, the type for dest is Array<? super T>.

This rule can be remembered with the mnemonic PECS, or "Producer Extends; Consumer Super".

Unbounded Wildcards

It is also possible to have unbounded wildcards, such as Array<?>. Array<?> is the supertype of every parameterized type of Array<T>. Recall that Object is the supertype of all reference types. When we want to write a method that takes in a reference type, but we want the method to be flexible enough, we can make the method accept a parameter of type Object. Similarly, Array<?> is useful when you want to write a method that takes in an array of some specific type, and you want the method to be flexible enough to take in an array of any type. For instance, if we have:

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void foo(Array<?> array) {
}

We could call it with: 

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Array<Circle> ac;
Array<String> as;
foo(ac); // ok
foo(as); // ok

A method that takes in generic type with unbounded wildcard would be pretty restrictive, however. Consider this: 

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void foo(Array<?> array) {
   :
  x = array.get(0);
  array.set(0, y);

}

What should the type of the returned element x be? Since Array<?> is the supertype of all possible Array<T>, the method foo can receive an instance of Array<Circle>, Array<String>, etc. as an argument. The only safe choice for the type of x is Object.

The type for y is every more restrictive. Since there are many possibilities of what type of array it is receiving, we can only put null into array!

There is an important distinction to be made between Array, Array<?> and Array<Object>. Whilst Object is the supertype of all T, it does not follow that Array<Object> is the supertype of all Array<T> due to generics being invariant. Therefore, the following statements will fail to compile:

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Array<Object> a1 = new Array<String>(0); 
Array<Object> a2 = new Array<Integer>(0);

Whereas the following statements will compile:

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Array<?> a1 = new Array<String>(0); // Does compile
Array<?> a2 = new Array<Integer>(0); // Does compile

If we have a function 

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void bar(Array<Object> array) {
}

Then, the method bar is restricted to only takes in an Array<Object> instance as argument. 

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Array<Circle> ac;
Array<String> as;
bar(ac); // compilation error
bar(as); // compilation error

What about raw types? Suppose we write the method below that accepts a raw type 

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void qux(Array array) {
}

Then, the method qux is also flexible enough to take in any Array<T> as argument. 

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Array<Circle> ac;
Array<String> as;
qux(ac); 
qux(as);

Unlike Array<?>, however, the compiler does not have the information about the type of the component of the array, and cannot type check for us. It is up to the programmer to ensure type safety. For this reason, we must not use raw types.

Intuitively, we can think of Array<?>, Array<Object>, and Array as follows:

  • Array<?> is an array of objects of some specific, but unknown type;
  • Array<Object> is an array of Object instances, with type checking by the compiler;
  • Array is an array of Object instances, without type checking.

Back to contains

Now, let's simplify our contains methods with the help of wildcards. Recall that to add flexibility into the method parameter and allow us to search for a shape in an array of circles, we have modified our method into the following:

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class A {
  // version 0.6 (with generic array)
  public static <S,T extends S> boolean contains(Array<T> array, S obj) {
    for (int i = 0; i < array.getLength(); i++) {
      T curr = array.get(i);
      if (curr.equals(obj)) {
        return true;
      }
    }
    return false;
  }
}

Can we make this simpler using wildcards? Since we want to search for an object of type S in an array of its subtype, we can remove the second parameter type T and change the type of array to Array<? extends S>:

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class A {
  // version 0.7 (with wild cards array)
  public static <S> boolean contains(Array<? extends S> array, S obj) {
    for (int i = 0; i < array.getLength(); i++) {
      S curr = array.get(i);
      if (curr.equals(obj)) {
        return true;
      }
    }
    return false;
  }
}

We can double-check that array is a producer (it produces curr on Line 5) and this follows the PECS rules. Now, we can search for a shape in an array of circles. 

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     A.<Shape>contains(circleArray, shape);

Revisiting Raw Types

In previous units, we said that you may use raw types only in two scenarios. Namely, when using generics and instanceof together, and when creating arrays. However, with unbounded wildcards, we can now see it is possible to remove both of these exceptions. We can now use instanceof in the following way:

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a instanceof A<?>

Recall that in the example above, instanceof checks of the run-time type of a. Previously, we said that we can't check for, say, 

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a instanceof A<String>

since the type argument String is not available during run-time due to erasure. Using <?> fits the purpose here because it explicitly communicates to the reader of the code that we are checking that a is an instance of A with some unknown (erased) type parameter.

Similarly, we can create arrays in the following way: 

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new Comparable<?>[10];

Previously, we said that we could not create an array using the expression new Comparable<String>[10] because generics and arrays do not mix well. Java insists that the array creation expression uses a reifiable type, i.e., a type where no type information is lost during compilation. Unlike Comparable<String>, however, Comparible<?> is reifiable. Since we don't know what is the type of ?, no type information is lost during erasure!

Going forward now in the module, we will not permit the use of raw types in any scenario.