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Java generics explained

巴扎黑

Apr 30, 2017 am 10:03 AM

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Introduction

Generics are a very important knowledge point in Java. Generics are widely used in the Java collection class framework. In this article, we will look at the design of Java generics from scratch, which will involve wildcard processing and annoying type erasure.

Generic basics

Generic class

We first define a simple Box class:

public class Box {
    private String object;
    public void set(String object) { this.object = object; }
    public String get() { return object; }
}

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This is the most common approach. A disadvantage of this is that only String type elements can be loaded into Box now. If we need to load other types of elements such as Integer in the future, we must rewrite another Box, and the code will be complicated. When it comes to reuse, using generics can solve this problem very well.

public class Box<T> {
    // T stands for "Type"
    private T t;
    public void set(T t) { this.t = t; }
    public T get() { return t; }
}

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In this way, our Box class can be reused, and we can replace T with any type we want:

Box<Integer> integerBox = new Box<Integer>();
Box<Double> doubleBox = new Box<Double>();
Box<String> stringBox = new Box<String>();

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Generic method

After reading about generic classes, let’s take a look at generic methods. Declaring a generic method is very simple, just add a form like in front of the return type:

public class Util {
    public static <K, V> boolean compare(Pair<K, V> p1, Pair<K, V> p2) {
        return p1.getKey().equals(p2.getKey()) &&
               p1.getValue().equals(p2.getValue());
    }
}
public class Pair<K, V> {
    private K key;
    private V value;
    public Pair(K key, V value) {
        this.key = key;
        this.value = value;
    }
    public void setKey(K key) { this.key = key; }
    public void setValue(V value) { this.value = value; }
    public K getKey()   { return key; }
    public V getValue() { return value; }
}

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We can call generic methods like this:

Pair<Integer, String> p1 = new Pair<>(1, "apple");
Pair<Integer, String> p2 = new Pair<>(2, "pear");
boolean same = Util.<Integer, String>compare(p1, p2);

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Or use type inference in Java1.7/1.8 to let Java automatically deduce the corresponding type parameters:

Pair<Integer, String> p1 = new Pair<>(1, "apple");
Pair<Integer, String> p2 = new Pair<>(2, "pear");
boolean same = Util.compare(p1, p2);

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Boundary character

Now we want to implement such a function to find the number of elements in a generic array that is greater than a specific element. We can implement it like this:

public static <T> int countGreaterThan(T[] anArray, T elem) {
    int count = 0;
    for (T e : anArray)
        if (e > elem)  // compiler error
            ++count;
    return count;
}

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But this is obviously wrong, because except for primitive types such as short, int, double, long, float, byte, char, etc., other classes may not be able to use the operator >, so the compiler reports an error, so how to solve this problem Woolen cloth? The answer is to use boundary characters.

public interface Comparable<T> {
    public int compareTo(T o);
}

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Making a statement similar to the following is equivalent to telling the compiler that the type parameter T represents classes that implement the Comparable interface, which is equivalent to telling the compiler that they all implement at least the compareTo method.

public static <T extends Comparable<T>> int countGreaterThan(T[] anArray, T elem) {
    int count = 0;
    for (T e : anArray)
        if (e.compareTo(elem) > 0)
            ++count;
    return count;
}

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Wildcard

Before understanding wildcards, we must first clarify a concept, or borrow the Box class we defined above, assuming we add a method like this:

public void boxTest(Box<Number> n) { /* ... */ }

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So what types of parameters does Box n allow to accept now? Can we pass in Box or Box? The answer is no. Although Integer and Double are subclasses of Number, there is no relationship between Box or Box and Box in generics. This is very important. Next, let’s deepen our understanding through a complete example.

First we define a few simple classes, which we will use below:

class Fruit {}
class Apple extends Fruit {}
class Orange extends Fruit {}

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In the following example, we create a generic class Reader, and then in f1() when we try Fruit f = fruitReader.readExact(apples); the compiler will report an error because there is a gap between List and List It doesn't have any relationship.

public class GenericReading {
    static List<Apple> apples = Arrays.asList(new Apple());
    static List<Fruit> fruit = Arrays.asList(new Fruit());
    static class Reader<T> {
        T readExact(List<T> list) {
            return list.get(0);
        }
    }
    static void f1() {
        Reader<Fruit> fruitReader = new Reader<Fruit>();
        // Errors: List<Fruit> cannot be applied to List<Apple>.
        // Fruit f = fruitReader.readExact(apples);
    }
    public static void main(String[] args) {
        f1();
    }
}

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But according to our usual thinking habits, there must be a connection between Apple and Fruit, but the compiler cannot recognize it. So how to solve this problem in generic code? We can solve this problem by using wildcard characters:

static class CovariantReader<T> {
    T readCovariant(List<? extends T> list) {
        return list.get(0);
    }
}
static void f2() {
    CovariantReader<Fruit> fruitReader = new CovariantReader<Fruit>();
    Fruit f = fruitReader.readCovariant(fruit);
    Fruit a = fruitReader.readCovariant(apples);
}
public static void main(String[] args) {
    f2();
}

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This is equivalent to telling the compiler that the parameters accepted by the readCovariant method of fruitReader only need to be a subclass of Fruit (including Fruit itself), so that the relationship between the subclass and the parent class is also related.

PECS principle

Above we saw usage similar to , using which we can get elements from the list, so can we add elements to the list? Let’s try it:

public class GenericsAndCovariance {
    public static void main(String[] args) {
        // Wildcards allow covariance:
        List<? extends Fruit> flist = new ArrayList<Apple>();
        // Compile Error: can&#39;t add any type of object:
        // flist.add(new Apple())
        // flist.add(new Orange())
        // flist.add(new Fruit())
        // flist.add(new Object())
        flist.add(null); // Legal but uninteresting
        // We Know that it returns at least Fruit:
        Fruit f = flist.get(0);
    }
}

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The answer is no, the Java compiler does not allow us to do this, why? We might as well consider this issue from the perspective of the compiler. Because List flist itself can have multiple meanings:

List<? extends Fruit> flist = new ArrayList<Fruit>();
List<? extends Fruit> flist = new ArrayList<Apple>();
List<? extends Fruit> flist = new ArrayList<Orange>();

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When we try to add an Apple, flist may point to new ArrayList();
## When we try to add an Orange, flist may point to new ArrayList();

What should we do if we want to add elements? You can use :

public class GenericWriting {
    static List<Apple> apples = new ArrayList<Apple>();
    static List<Fruit> fruit = new ArrayList<Fruit>();
    static <T> void writeExact(List<T> list, T item) {
        list.add(item);
    }
    static void f1() {
        writeExact(apples, new Apple());
        writeExact(fruit, new Apple());
    }
    static <T> void writeWithWildcard(List<? super T> list, T item) {
        list.add(item)
    }
    static void f2() {
        writeWithWildcard(apples, new Apple());
        writeWithWildcard(fruit, new Apple());
    }
    public static void main(String[] args) {
        f1(); f2();
    }
}

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　　这样我们可以往容器里面添加元素了，但是使用super的坏处是以后不能get容器里面的元素了，原因很简单，我们继续从编译器的角度考虑这个问题，对于List list，它可以有下面几种含义：

List<? super Apple> list = new ArrayList<Apple>();
List<? super Apple> list = new ArrayList<Fruit>();
List<? super Apple> list = new ArrayList<Object>();

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　　当我们尝试通过list来get一个Apple的时候，可能会get得到一个Fruit，这个Fruit可以是Orange等其他类型的Fruit。

　　根据上面的例子，我们可以总结出一条规律，”Producer Extends, Consumer Super”：

“Producer Extends” – 如果你需要一个只读List，用它来produce T，那么使用? extends T。
“Consumer Super” – 如果你需要一个只写List，用它来consume T，那么使用? super T。
如果需要同时读取以及写入，那么我们就不能使用通配符了。

　　如何阅读过一些Java集合类的源码，可以发现通常我们会将两者结合起来一起用，比如像下面这样：

public class Collections {
    public static <T> void copy(List<? super T> dest, List<? extends T> src) {
        for (int i=0; i<src.size(); i++)
            dest.set(i, src.get(i));
    }
}

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　类型擦除

　　Java泛型中最令人苦恼的地方或许就是类型擦除了，特别是对于有C++经验的程序员。类型擦除就是说Java泛型只能用于在编译期间的静态类型检查，然后编译器生成的代码会擦除相应的类型信息，这样到了运行期间实际上JVM根本就知道泛型所代表的具体类型。这样做的目的是因为Java泛型是1.5之后才被引入的，为了保持向下的兼容性，所以只能做类型擦除来兼容以前的非泛型代码。对于这一点，如果阅读Java集合框架的源码，可以发现有些类其实并不支持泛型。

　　说了这么多，那么泛型擦除到底是什么意思呢？我们先来看一下下面这个简单的例子：

public class Node<T> {
    private T data;
    private Node<T> next;
    public Node(T data, Node<T> next) }
        this.data = data;
        this.next = next;
    }
    public T getData() { return data; }
    // ...
}

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　　编译器做完相应的类型检查之后，实际上到了运行期间上面这段代码实际上将转换成：

public class Node {
    private Object data;
    private Node next;
    public Node(Object data, Node next) {
        this.data = data;
        this.next = next;
    }
    public Object getData() { return data; }
    // ...
}

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　　这意味着不管我们声明Node还是Node，到了运行期间，JVM统统视为Node