扩展

List<?数字> foo3意味着这些都是合法的赋值:

List<? extends Number> foo3 = new ArrayList<Number>();  // Number "extends" Number (in this context)
List<? extends Number> foo3 = new ArrayList<Integer>(); // Integer extends Number
List<? extends Number> foo3 = new ArrayList<Double>();  // Double extends Number

Reading - Given the above possible assignments, what type of object are you guaranteed to read from List foo3: You can read a Number because any of the lists that could be assigned to foo3 contain a Number or a subclass of Number. You can't read an Integer because foo3 could be pointing at a List<Double>. You can't read a Double because foo3 could be pointing at a List<Integer>. Writing - Given the above possible assignments, what type of object could you add to List foo3 that would be legal for all the above possible ArrayList assignments: You can't add an Integer because foo3 could be pointing at a List<Double>. You can't add a Double because foo3 could be pointing at a List<Integer>. You can't add a Number because foo3 could be pointing at a List<Integer>.

你不能向List<?extends t>因为你不能保证它真正指向的是什么样的List，所以你不能保证对象被允许在那个List中。唯一的“保证”是你只能读取它，你会得到一个T或T的子类。

超级

现在考虑List <?超级T >。

List<?super Integer> foo3表示这些都是合法的赋值:

List<? super Integer> foo3 = new ArrayList<Integer>();  // Integer is a "superclass" of Integer (in this context)
List<? super Integer> foo3 = new ArrayList<Number>();   // Number is a superclass of Integer
List<? super Integer> foo3 = new ArrayList<Object>();   // Object is a superclass of Integer

Reading - Given the above possible assignments, what type of object are you guaranteed to receive when you read from List foo3: You aren't guaranteed an Integer because foo3 could be pointing at a List<Number> or List<Object>. You aren't guaranteed a Number because foo3 could be pointing at a List<Object>. The only guarantee is that you will get an instance of an Object or subclass of Object (but you don't know what subclass). Writing - Given the above possible assignments, what type of object could you add to List foo3 that would be legal for all the above possible ArrayList assignments: You can add an Integer because an Integer is allowed in any of above lists. You can add an instance of a subclass of Integer because an instance of a subclass of Integer is allowed in any of the above lists. You can't add a Double because foo3 could be pointing at an ArrayList<Integer>. You can't add a Number because foo3 could be pointing at an ArrayList<Integer>. You can't add an Object because foo3 could be pointing at an ArrayList<Integer>.

PECS

记住PECS:“生产者延伸，消费者至上”。

"Producer Extends" - If you need a List to produce T values (you want to read Ts from the list), you need to declare it with ? extends T, e.g. List<? extends Integer>. But you cannot add to this list. "Consumer Super" - If you need a List to consume T values (you want to write Ts into the list), you need to declare it with ? super T, e.g. List<? super Integer>. But there are no guarantees what type of object you may read from this list. If you need to both read from and write to a list, you need to declare it exactly with no wildcards, e.g. List<Integer>.

例子

请注意这个来自Java泛型常见问题解答的例子。注意源列表src(生产列表)如何使用extends，而目标列表dest(消费列表)如何使用super:

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)); 
  } 
}

也看到 如何添加到List<?extends数字>数据结构?

我想把区别形象化。假设我们有:

class A { }
class B extends A { }
class C extends B { }

列表< ?扩展T> -读取和赋值:

|-------------------------|-------------------|---------------------------------|
|         wildcard        |        get        |              assign             |
|-------------------------|-------------------|---------------------------------|
|    List<? extends C>    |    A    B    C    |                       List<C>   |
|-------------------------|-------------------|---------------------------------|
|    List<? extends B>    |    A    B         |             List<B>   List<C>   |
|-------------------------|-------------------|---------------------------------|
|    List<? extends A>    |    A              |   List<A>   List<B>   List<C>   |
|-------------------------|-------------------|---------------------------------|

列表< ?超级T> -写和分配:

|-------------------------|-------------------|-------------------------------------------|
|         wildcard        |        add        |                   assign                  |
|-------------------------|-------------------|-------------------------------------------|
|     List<? super C>     |              C    |  List<Object>  List<A>  List<B>  List<C>  |
|-------------------------|-------------------|-------------------------------------------|
|     List<? super B>     |         B    C    |  List<Object>  List<A>  List<B>           |
|-------------------------|-------------------|-------------------------------------------|
|     List<? super A>     |    A    B    C    |  List<Object>  List<A>                    |
|-------------------------|-------------------|-------------------------------------------|

在所有的情况下:

不管通配符是什么，你总是可以从列表中获得Object。 不管通配符是什么，你总是可以在可变列表中添加null。

Super是下界，extends是上界。

根据http://download.oracle.com/javase/tutorial/extra/generics/morefun.html:

解决方法是使用一种形式的 有界通配符我们还没见过 具有下界的通配符。的 语法呢?超级T表示未知 类型是T(或T 本身;记住超类型 关系是自反的)。它是双重的 有界通配符 使用，我们在哪里使用?将T延伸到 表示未知类型 T的子型。

扩展

List<?数字> foo3意味着这些都是合法的赋值:

List<? extends Number> foo3 = new ArrayList<Number>();  // Number "extends" Number (in this context)
List<? extends Number> foo3 = new ArrayList<Integer>(); // Integer extends Number
List<? extends Number> foo3 = new ArrayList<Double>();  // Double extends Number

Reading - Given the above possible assignments, what type of object are you guaranteed to read from List foo3: You can read a Number because any of the lists that could be assigned to foo3 contain a Number or a subclass of Number. You can't read an Integer because foo3 could be pointing at a List<Double>. You can't read a Double because foo3 could be pointing at a List<Integer>. Writing - Given the above possible assignments, what type of object could you add to List foo3 that would be legal for all the above possible ArrayList assignments: You can't add an Integer because foo3 could be pointing at a List<Double>. You can't add a Double because foo3 could be pointing at a List<Integer>. You can't add a Number because foo3 could be pointing at a List<Integer>.

你不能向List<?extends t>因为你不能保证它真正指向的是什么样的List，所以你不能保证对象被允许在那个List中。唯一的“保证”是你只能读取它，你会得到一个T或T的子类。

超级

现在考虑List <?超级T >。

List<?super Integer> foo3表示这些都是合法的赋值:

List<? super Integer> foo3 = new ArrayList<Integer>();  // Integer is a "superclass" of Integer (in this context)
List<? super Integer> foo3 = new ArrayList<Number>();   // Number is a superclass of Integer
List<? super Integer> foo3 = new ArrayList<Object>();   // Object is a superclass of Integer

Reading - Given the above possible assignments, what type of object are you guaranteed to receive when you read from List foo3: You aren't guaranteed an Integer because foo3 could be pointing at a List<Number> or List<Object>. You aren't guaranteed a Number because foo3 could be pointing at a List<Object>. The only guarantee is that you will get an instance of an Object or subclass of Object (but you don't know what subclass). Writing - Given the above possible assignments, what type of object could you add to List foo3 that would be legal for all the above possible ArrayList assignments: You can add an Integer because an Integer is allowed in any of above lists. You can add an instance of a subclass of Integer because an instance of a subclass of Integer is allowed in any of the above lists. You can't add a Double because foo3 could be pointing at an ArrayList<Integer>. You can't add a Number because foo3 could be pointing at an ArrayList<Integer>. You can't add an Object because foo3 could be pointing at an ArrayList<Integer>.

PECS

记住PECS:“生产者延伸，消费者至上”。

"Producer Extends" - If you need a List to produce T values (you want to read Ts from the list), you need to declare it with ? extends T, e.g. List<? extends Integer>. But you cannot add to this list. "Consumer Super" - If you need a List to consume T values (you want to write Ts into the list), you need to declare it with ? super T, e.g. List<? super Integer>. But there are no guarantees what type of object you may read from this list. If you need to both read from and write to a list, you need to declare it exactly with no wildcards, e.g. List<Integer>.

例子

请注意这个来自Java泛型常见问题解答的例子。注意源列表src(生产列表)如何使用extends，而目标列表dest(消费列表)如何使用super:

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)); 
  } 
}

也看到 如何添加到List<?extends数字>数据结构?

想象一下有这样的层次结构

1. 扩展

通过编写

    List<? extends C2> list;

你是说，list将能够引用类型为ArrayList的对象，其泛型类型是C2的7个子类型之一(包括C2):

C2: new ArrayList<C2>();，(可以存储C2或子类型的对象)或 D1: new ArrayList<D1>();，(可以存储D1或子类型的对象)或 D2: new ArrayList<D2>();，(可以存储D2或子类型的对象)or…

等等。七种不同的情况:

    1) new ArrayList<C2>(): can store C2 D1 D2 E1 E2 E3 E4
    2) new ArrayList<D1>(): can store    D1    E1 E2  
    3) new ArrayList<D2>(): can store       D2       E3 E4
    4) new ArrayList<E1>(): can store          E1             
    5) new ArrayList<E2>(): can store             E2             
    6) new ArrayList<E3>(): can store                E3             
    7) new ArrayList<E4>(): can store                   E4             

对于每种可能的情况，我们都有一组“可存储”类型:这里图形化地表示了7个(红色)集

正如你所看到的，并没有一种安全的类型适用于所有情况:

你不能列举。添加(新的C2 () {});因为可以是list = new ArrayList<D1>(); 你不能列举。添加(新D1 () {});因为可以是list = new ArrayList<D2>();

等等。

2. 超级

通过编写

    List<? super C2> list;

你是说，list将能够引用类型为ArrayList的对象，其泛型类型是C2的7个超类型之一(包括C2):

A1: new ArrayList<A1>();，(可以存储A1或子类型的对象)或 A2: new ArrayList<A2>();，(可以存储A2或子类型的对象)或 A3: new ArrayList<A3>();，(可以存储A3或子类型的对象)or…

等等。七种不同的情况:

    1) new ArrayList<A1>(): can store A1          B1 B2       C1 C2    D1 D2 E1 E2 E3 E4
    2) new ArrayList<A2>(): can store    A2          B2       C1 C2    D1 D2 E1 E2 E3 E4
    3) new ArrayList<A3>(): can store       A3          B3       C2 C3 D1 D2 E1 E2 E3 E4
    4) new ArrayList<A4>(): can store          A4       B3 B4    C2 C3 D1 D2 E1 E2 E3 E4
    5) new ArrayList<B2>(): can store                B2       C1 C2    D1 D2 E1 E2 E3 E4
    6) new ArrayList<B3>(): can store                   B3       C2 C3 D1 D2 E1 E2 E3 E4
    7) new ArrayList<C2>(): can store                            C2    D1 D2 E1 E2 E3 E4

对于每种可能的情况，我们都有一组“可存储”类型:这里图形化地表示了7个(红色)集

正如你所看到的，这里有7种安全类型，在每种情况下都是常见的:C2, D1, D2, E1, E2, E3, E4。

你可以列出。添加(新的C2 () {});因为，不管我们引用的是哪种List, C2都是允许的 你可以列出。添加(新D1 () {});因为，不管我们引用的是哪种List, D1都是允许的

等等。您可能注意到，这些类型对应于从类型C2开始的层次结构。

笔记

如果您希望进行一些测试，这里是完整的层次结构

interface A1{}
interface A2{}
interface A3{}
interface A4{}

interface B1 extends A1{}
interface B2 extends A1,A2{}
interface B3 extends A3,A4{}
interface B4 extends A4{}

interface C1 extends B2{}
interface C2 extends B2,B3{}
interface C3 extends B3{}

interface D1 extends C1,C2{}
interface D2 extends C2{}

interface E1 extends D1{}
interface E2 extends D1{}
interface E3 extends D2{}
interface E4 extends D2{}

的例子, 假设继承顺序为O > S > T > U > V

使用extends关键字，

正确的:

List<? extends T> Object = new List<T>();
List<? extends T> Object = new List<U>();
List<? extends T> Object = new List<V>();

不正确的:

List<? extends T> Object = new List<S>();
List<? extends T> Object = new List<O>();

超级关键字:

正确的:

List<? super T> Object = new List<T>();
List<? super T> Object = new List<S>();
List<? super T> Object = new List<O>();

不正确的:

List<? super T> Object = new List<U>();
List<? super T> Object = new List<V>();

添加对象: List对象= new List();

Object.add(new T()); //error

但是为什么会出错呢? 让我们看看初始化列表对象的可能性

List<? extends T> Object = new List<T>();
List<? extends T> Object = new List<U>();
List<? extends T> Object = new List<V>();

如果我们使用Object。添加(新T ());那么只有当

List<? extends T> Object = new List<T>(); 

但还有另外两种可能

List对象= new List(); List对象= new List(); 如果我们尝试将(new T())添加到上述两个可能性中，它将会给出一个错误，因为T是U和V的高级类。我们尝试添加一个T对象[它是(new T())]到类型为U和V的列表。高级类对象(基类)不能传递给低级对象(子类)。

由于额外的两种可能性，即使你使用了正确的可能性，Java也会给你错误，因为Java不知道你所指的对象是什么，所以你不能向List中添加对象Object = new List();因为有不成立的可能性。

添加对象: List对象= new List();

Object.add(new T()); // compiles fine without error
Object.add(new U()); // compiles fine without error
Object.add(new V()); // compiles fine without error

Object.add(new S()); //  error
Object.add(new O()); //  error

但是为什么上面两个会出现错误呢? 我们可以使用Object。添加(新T ());只有在以下的可能性中，

List<? super T> Object = new List<T>();
List<? super T> Object = new List<S>();
List<? super T> Object = new List<O>();

如果我们尝试使用Object。add(new T()) in List对象= new List(); 而且 List对象= new List(); 那么它就会给出错误 这是因为 我们不能将T对象[which is new T()]添加到List object = new List();因为它是U类型的对象。我们不能给U对象添加一个T对象[它是新的T()]，因为T是一个基类，U是一个子类。我们不能将基类添加到子类，这就是发生错误的原因。另一种情况也是一样的。