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?

根据Bert F的回答，我想解释一下我的理解。

假设我们有3个类

public class Fruit{}

public class Melon extends Fruit{}

public class WaterMelon extends Melon{}

这里我们有

List<? extends Fruit> fruitExtendedList = …

//Says that I can be a list of any object as long as this object extends Fruit.

好的，现在让我们尝试从fruitExtendedList中获取一些值

Fruit fruit = fruitExtendedList.get(position)

//This is valid as it can only return Fruit or its subclass.

让我们再试一次

Melon melon = fruitExtendedList.get(position)

//This is not valid because fruitExtendedList can be a list of Fruit only, it may not be 
//list of Melon or WaterMelon and in java we cannot assign sub class object to 
//super class object reference without explicitly casting it.

的情况也是一样

WaterMelon waterMelon = fruitExtendedList.get(position)

现在让我们尝试在fruitExtendedList中设置一些对象

添加水果对象

fruitExtendedList.add(new Fruit())

//This in not valid because as we know fruitExtendedList can be a list of any 
//object as long as this object extends Fruit. So what if it was the list of  
//WaterMelon or Melon you cannot add Fruit to the list of WaterMelon or Melon.

添加甜瓜对象

fruitExtendedList.add(new Melon())

//This would be valid if fruitExtendedList was the list of Fruit but it may 
//not be, as it can also be the list of WaterMelon object. So, we see an invalid 
//condition already.

最后让我们尝试添加西瓜对象

fruitExtendedList.add(new WaterMelon())

//Ok, we got it now we can finally write to fruitExtendedList as WaterMelon 
//can be added to the list of Fruit or Melon as any superclass reference can point 
//to its subclass object.

但是等等，如果有人决定制造一种新的柠檬，让我们说，为了争论，SaltyLemon作为

public class SaltyLemon extends Lemon{}

现在fruitExtendedList可以是Fruit, Melon, WaterMelon或SaltyLemon的列表。

因此，我们的声明

fruitExtendedList.add(new WaterMelon())

也是无效的。

基本上，我们可以说不能向fruitExtendedList中写入任何内容。

这对List<?扩展了水果>

现在让我们看看

List<? super Melon> melonSuperList= …

//Says that I can be a list of anything as long as its object has super class of Melon.

现在让我们尝试从melonSuperList中获取一些值

Fruit fruit = melonSuperList.get(position)

//This is not valid as melonSuperList can be a list of Object as in java all 
//the object extends from Object class. So, Object can be super class of Melon and 
//melonSuperList can be a list of Object type

同样地，Melon、Melon或其他任何对象都不能读取。

但请注意，我们可以读取对象类型实例

Object myObject = melonSuperList.get(position)

//This is valid because Object cannot have any super class and above statement 
//can return only Fruit, Melon, WaterMelon or Object they all can be referenced by
//Object type reference.

现在，让我们尝试从melonSuperList中设置一些值。

添加对象类型对象

melonSuperList.add(new Object())

//This is not valid as melonSuperList can be a list of Fruit or Melon.
//Note that Melon itself can be considered as super class of Melon.

添加水果类型对象

melonSuperList.add(new Fruit())

//This is also not valid as melonSuperList can be list of Melon

添加Melon类型对象

melonSuperList.add(new Melon())

//This is valid because melonSuperList can be list of Object, Fruit or Melon and in 
//this entire list we can add Melon type object.

添加西瓜类型对象

melonSuperList.add(new WaterMelon())

//This is also valid because of same reason as adding Melon

综上所述，我们可以在melonSuperList中添加Melon或其子类，并且只读取Object类型的对象。

这里最令人困惑的是，无论我们指定了什么类型限制，赋值只能以一种方式工作:

baseClassInstance = derivedClassInstance;

您可能认为Integer扩展了Number，并且Integer可以作为<?扩展数字>，但是编译器会告诉你<?extends Number>不能转换为Integer(也就是说，在人类的说法中，任何扩展Number的东西都可以转换为Integer是错误的):

class Holder<T> {
    T v;
    T get() { return v; }
    void set(T n) { v=n; }
}
class A {
    public static void main(String[]args) {
        Holder<? extends Number> he = new Holder();
        Holder<? super Number> hs = new Holder();

        Integer i;
        Number n;
        Object o;

        // Producer Super: always gives an error except
        //       when consumer expects just Object
        i = hs.get(); // <? super Number> cannot be converted to Integer
        n = hs.get(); // <? super Number> cannot be converted to Number
                      // <? super Number> cannot be converted to ... (but
                      //       there is no class between Number and Object)
        o = hs.get();

        // Consumer Super
        hs.set(i);
        hs.set(n);
        hs.set(o); // Object cannot be converted to <? super Number>

        // Producer Extends
        i = he.get(); // <? extends Number> cannot be converted to Integer
        n = he.get();
        o = he.get();

        // Consumer Extends: always gives an error
        he.set(i); // Integer cannot be converted to <? extends Number>
        he.set(n); // Number cannot be converted to <? extends Number>
        he.set(o); // Object cannot be converted to <? extends Number>
    }
}

hs.set(我);是可以的，因为Integer可以转换为Number的任何超类(而不是因为Integer是Number的超类，这不是真的)。

EDIT添加了一条关于消费者扩展和生产者超级的注释——它们没有意义，因为它们相应地指定了什么，而只是对象。建议您记住PECS，因为CEPS从来都没有用。

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{}

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?

我喜欢来自@Bert F的答案，但这是我大脑看到的方式。

我手里有个X。如果我想把我的X写进一个列表，这个列表需要是一个X的列表，或者是一个我写X时可以向上转换的列表，即X的任何超类…

List<? super   X>

如果我得到一个列表，我想从这个列表中读取一个X，那最好是一个X的列表，或者是一个可以在我读取它们时向上转换为X的列表，即任何扩展X的东西

List<? extends X>