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?

使用只能从集合中获得的扩展。你不能投入进去。此外，虽然超级允许获取和放置，在获取期间的返回类型是?超级T。

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。

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。

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。

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>