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。

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?

何时使用extends和super

通配符在方法参数中最有用。它们允许方法接口中必要的灵活性。

人们经常混淆什么时候使用扩展，什么时候使用超边界。经验法则是“得到-放”原则。如果您从参数化容器中获取某些内容，请使用extends。

int totalFuel(List<? extends Vehicle> list) {
int total = 0;
for(Vehicle v : list) {
    total += v.getFuel();
}
return total;}

totalFuel方法从列表中获取车辆，询问它们有多少燃料，并计算总数。 如果将对象放入参数化容器中，请使用super。

int totalValue(Valuer<? super Vehicle> valuer) {
int total = 0;
for(Vehicle v : vehicles) {
    total += valuer.evaluate(v);
}
return total;}

totalValue方法将车辆放入Valuer中。 知道扩展界比super更常见是很有用的。

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。

你可以通过上面所有的答案来理解为什么.add()被限制为'<?>”、“< ?扩展>'，部分扩展到'<?超级>”。

但如果你想记住它，而不想每次都去探索答案，这里是所有问题的结论:

列表< ?extends A>表示它将接受A的任何List和A的子类。 但是你不能在这个列表中添加任何东西。甚至不是A类型的对象。

列表< ?super A>表示它将接受A的任何列表和A的超类。 您可以添加类型A的对象及其子类。