问题描述
我知道关于这个话题有很多问题，但不幸的是，他们无法帮助我消除我的隐晦。首先，看下面的例子。我不明白，为什么下面的添加 - 方法 someCage.add（rat1）不起作用，并且以下例外情况中止：
这是为什么 Cage< Rat> 不是 Cage< Animal> ？如果是的话，我不明白这个例子，所以我不确定编译器究竟做了什么。代码示例如下：
package exe;
导入cage.Cage;
进口动物。动物;
进口动物.Ape;
进口动物。狮子;
进口animals.Rat;
public class Main {
public static void main（String [] args）{
Lion lion1 = new Lion（true，4，Lion King， 8）;
狮子狮子2 =新狮子（true，4，动物之王，9）;
Ape ape1 =新猿（true，2，Gimpanse，true）;
Ape ape2 =新猿（true，2，Orang Utan，true）;
大鼠rat1 =新大鼠（true，4，RatBoy，true）;
大鼠rat2 =新大鼠（true，4，RatGirl，true）;
大鼠rat3 =新大鼠（true，4，RatChild，true）;
笼子<动物> animalCage = new Cage< Animal>（）;
animalCage.add（rat2）;
animalCage.add（lion2）;
笼子< Rat> ratCage = new Cage< Rat>（）;
ratCage.add（rat3）;
ratCage.add（rat1）;
ratCage.add（rat2）;
// ratCage.add（lion1）; //不可能。狮子不是老鼠
笼子<狮子> lionCage = new Cage< Lion>（）;
lionCage.add（lion2）;
lionCage.add（lion1）;
笼子 someCage = ratCage; // OK
// someCage = animalCage; // OK
someCage.add（rat1）; //不可能，但是为什么？
animalCage.showAnimals（）;
System.out.println（\\\
RatCage ........）;
ratCage.showAnimals（）;
System.out.println（\\\
LionCage ........）;
lionCage.showAnimals（）;
System.out.println（\\\
SomeCage ........）;
someCage.showAnimals（）;
这是笼子类：
包装箱;
import java.util.HashSet;
import java.util.Set;
进口动物。动物;
公开课笼子< T延伸动物> {//一些动物类型的笼子
private Set< T> cage = new HashSet< T>（）;
public void add（T animal）{
cage.add（animal）;
$ b $ public void showAnimals（）{
for（T animal：cage）{
System.out.println（animal.getName（））;
}
}
}
此外，我会很高兴如果你能给我一个有意义的超级的例子，这个动物笼代码。直到现在我还没有理解如何使用它。有很多理论的例子，我读了PECS的概念，但无论如何，我还没有把它用在一个有意义的事情上。在这个例子中，有一个消费者（与超级）是什么意思？ 超级界限 >
引入的 transferTo（）方法接受 Cage - 一个拥有 T 超类的笼子。由于T是其超类的一个实例，因此可以将 T 放入 Cage中。
public static class Cage< T extends Animal> {
私人套餐< T> pen = new HashSet< T>（）;
public void add（T animal）{
pen.add（animal）;
}
/ *将子类放入超类笼子* /
public void transferTo（Cage< .pen.addAll（this.pen）;
}
public void showAnimals（）{
System.out.println（pen）;
$ / code $ / pre
现在我们来看看< ？
public static class Animal {
public String toString（ ）{
return getClass（）。getSimpleName（）;
public static class Rat extends Animal {}
public static class Lion extends Animal {}
public static class Cage< T extends Animal> {/ * above * /}
public static void main（String [] args）{
Cage< Animal>动物=新Cage< Animal>（）;
笼< Lion> lions = new Cage< Lion>（）;
animals.add（new Rat（））; //确定将鼠放入笼中<动物>
lions.add（new Lion（））;
lions.transferTo（动物）; //调用超类方法
animals.showAnimals（）;
}
输出：
[Rat，Lion]
另一个重要的概念是，尽管这是真的：
Lion instanceof Animal // true
不是 真的
笼<狮GT; instanceof Cage< Animal> // false
它不是这种情况，这段代码可以编译：
笼子<动物>动物;
笼< Lion>狮子。
动物=狮子; //这个任务是不允许的
animals.add（rat）; //如果执行此操作，我们会在笼子<狮子>
I know that there are a lot of questions about this topic, but unfortunately they couldn't help me to eliminate my obscurities. First of all, look at the following example. I don't understand, why the following "add"-method someCage.add(rat1) doesn't work and aborts with the following exception:
Is this the same reason why Cage<Rat> is not a Cage<Animal>? If yes, I don't understand it in this example, so I'm not sure what the compiler exactly does. Here is the code example:
package exe;
import cage.Cage;
import animals.Animal;
import animals.Ape;
import animals.Lion;
import animals.Rat;
public class Main {
public static void main(String[] args) {
Lion lion1 = new Lion(true, 4, "Lion King", 8);
Lion lion2 = new Lion(true, 4, "King of Animals", 9);
Ape ape1 = new Ape(true, 2, "Gimpanse", true);
Ape ape2 = new Ape(true, 2, "Orang Utan", true);
Rat rat1 = new Rat(true, 4, "RatBoy", true);
Rat rat2 = new Rat(true, 4, "RatGirl", true);
Rat rat3 = new Rat(true, 4, "RatChild", true);
Cage<Animal> animalCage = new Cage<Animal>();
animalCage.add(rat2);
animalCage.add(lion2);
Cage<Rat> ratCage = new Cage<Rat>();
ratCage.add(rat3);
ratCage.add(rat1);
ratCage.add(rat2);
// ratCage.add(lion1); //Not Possible. A Lion is no rat
Cage<Lion> lionCage = new Cage<Lion>();
lionCage.add(lion2);
lionCage.add(lion1);
Cage<? extends Animal> someCage = new Cage<Animal>(); //? = "unknown type that is a subtype of Animal, possibly Animal itself"
someCage = ratCage; //OK
// someCage = animalCage; //OK
someCage.add(rat1); //Not Possible, but why?
animalCage.showAnimals();
System.out.println("\nRatCage........");
ratCage.showAnimals();
System.out.println("\nLionCage........");
lionCage.showAnimals();
System.out.println("\nSomeCage........");
someCage.showAnimals();
}
}
This is the cage class:
package cage;
import java.util.HashSet;
import java.util.Set;
import animals.Animal;
public class Cage<T extends Animal> { //A cage for some types of animals
private Set<T> cage = new HashSet<T>();
public void add(T animal) {
cage.add(animal);
}
public void showAnimals() {
for (T animal : cage) {
System.out.println(animal.getName());
}
}
}
Moreover, I would be pleased if you could give me a meaningful "super" example with this animal-cage-code. Until now I haven't understood how to use it. There are a lot of theoretical examples and I read about the PECS concept but anyhow I wasn't able to employ it in a meaningful matter yet. What would it mean to have a "consumer" (with super) in this example?
解决方案 Example of super bound
The introduced transferTo() method accepts Cage<? super T> - a Cage that holds a superclass of T. Because T is an instanceof its superclass, it's OK to put a T in a Cage<? super T>.
public static class Cage<T extends Animal> {
private Set<T> pen = new HashSet<T>();
public void add(T animal) {
pen.add(animal);
}
/* It's OK to put subclasses into a cage of super class */
public void transferTo(Cage<? super T> cage) {
cage.pen.addAll(this.pen);
}
public void showAnimals() {
System.out.println(pen);
}
}
Now let's see <? super T> in action:
public static class Animal {
public String toString() {
return getClass().getSimpleName();
}
}
public static class Rat extends Animal {}
public static class Lion extends Animal {}
public static class Cage<T extends Animal> { /* above */ }
public static void main(String[] args) {
Cage<Animal> animals = new Cage<Animal>();
Cage<Lion> lions = new Cage<Lion>();
animals.add(new Rat()); // OK to put a Rat into a Cage<Animal>
lions.add(new Lion());
lions.transferTo(animals); // invoke the super generic method
animals.showAnimals();
}
Output:
[Rat, Lion]
Another important concept is that while it is true that:
Lion instanceof Animal // true
it is not true that
Cage<Lion> instanceof Cage<Animal> // false
It this were not the case, this code would compile:
Cage<Animal> animals;
Cage<Lion> lions;
animals = lions; // This assignment is not allowed
animals.add(rat); // If this executed, we'd have a Rat in a Cage<Lion>
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