Java 泛型 - 超类型引用

发布于 2024-12-20 06:55:14 字数 1028 浏览 1 评论 0原文

如果我正确理解了泛型，那么参数声明为将接受类型 T 或超类型 T 的任何引用。我试图用下面的代码来测试它，但编译器不喜欢它。

class Animal{}
class Dog extends Animal{}
class Cat extends Animal{}

class ZiggyTest2{

    public static void main(String[] args){                 

        List<Animal> anim2 = new ArrayList<Animal>();
        anim2.add(new Animal());
        anim2.add(new Dog());
        anim2.add(new Cat());   

        testMethod(anim2);
    }

    public static void testMethod(ArrayList<? super Dog> anim){
        System.out.println("In TestMethod");
        anim.add(new Dog());
        //anim.add(new Animal());
    }
}

编译器错误是：

ZiggyTest2.java:16: testMethod(java.util.ArrayList<? super Dog>) in ZiggyTest2 cannot be applied to (java.util.List<Animal>)
                testMethod(anim2);
                ^
1 error

我不明白为什么我不能传入 anim2，因为它的类型是并且 Animal 是 Dog 的超类型。

谢谢

原文

If i have understood generics correctly, a method with parameters declared as <? super T> will accept any reference that is either of Type T or a super type of T. I am trying to test this with the following bit of code but the compiler does not like it.

class Animal{}
class Dog extends Animal{}
class Cat extends Animal{}

class ZiggyTest2{

    public static void main(String[] args){                 

        List<Animal> anim2 = new ArrayList<Animal>();
        anim2.add(new Animal());
        anim2.add(new Dog());
        anim2.add(new Cat());   

        testMethod(anim2);
    }

    public static void testMethod(ArrayList<? super Dog> anim){
        System.out.println("In TestMethod");
        anim.add(new Dog());
        //anim.add(new Animal());
    }
}

Compiler error is :

ZiggyTest2.java:16: testMethod(java.util.ArrayList<? super Dog>) in ZiggyTest2 cannot be applied to (java.util.List<Animal>)
                testMethod(anim2);
                ^
1 error

I dont understand why i cant pass in anim2 since it is of type <Animal> and Animal is a super type of Dog.

Thanks

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白云不回头 2024-12-27 06:55:14

您正在尝试将 List<> 类型的表达式传递到 ArrayList<> 类型的参数中。不去上班。

这条线要么

public static void testMethod(ArrayList<? super Dog> anim){

应该是

public static void testMethod(List<? super Dog> anim){

要么

    List<Animal> anim2 = new ArrayList<Animal>();

应该是

    ArrayList<Animal> anim2 = new ArrayList<Animal>();

You are trying to pass an expression of type List<> into a parameter of type ArrayList<>. Not going to work.

Either this line

public static void testMethod(ArrayList<? super Dog> anim){

should be

public static void testMethod(List<? super Dog> anim){

    List<Animal> anim2 = new ArrayList<Animal>();

should be

    ArrayList<Animal> anim2 = new ArrayList<Animal>();

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断舍离 2024-12-27 06:55:14

您尝试过 < 吗？扩展狗> ？

另外我可以问你为什么使用 ArrayList< 吗？ super Dog> 而不仅仅是 ArrayList？也许这个例子只是您想要做的事情的一个简单形式，但它似乎过于复杂。

这是泛型的基本示例。希望有帮助。

class Animal{
public sleep(){}
}
class Dog extends Animal{
public sleep(){
    log("Dog sleeps");
}
}
class Rabbit extends Animal{
public sleep(){
    log("Rabbit sleeps");
}
}

class Place<T>{
T animal;
}
class Kennel extends Place<Dog>{
public Kennel(Dog dog){
    super();
    this.animal = dog;
}
}
class Hutch extends Place<Rabbit>{
public Kennel(Rabbit rabbit){
    super();
    this.animal = rabbit;
}
}

Have you tried <? extends Dog> ?

Also can I ask why you are using ArrayList<? super Dog> rather than just ArrayList<Animal>? Perhaps this example is just a simple form of what you are trying to do but it seems inexplicably over complicated.

This is a basic example of generics. Hope it helps.

class Animal{
public sleep(){}
}
class Dog extends Animal{
public sleep(){
    log("Dog sleeps");
}
}
class Rabbit extends Animal{
public sleep(){
    log("Rabbit sleeps");
}
}

class Place<T>{
T animal;
}
class Kennel extends Place<Dog>{
public Kennel(Dog dog){
    super();
    this.animal = dog;
}
}
class Hutch extends Place<Rabbit>{
public Kennel(Rabbit rabbit){
    super();
    this.animal = rabbit;
}
}

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