静态绑定和动态绑定

Question

我对动态绑定和静态绑定感到困惑.我已经读过在编译时确定对象的类型称为静态绑定,在运行时确定它称为动态绑定.

以下代码中会发生什么:

静态绑定还是动态绑定？
这显示了什么样的多态性？

class Animal
{
    void eat()
    {
        System.out.println("Animal is eating");
    }
}

class Dog extends Animal
{
    void eat()
    {
        System.out.println("Dog is eating");
    }
}

public static void main(String args[])
{
    Animal a=new Animal();
    a.eat();
}

Answer 1

您的示例是动态绑定,因为在运行时确定类型a是什么,并调用适当的方法.

现在假设您还有以下两种方法:

public static void callEat(Animal animal) {
    System.out.println("Animal is eating");
}
public static void callEat(Dog dog) {
    System.out.println("Dog is eating");
}

即使你改变你main的

public static void main(String args[])
{
    Animal a = new Dog();
    callEat(a);
}

这将打印Animal is eating,因为调用callEat使用静态绑定,并且编译器只知道它a是类型Animal.

Answer 2

如果你做了这样的事情,这真的取决于重载和覆盖:

public class Animal{}


public class Dog extends Animal{}

public class AnimalActivity{

    public void eat(Animal a){
        System.out.println("Animal is eating");
    }

    public void eat(Dog d){
        System.out.println("Dog is eating");
    }
}

然后在主类:

public static void main(String args[])
{
    Animal a=new Animal();
    Animal d=new Dog();
    AnimalActivity aa=new AnimalActivity();
    aa.eat(a);
    aa.eat(d);
}

这两个案例的结果将是: Animal is eating

但是让它扭曲一下,让我们这样:

public class Animal{
    public void eat(){
        System.out.println("Animal is eating");
    }
}

然后:

public class Dog extends Animal{
    public void eat(){
        System.out.println("Dog is eating");
    }
}

然后在主类:

public static void main(String args[]){
    Animal d=new Dog();
    Animal a=new Animal();
    a.eat();
    d.eat();
}

现在的结果应该是:

Animal is eating
Dog is eating

这是因为重载在编译时绑定"静态绑定",而在运行时覆盖绑定"动态绑定"