Question

我对构建对象类型有点困惑。鉴于以下示例：

public class Test() {

 public void method1() { System.out.println("Method 1");
 public void method2() { System.out.println("Method 2");
 }

public class Test1() extends Test{

  @Override public void method1() { System.out.println("Method 11");
  @Override public void method2() { System.out.println("Method 22");
   public void method 3() { System.out.println("Method 3");

   public static void main(String[] args) {
       Test a = new Test1(); 
       a.method1(); //method invokes the overridden method1() of Test1, not Test the superclass
       a.method2(); //method invokes the overridden method2() of Test1, not Test the superclass
       a.method3(); //error, must cast == ((Test1)a).method3();
    }
}

当我调用method1＆amp; amp时，我感到困惑的是什么。 method2，编译器或JVM能够调用派生或子类的方法，那么为什么还需要下调/转换它来调用method3（）？

我尝试重载超类的method1和method2而不是覆盖它;在允许您调用重载方法之前，编译器/ JVM将要求您向下转换对象引用变量。那么，这意味着如果没有向下转换，你只能调用超类的方法以及子类中定义的重写方法？

Answer 1

您的参考a属于Test，但method3中不存在Test。

Test a = new Test1();
// ^-reference   ^-object 
//   type          type

规则：

参考类型告诉我们哪些方法可见。
对象类型告诉我们可以考虑哪些可见方法的实现。

<小时/> 这是我能想到的最基本的例子之一：

public class Animal
{
     public void speak()
     {
          System.out.println("BLARGHGH");
     }
}

public class Dog extends Animal
{
     @Override
     public void speak()
     {
          System.out.println("Woof");
     }
}

public class Cat extends Animal
{
     @Override 
     public void speak()
     {
          System.out.println("Meow");
     }

     public void throwUpFurball()
     {
          System.out.println("So fluffy!");
     }
}

public class Test()
{
     public static void main(final String args[])
     {
          Animal animal1 = new Animal(); 
          animal1.speak(); // BLARGHGH
          animal1.throwUpFurball(); // Compilation error - method not found. It will ask for casting

          Animal animal2 = new Cat();
          animal2.speak(); // Meow
          ((Cat)animal2).throwUpFurball(); // Must be cast, because throwUpFurball does not exist in Animal

          Animal animal3 = new Dog();
          animal3.speak(); // Woof
          ((Cat)animal3).throwUpFurball(); // Compiles, but throws ClassCastException at runtime, because the object type of animal3 is Dog

          Cat animal4 = new Cat();
          animal4.speak(); // Meow
          animal4.throwUpFurball(); // No casting necessary, because animal4 is of type Cat

          Object animal5 = new Animal();
          animal5.speak(); // Again, compilation problem. The type Object does not contain the speak() API, so it will require casting.
     }
}

Answer 2

您可以尝试Test1 a = new Test1(); - 参考文献应该清楚！

ps：用}}关闭你的方法;只是因为... AND方法3（）应该是method3（）

:)

希望我明白你的意思！

public class Test1() extends Test{

  @Override public void method1() { System.out.println("Method 11");}
  @Override public void method2() { System.out.println("Method 22");}
   public void method3() { System.out.println("Method 3");}

   public static void main(String[] args) {
       Test1 a = new Test1(); 
       a.method1(); //method invokes the overridden method1() of Test1, not Test the superclass
       a.method2(); //method invokes the overridden method2() of Test1, not Test the superclass
       a.method3(); //error, must cast == ((Test1)a).method3();
    }
}

转换对象类型

2 个答案: