内部类与外部类3

关于重名的小问题：

如果我们在外部类声明了一个age,在内部类也声明了一个age，那么系统会根据就近原则来确定到底是哪个age：

package com.hw.InnerClassDemo;
public class InnerDemo {
    public static void main(String[] args){
        Outer.Inner inn = new Outer(20).new Inner("瀚洋",69,18);
        inn.show();
    }
}

class Outer{
    private int age;
    public Outer(int age){
        this.age = age;
    }

    public class Inner{
        private String name;
        private double weight;
        private int age;

        public Inner(String name,double weight,int age){
            this.name = name;
            this.weight = weight;
            this.age = age;
        }

        public void show(){
            System.out.println("姓名是:"+name+"，体重是:"+weight+"，年龄是:"+age);
        }
    }
}

 

以上代码中，离输出语句中的age近的是内部类的age，所以输出是18：

---

 

如果两个age都想访问，可以通过下面这种方式：

package com.hw.InnerClassDemo;
public class InnerDemo {
    public static void main(String[] args){
        Outer.Inner inn = new Outer(20).new Inner("瀚洋",69,18);
        inn.show();
    }
}

class Outer{         
    private int age;
    public Outer(int age){
        this.age = age;
    }
    
    public class Inner{       
        private String name;
        private double weight;
        private int age;
        
        public Inner(String name,double weight,int age){
            this.name = name;
            this.weight = weight;
            this.age = age;
        }
        
        public void show(){
            System.out.println("姓名是:"+name+"，体重是:"+weight+"，年龄是:"+this.age+" "+Outer.this.age);
        }
    }
}

运行效果如下：