kotlin类继承的演示

简书链接:kotlin类继承的演示
文章字数：94,阅读全文大约需要1分钟

open class Person(name: String, age: Int) {
     var currentName = name;
     var age = age;

     init {
         "init call ${currentName} name:${name}"
     }

     fun hello() {
         println("hello method call  my name is:${this.currentName} ,i'm ${this.age} years old");
     }
 }

 class Zhang(name: String, age: Int) : Person(name, age) {

 }

 class Luo : Person {
     var interect: String? = null;

     constructor(name: String) : super(name, 0);
     constructor(name: String, birthday: Long, age: Int) : super(name, age);
     constructor(name: String, birthday: Long, interect: String, age: Int) : super(name, age) {
         this.interect = interect;

     }

     init {
         println("luo init call this.interect  ${this.interect}");
     }
 }

 class Li : Person {
     constructor(deathTime:Long, age: Int=1,name: String = "i am zhang") : super(name, age) {
         println(" Li constructor ${name}");
     }
 }

 class SuperMan : Person {
     fun productSuerMan(): SuperMan {
         return SuperMan();
     }

     private constructor() : super("qssq", 1000);
 }



 var person = Li(100);
 var person3 = Li(100, 1);
 var person2 = Li(100, 1, "hello");
 person.hello();
 person3.hello();
 person2.hello();
 val obj = TestStatic.getInstanceXXX()
 TestStatic.main()
 println("hello obj${obj}")

由于有默认值，应该是不能进行重载的，另外 没有默认值的参数必须放前面，否则报错
无论什么构造 在构造完成之后 都会走init方法，因此init里面 可以拿构造方法的值。