题目内容

请说出E类中【代码1】~【代码4】的输出结果。class A {double f(double x,double y) {return x+y;}static int g(int n) {return nn;}}class B extends A {double f(double x,double y) {double m = super.f(x,y);return m+xy;}static int g(int n) {int m = A.g(n);return m+n;}}public class E {public static void main(String args[]) {B b = new B();System.out.println(b.f(10.0,8.0));//【代码1】System.out.println(b.g(3));//【代码2】A a = new B();System.out.println(a.f(10.0,8.0));//【代码3】System.out.println(a.g(3));//【代码4】}}【代码1】的输出结果【代码2】的输出结果【代码3】的输出结果【代码4】的输出结果

查看答案

搜索结果不匹配？点我反馈

更多问题

请说出E类中【代码1】~【代码3】的输出结果。class A {int m;int getM() {return m;}int seeM() {return m;}}class B extends A {int m ;int getM() {return m+100;}}public class E {public static void main(String args[]) {B b = new B();b.m = 20;System.out.println(b.getM()); //【代码1】A a = b;a.m = -100;// 上转型对象访问的是被隐藏的mSystem.out.println(a.getM()); //【代码2】上转型对象调用的一定是子类重写的getM()方法System.out.println(b.seeM()); //【代码3】子类继承的seeM()方法操作的m是被子类隐藏的m}}【代码1】的输出结果【代码2】的输出结果【代码3】的输出结果

public class A {public void start() {System.out.println("TestA");}}public class B extends A {public void start() {System.out.println("TestB");}public static void main(String[] args) {B b=new B();A a=b;a.start();}}程序运行的结果是

class A{double f(double x,double y){return x+y;}}class B extends A{double f(int x,int y){return x*y;}}public class E{public static void main(String args[]) {B b=new B();System.out.println(b.f(8.0,12.0));}}运行结果是

Java语言类不支持多继承，但可以通过实现多继承。