问题描述：eclipse的版本是oxygen的，JDK和jre的版本是10；在eclipse中使用clone()方法进行浅克隆的时候，出现如下提示的编译错误。提示“返回类型与Object.clone()不兼容”，具体位置如下图所示：但是我在使用JDK编译运行的时候是正常的，所以基本上能确定是eclipse哪里出现问题，所以想请问各位大佬这个是什么原因造成的问题（我猜测是eclipse的版本和JDK的版本不兼容的问题，但是不敢确定）源码附上：public class Address {
private String country;
private String province;
private String city;
public void setCountry(String country) {
this.country=country;
}
public void setProvince(String province) {
this.province=province;
}
public void setCity(String city) {
this.city=city;
}
public String toString() {
//return "地址："+"国家："+country+","+"省:"+province+","+"市："+city;
StringBuilder sb=new StringBuilder();
sb.append("国家："+country+",");
sb.append("省:"+province+",");
sb.append("市："+city);
return sb.toString();
}
}public class Employee implements Cloneable{
private String name;
private int age;
private Address address;
public void setName(String name){
this.name=name;
}
public void setAge(int age){
this.age=age;
}
public void setAddress(Address address) {
this.address=address;
}
public Address getAddress(){
return this.address;
}
public Employee clone() {
Employee employee=null;
try {
employee=(Employee)super.clone();
}catch(CloneNotSupportedException e) {
e.printStackTrace();
}
return employee;
}
public String toString() { //重写toString方法，使用字符串输出对象
StringBuilder sb=new StringBuilder();
sb.append("姓名："+name+",");
sb.append("年龄:"+age+'\n');
sb.append(address);
return sb.toString();
}
}public class Test {
public static void main(String[] args) {
System.out.println("克隆之前：");
Address address=new Address();
address.setCountry("中国");
address.setProvince("四川");
address.setCity("成都");
Employee employee1=new Employee();
employee1.setName("张XX");
employee1.setAge(30);
employee1.setAddress(address);
System.out.println("员工1的信息");
System.out.println(employee1);
System.out.println("克隆之后：");
Employee employee2=employee1.clone(); //使用浅克隆
employee2.setName("陈XX");
employee2.setAge(20);
employee2.getAddress().setCountry("中国");
employee2.getAddress().setProvince("山东");
employee2.getAddress().setCity("济南");
System.out.println("员工1的信息");
System.out.println(employee1);
System.out.println("员工2的信息");
System.out.println(employee2);
}
}
添加回答
举报
0/150
提交
取消