如何解決java聚合中的錯誤輸出錯誤？

Question

我的代碼運行良好，但在輸出中，兩名員工的地址相同。 為什么會這樣，我該如何解決？

package practice;

class address{
    static String country,state,cityname;
    public address(String country, String state, String cityname) {
    this.country=country;
    this.state=state;
    this.cityname=cityname;
    }
}
class employee{
    String name;
    int id;
    int age;
    address add;
    public employee(String name, int id, int age,address add) {
    this.name=name;
    this.id=id;
    this.age=age;
    this.add=add;
    }
    void display() {
        System.out.println(name+" "+id+" "+age);
        System.out.println("the employee stays at"+ address.country+" "+ 
        address.state+" "+address.cityname);
    }
}
public class Document {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        address a2 = new address("A","B","C");
        address a1 = new address("D","E","F");
        employee e1 = new employee("lmn",123,20,a2);
        employee e2 = new employee("pqr", 456,24,a1);
        e1.display();
        e2.display();
    }
}

Answer 1

問題是address的靜態變量：

class address{
    static String country,state,cityname;
...

刪除static關鍵字。

還將您的屬性設為私有並添加 getter 和 setter。

注意java命名約定。 類名應以大寫字符開頭