为什么在HashMap中使用键检索这些值？(Why are these values retrieved with keys in HashMap? [duplicate])

 
 
 这个问题在这里已有答案： 
 
 
 
  
 可变的hashmap键是危险的做法吗？ 7个答案 
 
 
 
 
 当我在HashMap上运行以下代码时，我正在努力解释为什么我将第二行输出作为“Line2：null”： 
 
import java.util.*;

class Dog {
  public Dog(String n) {name = n;}
  public String name;
  public boolean equals(Object o) {
    if((o instanceof Dog) && (((Dog)o).name == name)) {
        return true;
    }
    else {return false;}
  }
  public int hashCode() {return name.length();}
}

public class HelloWorld{

 public static void main(String []args){
    Map<Object, Object> m = new HashMap<Object, Object>();
    Dog d1 = new Dog("clover");
    m.put(d1, "Dog key");
    System.out.println("Line1: " + m.get(d1));
    d1.name = "magnolia";
    System.out.println("Line2: " + m.get(d1));
    d1.name = "clover";
    System.out.println("Line3: " + m.get(new Dog("clover")));
    d1.name = "arthur";
    System.out.println("Line4: " + m.get(new Dog("clover")));
 }
}
 
 显示的输出是： 
 
 Line1：狗钥匙 
 
 第2行：null 
 
 Line3：狗钥匙 
 
 第4行：null 
 
 是的，我确实意识到，由于我计算哈希码的方式，修改实例变量名反过来会影响Dog实例的哈希码。 但是，我使用相同的实例作为密钥！ 那么，为什么get（）方法不能找到相应的值呢？ 看起来，一旦将一对推入HashMap，关键就是永久硬编码！ 这是它应该如何工作，这意味着，一旦在HashMap中放置该对之前确定了一个值的哈希码，就不能再次修改哈希码？ 

 
 
This question already has an answer here:
 
 
 
  
 Are mutable hashmap keys a dangerous practice?  7 answers  
 
 
 
 
I am struggling with the reason why I am getting the second line of outputs as "Line2: null" when running the following piece of code on HashMap:
 
import java.util.*;

class Dog {
  public Dog(String n) {name = n;}
  public String name;
  public boolean equals(Object o) {
    if((o instanceof Dog) && (((Dog)o).name == name)) {
        return true;
    }
    else {return false;}
  }
  public int hashCode() {return name.length();}
}

public class HelloWorld{

 public static void main(String []args){
    Map<Object, Object> m = new HashMap<Object, Object>();
    Dog d1 = new Dog("clover");
    m.put(d1, "Dog key");
    System.out.println("Line1: " + m.get(d1));
    d1.name = "magnolia";
    System.out.println("Line2: " + m.get(d1));
    d1.name = "clover";
    System.out.println("Line3: " + m.get(new Dog("clover")));
    d1.name = "arthur";
    System.out.println("Line4: " + m.get(new Dog("clover")));
 }
}
 
The outputs displayed are:
 
Line1: Dog key
 
Line2: null
 
Line3: Dog key
 
Line4: null
 
Yes, I do realize that modifying the instance variable name, in turn, affects the hashcode of the instance of Dog because of the way I calculate the hashcode. But, I am using the same instance as the key! So, why cannot the get() method find the corresponding value? It seems like once a pair is pushed into a HashMap, the key is hardcoded with the value forever! Is this how it is supposed to work, meaning that, once a hashcode has been determined for a value before placing the pair in HashMap, the hashcode can never be modified again?

原文：https://stackoverflow.com/questions/30955953