正确实现hashCode（）方法

Question

我有一个POJO，在我所定义的hashCode方法我认为是的..

public int hashCode() 
    {  
    return name.hashCode()+job.hashCode()+salary;  

} 

但因为我使用的Eclipse IDE，它也提供了我的自动生成散列码，其是..

 @Override 
public int hashCode() { 
    final int prime = 31; 
    int result = 1; 
    result = prime * result + ((job == null) ? 0 : job.hashCode()); 
    result = prime * result + ((name == null) ? 0 : name.hashCode()); 
    result = prime * result + salary; 
    return result; 
} 

现在我的疑问是，这两个之间的差异，哪一个实现更好.. ..！我完全POJO是...

enter codepackage CollectionsPrac; 

公共类Employee {

String name,job; 
int salary; 


public Employee(String n , String j, int t) 
{ 
    this.name= n; 
    this.job=j; 
    this.salary= t;   
} 


@Override 
public int hashCode() { 
    final int prime = 31; 
    int result = 1; 
    result = prime * result + ((job == null) ? 0 : job.hashCode()); 
    result = prime * result + ((name == null) ? 0 : name.hashCode()); 
    result = prime * result + salary; 
    return result; 
} 


/*@Override 
public boolean equals(Object obj) { 
    if (this == obj) 
     return true; 
    if (obj == null) 
     return false; 
    if (getClass() != obj.getClass()) 
     return false; 
    Employee other = (Employee) obj; 
    if (job == null) { 
     if (other.job != null) 
      return false; 
    } else if (!job.equals(other.job)) 
     return false; 
    if (name == null) { 
     if (other.name != null) 
      return false; 
    } else if (!name.equals(other.name)) 
     return false; 
    if (salary != other.salary) 
     return false; 
    return true; 
} 
*/ 

/* @Override 
public int hashCode() 
    {  
    return name.hashCode()+job.hashCode()+salary;  

}*/ 

@Override 
    public boolean equals(Object obj) { 
    if (this == obj) 
    { 
     return true; 
    } 
    // make sure o can be cast to this class 
    if (obj == null || obj.getClass() != getClass()) 
    { 
     // cannot cast 
     return false; 
    }   

    Employee e = (Employee) obj; 
    return this.name.equals(e.name)&&this.job.equals(e.job)&&this.salary==e.salary; 
} 

@Override 
public String toString() { 
     return name+"\t" +"\t"+ job +"\t"+ salary; 
    } 

} 这里

Answer 1

基于Eclipse的产生hashCode()敏感得多，当谈到在你的POJO的微小变化。例如。如果切换job和name值彼此，你hashCode()将返回相同的值（加法是可交换的），而花哨的Eclipse版本将返回完全不同的东西：

System.out.println(new Employee("John", "Blacksmith", 100).hashCode()); 
System.out.println(new Employee("Blacksmith", "John", 100).hashCode()); 

//your version of hashCode() produces identical result: 
376076563 
376076563 

//Eclipse version: 
-1520263300 
926019626 

又见

• Why does Java's hashCode() in String use 31 as a multiplier?

Answer 2

最重要的区别是，如果您的实现将抛出NullPointerException如果job或name是null。

此外，方法eclipse生成更不规则哈希码的结果，这在理论上意味着哈希表退化和性能较差的可能性较低，但实际上可能无关紧要，因为java.util.HashMap使用辅助哈希函数在使用之前扰乱哈希码。