Question

重写hashcode和equals 有什么作用

Answer 1

为了阐明其作用，先来假设有如下一个Person类。
classPerson
{
publicPerson(String name, intage)
{
this.name = name;
this.age = age;
}

privateString name;
privateint age;

publicString getName()
{
returnname;
}

publicvoid setName(String name)
{
this.name = name;
}

publicint getAge()
{
returnage;
}

publicvoid setAge(intage)
{
this.age = age;
}

publicString toString()
{
return"{" + name + ", " + age + "}";
}
}
现在有很多Person类的对象需要存储，很自然联想到用HashSet来存储，于是乎，写了下面的程序来测试一下：

importjava.util.*;

publicclass HashSetDemo
{
publicstatic void main(String[] args)
{
Collection<person> set = newHashSet<person>();
set.add(newPerson("张三",21));
set.add(newPerson("李四",19));
set.add(newPerson("王五",22));
set.add(newPerson("张三",21));

sop(set);
}

privatestatic void sop(Collection<person> set)
{
Iterator<person> it = set.iterator();
while(it.hasNext())
{
Person p = it.next();
System.out.println(p.toString());
}
}
}
其实，在往HashSet集合放置元素时，会根据其hashCode来判断两个元素是否一样，如果是一样，这后者覆盖前者。而hashCode默认是比较其地址值。
为了达到只存放一个的目的，我们必须重写hashCode方法，当然与其紧密联系的equals方法也要重写，新的Person类如下：
classPerson
{
publicPerson(String name, intage)
{
this.name = name;
this.age = age;
}

privateString name;
privateint age;

publicString getName()
{
returnname;
}

publicvoid setName(String name)
{
this.name = name;
}

publicint getAge()
{
returnage;
}

publicvoid setAge(intage)
{
this.age = age;
}

publicString toString()
{
return"{" + name + ", " + age + "}";
}

publicint hashCode()
{
returnname.hashCode() + age * 10;
}

publicboolean equals(Object obj)
{
if(!(obj instanceofPerson))
thrownew ClassCastException("类型不匹配");
Person p = (Person)obj;
returnthis.name.equals(p.getName()) && this.age == p.getAge();
}
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