重写equals和hashCode
如果X.equals(Y)，那么它们的X和Y的hashCode值一定相同，why？ 如果重写了equals，但没有重写hashCode，两个不是也可能不同吗？初学者。
import java.util.*;
public class test2{
public static void main(String args[]){
HashSet<Demo> hSet=new HashSet<Demo>();
Demo d1=new Demo(1,"abc");
Demo d2=new Demo(1,"abc");
hSet.add(d1);
hSet.add(d2);
System.out.println(hSet.size());//输出2
System.out.println(d1.equals(d2)); //true
}
}
class Demo{
int a;
String b;
public Demo(int a,String b){
this.a=a;
this.b=b;
}
// public int hashCode(){
// return a*(b.hashCode());
// }
public boolean equals(Object o){
Demo d=(Demo)o;
return (this.a==d.a)&&(this.b.equals(d.b));
}
}
d1.equals(d2)返回ture，为什么还能加入？不是比较的equals吗？这个必须hashCode也相等，然后再判断equals()?
[解决办法]
参考Java文档
public boolean equals(Object obj)
当此方法被重写时，通常有必要重写 hashCode 方法，以维护 hashCode 方法的常规协定，该协定声明相等对象必须具有相等的哈希码。
只是个约定而已。你算里 X.equals(X) 返回 false都是可以的，只是不符合一般人的正常思维而已
[解决办法]
set存储的是不重复的值，即obj1.equals(obj2)==true话，则只存一份。
虽然上面相等时，能够加入，但是并未真正加入，加入后set.size()不便的
比如加入N个相等的值，size()仍为1.
至于使hashCode()和equals尽可能一致是性能考虑。
[解决办法]
重写equals 只有在使用equals操作符时菜体现用处 你可以不从写hashCode 但是如果你要把这个对象放到hashmap等散列集合中为了保持不冲突 所以你要重写hashCode方法
[解决办法]
hashCode在不运用在hash算法的集合下面是不需要重写的.. 这个无所谓.!
但是重写equals的时候顺带重写下hashCode没什么不好, 如果我没记错的话, 规范上面应该有如果两个对象相等,则他们必须要有相同的hashCode, 不知道记错没
[解决办法]
楼上都提到了
如果使用了equals来判断两个对象是否相等，那么条件是：equals为true且它们hashcode值也相等
反之判断不等equals为false之后，不强制hashcode值须如何，虽然Object的hashCode方法的默认实现是不是同一个对象就返回不同整数（因为是根据对象的内部地址）
[解决办法]
jdk1.5 Object的hashCode()方法原型
/** * Returns a hash code value for the object. This method is * supported for the benefit of hashtables such as those provided by * <code>java.util.Hashtable</code>. * <p> * The general contract of <code>hashCode</code> is: * <ul> * <li>Whenever it is invoked on the same object more than once during * an execution of a Java application, the <tt>hashCode</tt> method * must consistently return the same integer, provided no information * used in <tt>equals</tt> comparisons on the object is modified. * This integer need not remain consistent from one execution of an * application to another execution of the same application. * <li>If two objects are equal according to the <tt>equals(Object)</tt> * method, then calling the <code>hashCode</code> method on each of * the two objects must produce the same integer result. * <li>It is <em>not</em> required that if two objects are unequal * according to the {@link java.lang.Object#equals(java.lang.Object)} * method, then calling the <tt>hashCode</tt> method on each of the * two objects must produce distinct integer results. However, the * programmer should be aware that producing distinct integer results * for unequal objects may improve the performance of hashtables. * </ul> * <p> * As much as is reasonably practical, the hashCode method defined by * class <tt>Object</tt> does return distinct integers for distinct * objects. (This is typically implemented by converting the internal * address of the object into an integer, but this implementation * technique is not required by the * Java<font size="-2"><sup>TM</sup></font> programming language.) * * @return a hash code value for this object. * @see java.lang.Object#equals(java.lang.Object) * @see java.util.Hashtable */ public native int hashCode();