在重新equals方法時爲啥要重寫hashCode方法?
重寫的原則是:兩個對象根據equals方法相等,則兩個對象hashCode產生同樣的整數結
果。
其實重寫hashCode方法是爲了保證一些基於散列的集合能正常工作,這樣集合包括
(HahsMap,HashSet,HashTable)。因爲此類集合是利用hash算法與equals來區分對象的等同性。比如:對類 Test重寫了equals方法,沒有重寫hashCode方法,意味着當Test類的兩個實例a和b,在邏輯上相同(a.equals(b)==true),但是兩個實例的hashCode不同。當定義一個Map對象:
Map<Test,String> map=new HashMap<Test,String>();
map.put(a,"this is Test object");
此時若系統通過map.get(b);來獲得“this is Test object”,但返回的確是null。這是因爲
map對象將a對象放入一個散列桶中,而get方法卻在另一個散列桶中查詢,即便兩個散列碼同時在一個散列桶中,也會因爲散列碼不同而找不到。散列碼不匹配就沒有必要驗證對象的等同性,所以此處找不到返回null。由此可見,重寫hashCode方法是很有必要的,因爲你不知道你定義的類不會被別人放入hash機制的集合中。
下面是重寫equals和hashCode通用的方式:
此處摘錄《Effective java》 和轉載:http://zhangjunhd.blog.51cto.com/113473/71571/
示例:
package com.zj.unit;
import java.util.Arrays;
public class Unit {
private short ashort;
private char achar;
private byte abyte;
private boolean abool;
private long along;
private float afloat;
private double adouble;
private Unit aObject;
private int[] ints;
private Unit[] units;
public boolean equals(Object o) {
if (!(o instanceof Unit))
return false;
Unit unit = (Unit) o;
return unit.ashort == ashort
&& unit.achar == achar
&& unit.abyte == abyte
&& unit.abool == abool
&& unit.along == along
&& Float.floatToIntBits(unit.afloat) == Float
.floatToIntBits(afloat)
&& Double.doubleToLongBits(unit.adouble) == Double
.doubleToLongBits(adouble)
&& unit.aObject.equals(aObject)
&& equalsInts(unit.ints)
&& equalsUnits(unit.units);
}
private boolean equalsInts(int[] aints) {
return Arrays.equals(ints, aints);
}
private boolean equalsUnits(Unit[] aUnits) {
return Arrays.equals(units, aUnits);
}
public int hashCode() {
int result = 17;
result = 37 * result + (int) ashort;
result = 37 * result + (int) achar;
result = 37 * result + (int) abyte;
result = 37 * result + (abool ? 0 : 1);
result = 37 * result + (int) (along ^ (along >>> 32));
result = 37 * result + Float.floatToIntBits(afloat);
long tolong = Double.doubleToLongBits(adouble);
result = 37 * result + (int) (tolong ^ (tolong >>> 32));
result = 37 * result + aObject.hashCode();
result = 37 * result + intsHashCode(ints);
result = 37 * result + unitsHashCode(units);
return result;
}
private int intsHashCode(int[] aints) {
int result = 17;
for (int i = 0; i < aints.length; i++)
result = 37 * result + aints[i];
return result;
}
private int unitsHashCode(Unit[] aUnits) {
int result = 17;
for (int i = 0; i < aUnits.length; i++)
result = 37 * result + aUnits[i].hashCode();
return result;
}
}