例子
package cn.xy.test;
public class Point2
{
private int x;
private int y;
public Point2(int x, int y)
{
super();
this.x = x;
this.y = y;
}
@Override
public int hashCode()
{
final int prime = 31;
int result = 1;
result = prime * result + x;
result = prime * result + y;
return result;
}
@Override
public boolean equals(Object obj)
{
if (this == obj) return true;
if (obj == null) return false;
if (getClass() != obj.getClass()) return false;
Point2 other = (Point2) obj;
if (x != other.x) return false;
if (y != other.y) return false;
return true;
}
public int getX()
{
return x;
}
public void setX(int x)
{
this.x = x;
}
public int getY()
{
return y;
}
public void setY(int y)
{
this.y = y;
}
}
public class HashSetAndHashCode2
{
public static void main(String[] args)
{
HashSet<Point2> hs2 = new HashSet<Point2>();
Point2 p21 = new Point2(3, 3);
Point2 p22 = new Point2(3, 5);
hs2.add(p21);
hs2.add(p22);
p22.setY(7);
hs2.remove(p22);
System.out.println(hs2.size());
}
}
很多人认为打印出的结果是1。结果是2。为什么呢?当一个对象被存储在Hashset中后,如果修改参与计算hashcode有关的字段,那么修改后的hashcode的值就与一开始存储进来的hashcode的值不同了,这样contains无法通过hashcode找到该元素,所以无法删除。这就告诉我们,当一个对象被存储在Hashset中后,不要修改与计算hashcode有关的字段。
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