江苏龙网

  1. 首页 > 生活 > >

关于java源码HashMap中的Entry属性 final int hash的作用

是不同的key可能拥有同样的hash,所以先要比较hash，再看key是否完全一致。这种情况叫哈希碰撞
■网友
要想理解为什么这样做，还是回归到源码，hashmap 的hash实现在JDK中的实现如下：方法一：static final int hash(Object key) { int h; // h = key.hashCode() 为第一步 取hashCode值 // h ^ (h \u0026gt;\u0026gt;\u0026gt; 16) 为第二步 高位参与运算 return (key == null) ? 0 : (h = key.hashCode()) ^ (h \u0026gt;\u0026gt;\u0026gt; 16);}方法二：static int indexFor(int h, int length) { //jdk1.7的源码，jdk1.8没有这个方法 return h \u0026amp; (length-1); //第三步 取模运算}可以看到在对key取hash的时候先调用了hashCode方法，再来看看这个方法的说明：* \u0026lt;ul\u0026gt; * \u0026lt;li\u0026gt;Whenever it is invoked on the same object more than once during * an execution of a Java application, the {@code hashCode} method * must consistently return the same integer, provided no information * used in {@code equals} 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. * \u0026lt;li\u0026gt;If two objects are equal according to the {@code equals(Object)} * method, then calling the {@code hashCode} method on each of * the two objects must produce the same integer result. * \u0026lt;li\u0026gt;It is \u0026lt;em\u0026gt;not\u0026lt;/em\u0026gt; required that if two objects are unequal * according to the {@link java.lang.Object#equals(java.lang.Object)} * method, then calling the {@code hashCode} 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 hash tables. * \u0026lt;/ul\u0026gt; * \u0026lt;p\u0026gt; * As much as is reasonably practical, the hashCode method defined by * class {@code Object} 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\u0026amp;trade; programming language.) * * @return a hash code value for this object. * @see java.lang.Object#equals(java.lang.Object) * @see java.lang.System#identityHashCode */public native int hashCode();通过说明可以看出，俩个相同的对象，只会生成相同的hash值，然而并没有说明不同的对象一定不会生成相同的hash值，这不是充分不必要条件。所以如楼上所说，是可能存在hash碰撞的。附上一篇分析hashmap的很透彻的文章：Java 8系列之重新认识HashMap
■网友
我觉得楼主的意思是不是对于这个数组来说，同一个下标下存储的所有entry都应该是相同的hash值，为什么还需要再检查一次？我也有点疑问，看看有没有大神解答

■网友
if (e.hash == hash \u0026amp;\u0026amp; ((k = e.key) == key || key.equals(k))) { 这句代码后面是重点，如果是hash值相等了，然后key不相等，但是如果equals相等，虽然不是同一个对象。但是我们依然认为是同一个对象。比如Person p1 = new Person("小明")，Person p2 = new Person("小明")；如果在equals方法里面只是判断如果名字相同就是同一个人，就认为是相同。


推荐阅读


上一篇：为啥用win7“画图”软件剪切图片像素变小了图片占用反而变大了但是继续剪切的话就会变小

下一篇：你期望的婚纱摄影应该是啥样子的呢