我把纯源码放到了随笔: https://www.cnblogs.com/zhangxuezhi/p/11660818.html
public class HashMap<K, V> extends AbstractMap<K, V> implements Map<K, V>, Cloneable, Serializable
先来看put函数的源码:
/** * Associates the specified value with the specified key in this map. * If the map previously contained a mapping for the key, the old * value is replaced. * * @param key key with which the specified value is to be associated * @param value value to be associated with the specified key * @return the previous value associated with <tt>key</tt>, or * <tt>null</tt> if there was no mapping for <tt>key</tt>. * (A <tt>null</tt> return can also indicate that the map * previously associated <tt>null</tt> with <tt>key</tt>.) */ public V put(K key, V value) { return putVal(hash(key), key, value, false, true); }
该函数做的事情有两件,首先是计算key的hash值,然后调用putVal函数,先来看看hash计算的函数:
/** * Computes key.hashCode() and spreads (XORs) higher bits of hash * to lower. Because the table uses power-of-two masking, sets of * hashes that vary only in bits above the current mask will * always collide. (Among known examples are sets of Float keys * holding consecutive whole numbers in small tables.) So we * apply a transform that spreads the impact of higher bits * downward. There is a tradeoff between speed, utility, and * quality of bit-spreading. Because many common sets of hashes * are already reasonably distributed (so don't benefit from * spreading), and because we use trees to handle large sets of * collisions in bins, we just XOR some shifted bits in the * cheapest possible way to reduce systematic lossage, as well as * to incorporate impact of the highest bits that would otherwise * never be used in index calculations because of table bounds. */ static final int hash(Object key) { int h; return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16); }
如果key为null,返回0;
否则,取key的hashCode,高16位不变,然后将其高16位和低16位进行异或运算,用运算后的值替代原始值的低16位。(>>>为无符号右移,即右移时,在左边填充0。所以hashCode值右移16位之后,其左边16位为0,而与0进行异或运算,值不变。)
个人理解:
这样计算hash,最终计算出来的hash值,其低16位值同时也包含了高16位值的信息。这是因为一般使用hash值来求解数组下标时,一般使用到的是低位数据,很少会用到高位。除非数组的长度很大,达到2^16的时候,才会开始用到高位。因此一般情况下,这样更能充分利用到高位的信息。
计算数组下标的方法如下:
n = (tab = resize()).length;p = tab[i = (n - 1) & hash]
其中,n是容器table的size,而下标,即i,是通过 key的hash值和(n-1)进行&运算得到的。
因此,计算容器下标的方法,就涉及到key的hash值,以及容器本身的大小,
所以在容器的大小不超过2^16的情况下,其hash值因为同时包含了高低位的信息量,该值更能充分利用hashCode的整体信息,从一定程度上减少hash碰撞,计算出来的下标值,也可以减少群集现象。
有个问题,这类求容器下标i的时候,为什么用 & 运算?而不是使用%求余。这个另开一篇随笔来写吧:https://www.cnblogs.com/zhangxuezhi/p/11868058.html
接下来看putVal函数
/** * Implements Map.put and related methods * * @param hash hash for key * @param key the key * @param value the value to put * @param onlyIfAbsent if true, don't change existing value * @param evict if false, the table is in creation mode. * @return previous value, or null if none */ final V putVal(int hash, K key, V value, boolean onlyIfAbsent, boolean evict) { Node<K,V>[] tab; Node<K,V> p; int n, i; if ((tab = table) == null || (n = tab.length) == 0) //table 为空 //n被赋值为table的长度 n = (tab = resize()).length; if ((p = tab[i = (n - 1) & hash]) == null) //通过key的hash值获取到的下标,找到的桶里面是空的 //这里隐含的一个步骤是:p被赋值为key对应hash下标存放的node数据,i被赋值为桶下标 tab[i] = newNode(hash, key, value, null); else { //table 不为空,桶不为空 Node<K,V> e; K k; if (p.hash == hash && ((k = p.key) == key || (key != null && key.equals(k)))) //key重复 e = p; else if (p instanceof TreeNode) //桶里面的数据结构是红黑树 e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value); else { //桶里面的数据结构是链表 for (int binCount = 0; ; ++binCount) { if ((e = p.next) == null) { //到达链表末尾 p.next = newNode(hash, key, value, null); if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st //链表长度超过8,转化为红黑树 treeifyBin(tab, hash); break; } if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) //链表后续的数据中,有key值重复的 break; //p指向p.next p = e; } } if (e != null) { // existing mapping for key V oldValue = e.value; if (!onlyIfAbsent || oldValue == null) e.value = value; afterNodeAccess(e); return oldValue; } } ++modCount; if (++size > threshold) resize(); afterNodeInsertion(evict); return null; }
以上的关键步骤,我都加上了个人的理解,总体步骤如下:
1. 容器的table为空时,调用resize(),给table分配空间
2. 通过key的hash值,计算桶下标值i,
3. 通过i获取容器的桶,
如果为空,直接new node;
如果不为空,比较桶顶部元素的key值
如果key值重复,直接用新的value覆盖旧value,
如果key不重复,并且节点类型是树节点,则转到树的处理函数
如果key不重复,并且节点类型是不是树节点,遍历节点链表,
如果有重复的key值,则用新value替代旧value,
如果没有重复的key值,在桶尾部添加new node,然后根据桶的大小,如果超出8,则把链表重构成红黑树
看完put函数,接下来get的函数就比较好理解了:
/** * Returns the value to which the specified key is mapped, * or {@code null} if this map contains no mapping for the key. * * <p>More formally, if this map contains a mapping from a key * {@code k} to a value {@code v} such that {@code (key==null ? k==null : * key.equals(k))}, then this method returns {@code v}; otherwise * it returns {@code null}. (There can be at most one such mapping.) * * <p>A return value of {@code null} does not <i>necessarily</i> * indicate that the map contains no mapping for the key; it's also * possible that the map explicitly maps the key to {@code null}. * The {@link #containsKey containsKey} operation may be used to * distinguish these two cases. * * @see #put(Object, Object) */ public V get(Object key) { Node<K,V> e; return (e = getNode(hash(key), key)) == null ? null : e.value; } /** * Implements Map.get and related methods * * @param hash hash for key * @param key the key * @return the node, or null if none */ final Node<K,V> getNode(int hash, Object key) { Node<K,V>[] tab; Node<K,V> first, e; int n; K k; if ((tab = table) != null && (n = tab.length) > 0 && (first = tab[(n - 1) & hash]) != null) { if (first.hash == hash && // always check first node ((k = first.key) == key || (key != null && key.equals(k)))) return first; if ((e = first.next) != null) { if (first instanceof TreeNode) return ((TreeNode<K,V>)first).getTreeNode(hash, key); do { if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) return e; } while ((e = e.next) != null); } } return null; }
类似put,也是先计算key的hash值,然后调用其他函数来执行真正的get操作。
get的大概步骤如下:
当table不为空,并且通过key的hash计算出来的i值,获取到的桶也不为空时,执行以下步骤,否则返回null
判断桶的首元素,是否跟key相等,相等则返回该数据的value值
如果key不相等,
如果桶的数据结构是红黑树,遍历数结构获取数据
如果桶的数据结构是链表,遍历链表获取数据
注意,无论是get还是put,其key是否相等的判断如下:
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
即,key的hash值相等,并且key值本身也要相等。而其中,hash值的计算又依赖于其hashCode函数,因此,如果要自定义HashMap容器的key的对象,则必须要实现该对象的hashCode函数和equals函数,从而定义其相等的判断条件。