HashMap主要方法原始碼分析（JDK1.8）
阿新 • • 發佈：2020-04-21
本篇從HashMap的put、get、remove方法入手，分析原始碼流程
（不涉及紅黑樹的具體演算法）
jkd1.8中HashMap的結構為陣列、連結串列、紅黑樹的形式
（未轉化紅黑樹時）
putVal（）的整個流程如下 1、先判斷table是否為null，如果是，呼叫resize()
2、如果進入putVal（）判斷table不為null 利用hash值與（&）計算出陣列下標，並判斷是否為空 如果是，建立node物件並存入陣列 如果不是，從當前下標的連結串列第一位開始一個個往下對比 ①若hash和key值都相同，則break退出迴圈，之後進行value的替換，並返回oldValue ②若過程中遇到null，則建立node物件 判斷是否達到紅黑樹轉換條件：如果當前連結串列長度達到8，進入treeifyBin方法 判斷表長度（如果小於64，則呼叫resize()，判斷要不要增大陣列 反之replacementTreeNode，用紅黑樹代替當前連結串列
（轉化為紅黑樹時的情況）
一、關於HashMap需要了解的靜態常量
DEFAULT_INITIAL_CAPACITY 陣列預設初始容量 16 DEFAULT_LOAD_FACTOR 預設負載因子 0.75 MIN_TREEIFY_CAPACITY 最小樹容量 64 在下面的方法探究中將會提到這些靜態常量的用處二、方法探究
1、put
HashMap中的陣列是第一次呼叫put方法時才建立物件的
下面是從進入put到建立完陣列的全過程
將key、value作為引數傳入後，計算key的hash，再傳入putVal方法
/**
* 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)
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
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;
}putVal（）的整個流程如下 1、先判斷table是否為null，如果是，呼叫resize()
/**
* Initializes or doubles table size. If null, allocates in
* accord with initial capacity target held in field threshold.
* Otherwise, because we are using power-of-two expansion, the
* elements from each bin must either stay at same index, or move
* with a power of two offset in the new table.
*
* @return the table
*/
final Node<K,V>[] resize() {
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
if (oldCap > 0) {
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
}
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else { // zero initial threshold signifies using defaults
newCap = DEFAULT_INITIAL_CAPACITY;
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
}
if (newThr == 0) {
float ft = (float)newCap * loadFactor;
newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
(int)ft : Integer.MAX_VALUE);
}
threshold = newThr;
@SuppressWarnings({"rawtypes","unchecked"})
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
if (oldTab != null) {
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do {
next = e.next;
if ((e.hash & oldCap) == 0) {
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
}
else {
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
}
} while ((e = next) != null);
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}以上就是呼叫put方法時HashMap物件內陣列的建立過程
2、如果進入putVal（）判斷table不為null 利用hash值與（&）計算出陣列下標，並判斷是否為空 如果是，建立node物件並存入陣列 如果不是，從當前下標的連結串列第一位開始一個個往下對比 ①若hash和key值都相同，則break退出迴圈，之後進行value的替換，並返回oldValue ②若過程中遇到null，則建立node物件 判斷是否達到紅黑樹轉換條件：如果當前連結串列長度達到8，進入treeifyBin方法 判斷表長度（如果小於64，則呼叫resize()，判斷要不要增大陣列 反之replacementTreeNode，用紅黑樹代替當前連結串列
（向下遍歷陣列當前下標連結串列的操作）
（key相同時value的替換操作）
3、增加size（map中元素的數量）和modCount（對map的操作次數）
2、get
* 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;
}
get
將key和key的hash傳入getNode方法，並獲取返回的Node物件的value
先判斷陣列不為null，且長度大於0
利用hash值先找到node可能在的列的第一個元素（當前傳入key可能不存在）
再豎向遍歷連結串列對比hash和key值
（上面的first即為node可能存在的連結串列的第一個元素）
查詢時如果第一個即匹配則直接返回
否則往下遍歷直到匹配或node為null
3、removeNode
/**
* Implements Map.remove and related methods.
*
* @param hash hash for key
* @param key the key
* @param value the value to match if matchValue, else ignored
* @param matchValue if true only remove if value is equal
* @param movable if false do not move other nodes while removing
* @return the node, or null if none
*/
final Node<K,V> removeNode(int hash, Object key, Object value,
boolean matchValue, boolean movable) {
Node<K,V>[] tab; Node<K,V> p; int n, index;
if ((tab = table) != null && (n = tab.length) > 0 &&
(p = tab[index = (n - 1) & hash]) != null) {
Node<K,V> node = null, e; K k; V v;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
node = p;
else if ((e = p.next) != null) {
if (p instanceof TreeNode)
node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
else {
do {
if (e.hash == hash &&
((k = e.key) == key ||
(key != null && key.equals(k)))) {
node = e;
break;
}
p = e;
} while ((e = e.next) != null);
}
}
if (node != null && (!matchValue || (v = node.value) == value ||
(value != null && value.equals(v)))) {
if (node instanceof TreeNode)
((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
else if (node == p)
tab[index] = node.next;
else
p.next = node.next;
++modCount;
--size;
afterNodeRemoval(node);
return node;
}
}
return null;
}
removeNode
與getNode的邏輯類似，利用hash值查詢可能存在的位置
如果連結串列第一位就匹配：
對連結串列的遍歷：
判斷node的型別，如果是連結串列則用node.nest來代替當前位置
最後運算元+1，size-1，返回被remove的node物件
&nbs