HashMap get方法,根据key值相同取值,首先会比较 key值得hash码是否相等,再比较两个对象是否相等。Hash码相等两个对象不一定相等,但是两个对象相等Hash码一定相等。
(n - 1) & hash 是计算下标的公式
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
}
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) {// 判断tab是否为空
if (first.hash == hash &&
((k = first.key) == key || (key != null && key.equals(k))))// 先判断第一个Node的hash值和 参数的hash值 是否相等,在比较key值
return first;
if ((e = first.next) != null) {
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);// 如果是树节点则跳转树节点 get方法
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);// 遍历链表,看是否有相同的key值
}
}
return null;
}
HashMap 的扩容机制:1.判断当前内存大小是否大于0;2.判断当前扩容门限是否大于0;3.根据相应的条件设置新的容量和新的扩容门限 4.创建新的Node数据 5.赋值
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) {
// 旧内存大于0
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;//当前内存达到内存最大值,不进行扩容
}
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // 新扩容门限 扩大2倍,新内存扩大2倍
}
else if (oldThr > 0)
newCap = oldThr;// 当前内存大小为0,但是当前扩容门限大于0 ,新内存等于当前扩容门限
else {
// 当前内存和扩容门限都为 0
newCap = DEFAULT_INITIAL_CAPACITY;//新内存赋值 初始内存
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);//新扩容门限设值
}
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;//创建空Node数组,大小为新内存
if (oldTab != null) {
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {//遍历数组
oldTab[j] = null;
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;//只有一个元素,直接赋值
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);//如果是树节点,特殊处理
else {
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) {// 这里还是不太理解
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);// 遍历链表,赋值新tab,保持排序
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}
看了HashMap 的源码,发现二进制运算这一块很薄弱,基础功不扎实,有待提高!