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【HashMap源码分析】---- 源代码注释逐行分析

【HashMap源码分析】---- 源代码注释逐行分析

作者: qming_c | 来源:发表于2018-02-05 23:39 被阅读0次

    注意注意,hashmap里的红黑树的节点是node的一个子类,所以这个树节点也可以使用next,在构建数时他的next指针会保留,当需要的时候仍可以使用。

    TreeNode<K,V> extends LinkedHashMap.Entry<K,V>
    Entry<K,V> extends HashMap.Node<K,V>
    
    package java.util;
    
    import java.io.IOException;
    import java.io.InvalidObjectException;
    import java.io.Serializable;
    import java.lang.reflect.ParameterizedType;
    import java.lang.reflect.Type;
    import java.util.function.BiConsumer;
    import java.util.function.BiFunction;
    import java.util.function.Consumer;
    import java.util.function.Function;
    @jdk8
    
    jdk1.2引入
    
    public class HashMap<K,V> extends AbstractMap<K,V> implements Map<K,V>, Cloneable, Serializable {
    

    序列化ID
    private static final long serialVersionUID = 362498820763181265L;
    初始容量16
    static final int DEFAULT_INITIAL_CAPACITY = 1 << 4;
    最大容量2^30
    static final int MAXIMUM_CAPACITY = 1 << 30;
    默认负载因子(当hashmap中容量打到 负载因子当前容量时,扩容为当前容量2)
    static final float DEFAULT_LOAD_FACTOR = 0.75f;
    一个桶链表变换乘红黑树的临界值 不能小于2
    static final int TREEIFY_THRESHOLD = 8;
    红黑树退化成链表的临界值
    static final int UNTREEIFY_THRESHOLD = 6;
    static final int MIN_TREEIFY_CAPACITY = 64;
    key value 在hashmap中的存储形式
    static class Node<K,V> implements Map.Entry<K,V> {
    hash值
    final int hash;
    key值
    final K key;
    value值
    V value;
    Node<K,V> next;
    构造函数

    Node(int hash, K key, V value, Node<K,V> next) {
                this.hash = hash;
                this.key = key;
                this.value = value;
                this.next = next;
            }
    

    Entry中getKey
    public final K getKey() { return key; }
    Entry中getValue
    public final V getValue() { return value; }
    public final String toString() { return key + "=" + value; }
    计算哈希码 key的hash异或value的hash
    public final int hashCode() { return Objects.hashCode(key) ^ Objects.hashCode(value); }
    更改value,并返回原来的value
    public final V setValue(V newValue) { V oldValue = value; value = newValue; return oldValue; }
    先判断地址,如果地址不同判断类型和key,value

    public final boolean equals(Object o) {
                if (o == this)
                    return true;
                if (o instanceof Map.Entry) {
                    Map.Entry<?,?> e = (Map.Entry<?,?>)o;
                    if (Objects.equals(key, e.getKey()) &&
                        Objects.equals(value, e.getValue()))
                        return true;
                }
                return false;
            }
        }
    

    ??为啥要这么算有空再查查。

    static final int hash(Object key) {
            int h;
            return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
        }
    

    如果该类实现了Comparable了 返回x的class对,否则返回null

     static Class<?> comparableClassFor(Object x) {
            if (x instanceof Comparable) {
                Class<?> c; Type[] ts, as; Type t; ParameterizedType p;
                if ((c = x.getClass()) == String.class) // bypass checks
                    return c;
                if ((ts = c.getGenericInterfaces()) != null) {
                    for (int i = 0; i < ts.length; ++i) {
                        if (((t = ts[i]) instanceof ParameterizedType) &&
                            ((p = (ParameterizedType)t).getRawType() ==
                             Comparable.class) &&
                            (as = p.getActualTypeArguments()) != null &&
                            as.length == 1 && as[0] == c) // type arg is c
                            return c;
                    }
                }
            }
            return null;
        }
    

    如果x时kc 返回k.((Comparable)k).compareTo(x)) 否则返回0

        @SuppressWarnings({"rawtypes","unchecked"}) // for cast to Comparable
        static int compareComparables(Class<?> kc, Object k, Object x) {
            return (x == null || x.getClass() != kc ? 0 :
                    ((Comparable)k).compareTo(x));
        }
    

    用来扩容的,好巧妙啊,大神就时大神
    看了半天没看懂
    用了几个测试用例测试了一下
    cap = -1,0 1 ---> n = 1
    cap = 2 ---> n = 2
    cap = 3,4 ---> n = 4
    cap = 5,6,7,8 ---> 8
    cap = 9,10,11,12,13,14,15,16---> 16

        static final int tableSizeFor(int cap) {
            int n = cap - 1;
            n |= n >>> 1;
            n |= n >>> 2;
            n |= n >>> 4;
            n |= n >>> 8;
            n |= n >>> 16;
            return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
        }
    

    这应该就是哈希"表"了吧
    transient Node<K,V>[] table;
    所有的entry
    transient Set<Map.Entry<K,V>> entrySet;
    目前的size
    transient int size;
    iterator中用来防止篡改的
    transient int modCount;
    capacity * loadfactor
    int threshold;
    当前负载因子
    final float loadFactor;
    构造函数 如果initialCapacity 不是2^n 会向上进位

        public HashMap(int initialCapacity, float loadFactor) {
            if (initialCapacity < 0)
                throw new IllegalArgumentException("Illegal initial capacity: " +
                                                   initialCapacity);
            if (initialCapacity > MAXIMUM_CAPACITY)
                initialCapacity = MAXIMUM_CAPACITY;
            if (loadFactor <= 0 || Float.isNaN(loadFactor))
                throw new IllegalArgumentException("Illegal load factor: " +
                                                   loadFactor);
            this.loadFactor = loadFactor;
            this.threshold = tableSizeFor(initialCapacity);
        }
    
        public HashMap(int initialCapacity) {
            this(initialCapacity, DEFAULT_LOAD_FACTOR);
        }
    
        public HashMap() {
            this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
        }
    
        
        public HashMap(Map<? extends K, ? extends V> m) {
            this.loadFactor = DEFAULT_LOAD_FACTOR;
            putMapEntries(m, false);
        }
    

    将一个map加入到这个hashmap中 evict??,看看下面putVal咋实现的

        final void putMapEntries(Map<? extends K, ? extends V> m, boolean evict) {
            int s = m.size();
            if (s > 0) {
                if (table == null) { // pre-size
                    float ft = ((float)s / loadFactor) + 1.0F;
                    int t = ((ft < (float)MAXIMUM_CAPACITY) ?
                             (int)ft : MAXIMUM_CAPACITY);
                    if (t > threshold)
                        threshold = tableSizeFor(t);
                }
                else if (s > threshold)
                    resize();
                for (Map.Entry<? extends K, ? extends V> e : m.entrySet()) {
                    K key = e.getKey();
                    V value = e.getValue();
                    putVal(hash(key), key, value, false, evict);
                }
            }
        }
    

    size O(1)
    public int size() { return size; }
    isEmpty O(1)
    public boolean isEmpty() { return size == 0; }
    根据key返回val 时间复杂度取决于下面的getNode

    public V get(Object key) {
            Node<K,V> e;
            return (e = getNode(hash(key), key)) == null ? null : e.value;
        }
    

    第一个node的地址为(n - 1) & hash
    首先判断第一个节点与要找的key一样不一样

    • 判断的过程:先==,后equals

    然后判断该桶里时红黑树还是链表,然后根据对应的规则进行搜索

    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;
        }
    

    containskey通过getNode来实现

        public boolean containsKey(Object key) {
            return getNode(hash(key), key) != null;
        }
    

    put没啥说的调用putVal

        public V put(K key, V value) {
            return putVal(hash(key), key, value, false, true);
        }
    
    1. 首先判断实例中是否以及存在表了,如果不存在,通过resize创建一个Node数组
    2. 然后通过传入的key的【hash&(n -1)】计算索引查找所在的桶内是否以及存在元素了,
    3. 如果没有元素在这个桶里,根据传入的kv创建桶的头结点
      1. ++modCount;
      2. size++
      3. 判断是够扩容
      4. afterNodeInsertion(evict);
    4. 如果桶内已经存在元素了
      1. 如果发现桶内的头结点key与传入的key的值地址相同,或者key equals k, 令e等于当前节点
      2. 如果该桶已经进化成红黑数了,利用红黑树的api进行插入或者更新 e等于返回的节点
      3. 否则,向链表后面走,并更新e为当前节点,如果找到了与key对应的节点 p=e,如果没找到就在最后没新建一个,此时若发现打到了进化红黑树的阀值,将此链表进化
      4. 如果e中val为空插入val,不为空 根据onlyIfAbsent判断是否更新,如果false---更新,true---不更新e---
    • 执行 afterNodeAccess(e);
        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;
        }
    

    如果之前有实例中有table已经初始化并且桶数大于8 扩容2倍后小于2^30,进行扩容,并更新thr
    如果之前table为空,

    • 但oldThr有值,cap = oldThr,并更新thr
    • 如果oldThr等于0,cap = 0,thr = DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY
      根据新的容量新建一个数组,原来的顺序会被打乱,甚至一个桶内的元素会被散列到不同位置去,并将老数组的每一个位置更新为null,便于垃圾回收
     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;
        }
    

    把链表替换成红黑树,这个算法有空可以看一下

        final void treeifyBin(Node<K,V>[] tab, int hash) {
            int n, index; Node<K,V> e;
            if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
                resize();
            else if ((e = tab[index = (n - 1) & hash]) != null) {
                TreeNode<K,V> hd = null, tl = null;
                do {
                    TreeNode<K,V> p = replacementTreeNode(e, null);
                    if (tl == null)
                        hd = p;
                    else {
                        p.prev = tl;
                        tl.next = p;
                    }
                    tl = p;
                } while ((e = e.next) != null);
                if ((tab[index] = hd) != null)
                    hd.treeify(tab);
            }
        }
    

    将map复制到实例中,如果key存在会覆盖掉之前的值
    public void putAll(Map<? extends K, ? extends V> m) { putMapEntries(m, true); }

    public V remove(Object key) {
            Node<K,V> e;
            return (e = removeNode(hash(key), key, null, false, true)) == null ?
                null : e.value;
        }
    

    基本思路与put类似,然后先找到节点,然后删除

        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;
        }
    

    清空实例里所以的元素

    public void clear() {
            Node<K,V>[] tab;
            modCount++;
            if ((tab = table) != null && size > 0) {
                size = 0;
                for (int i = 0; i < tab.length; ++i)
                    tab[i] = null;
            }
        }
    

    判断包含value
    这里有的桶里已经时红黑树了,但是仍用next去寻找

    public boolean containsValue(Object value) {
            Node<K,V>[] tab; V v;
            if ((tab = table) != null && size > 0) {
                for (int i = 0; i < tab.length; ++i) {
                    for (Node<K,V> e = tab[i]; e != null; e = e.next) {
                        if ((v = e.value) == value ||
                            (value != null && value.equals(v)))
                            return true;
                    }
                }
            }
            return false;
        }
    

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