简单分析HashMap源码

雨点打透心脏的1/2处 2022-05-21 04:48 179阅读 0赞

在日常工作中,HashMap是我们经常用到的一个类,同时在面试过程中面试官也很喜欢问的一个对象!

因为面试的过程中被问的多了,所以就抽时间看了看源码^-^

在面试过程中,经常被问到的就是数据结构方面的问题,所以问到HashMap的时候的,难免就会问HashMap的底层结构是什么:

最简单的回答就是: HashMap底层采用的数据和链表的实现方式(现在加入了红黑树)

以下从日常使用HashMap的角度来看看HashMap的源码:

Map<Object, Object> map = new HashMap<>();   //这是个人最常用的一种方法

HashMap的空参构造方法如下:

/** * Constructs an empty <tt>HashMap</tt> with the default initial capacity * (16) and the default load factor (0.75). */ public HashMap() {
        this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
    }

/** * The load factor for the hash table. * * @serial */ final float loadFactor;

可以看到,空参的构造函数其实仅仅设置一个装载因子,默认是0.75f, 前面提到HashMap底层是一个数据加链表结构,此时这个0.75指的就是数据的长度的0.75(也就是HashMap的容量),HashMap的默认容量是16

Constructs an empty <tt>HashMap</tt> with the default initial capacity(16)

先提前做一下准备工作:

因为以下都要涉及到Node<K,V>这个类,所以下面先解释一下这个类

/** * Basic hash bin node, used for most entries. (See below for * TreeNode subclass, and in LinkedHashMap for its Entry subclass.) */ static class Node<K,V> implements Map.Entry<K,V> {
        final int hash;
        final K key;
        V value;
        Node<K,V> next;
    
        Node(int hash, K key, V value, Node<K,V> next) {
            this.hash = hash;
            this.key = key;
            this.value = value;
            this.next = next;
        }
    
        public final K getKey()        { return key; }
        public final V getValue()      { return value; }
        public final String toString() { return key + "=" + value; }
    
        public final int hashCode() {
            return Objects.hashCode(key) ^ Objects.hashCode(value);
        }
    
        public final V setValue(V newValue) {
            V oldValue = value;
            value = newValue;
            return oldValue;
        }
    
        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;
        }
    }

多次提到HashMap是一个数据加链表的结构,而这个Node<K,V>则是关键

HashMap实际上就是创建一个Node<K,V>的数组,再看Node<K,V>会发现,这个类是一个链表结构的类,简单介绍下其4个属性

hash : 保存的是要存储的key的哈希值

key: 传递进来key值

value: 传递进来要保存的value值

next: 下一个Node<K,V>对象 (链表结构的实现)

/** * 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);
    }

/** * 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;
    }

/** * The table, initialized on first use, and resized as * necessary. When allocated, length is always a power of two. * (We also tolerate length zero in some operations to allow * bootstrapping mechanics that are currently not needed.) */ transient Node<K,V>[] table;

从上面的代码中可以看到,在new HashMap()的时候,table其实是没有赋值的,也就是说第一次put的时候,table==null,所以会调用

n = (tab = resize()).length;

看resize()方法:

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;
    }

当我们第一次put的时候:执行如下代码(剥离一下)

Node<K,V>[] oldTab = table;
        int oldCap = (oldTab == null) ? 0 : oldTab.length;
        int oldThr = threshold;
        int newCap, newThr = 0;

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;

return newTab;

所以说其实HashMap的中数组的初始化,实际上是在我们第一次进行put操作的时候进行的!

当我们第二次进行put操作的时候(简单说一下过程,具体的代码还请读者自己看一下):

首先根据用户传入的key值计算得到key的hash值,然后再根据

(n - 1) & hash

可以获取到一个值: n为HashMap的容量(数组的长度),hash为key的哈希值,通过位运与算符(&)可以获得一个\[0-n)的值,也就是数组的下标,

然后再根据数组的下标去获取值,如果获取的值为null,这直接创建一个Node<K,V>,然后加入到数据即可,如果不为空,在进行其他逻辑处理,具体细看源码!(因为对红黑树不了解,所以..,等了解了红黑树之后再唠唠)!

再之后就是获取数据: 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;
    }

/** * 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;
    }

首先通过传入的key获取到所属数组的下标,如果有值则返回值,没有值则返回null

在这里需要提一个方法就是

getNode(hash(key), key)

中的hash()这个方法

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

可以看到获取hash值的时候用到了key的对象的hashCode的方法,所以如果你重写了hashCode方法,那就会出现问题

如下例子:

public class DemoController {
    
        public static void main(String[] args) {
      
            Map<Student, Object> map = new HashMap<>();
            Student s = new Student();
            s.setId(1L);
            s.setUsername("张三");
            map.put(s, 1);
            System.out.println(map.get(s));
            s.setId(2L);
            s.setUsername("李四");
            System.out.println(map.get(s));
    
        }
    }
    
    class Student {
    
        private Long id;
        private String username;
    
        public Long getId() {
            return id;
        }
    
        public void setId(Long id) {
            this.id = id;
        }
    
        public String getUsername() {
            return username;
        }
    
        public void setUsername(String username) {
            this.username = username;
        }
    
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
    
            Student student = (Student) o;
    
            if (id != null ? !id.equals(student.id) : student.id != null) return false;
            return username != null ? username.equals(student.username) : student.username == null;
        }
    
        @Override
        public int hashCode() {
            int result = id != null ? id.hashCode() : 0;
            result = 31 * result + (username != null ? username.hashCode() : 0);
            return result;
        }
    }

得到输出结果是:

null

所以这里需要注意一下!

随便提一下,在HashMap中是允许key 和 value值为null,但是在ConcurrentHashMap中是不允许(key 和 value 都不能为null)的:

final V putVal(K key, V value, boolean onlyIfAbsent) {
        if (key == null || value == null) throw new NullPointerException();
        return null;
    }

(小白日志)