HashMap 1.8 源码分析之put和get方法-一一网

hashcode & equals

hashcode 和equals在hash结构中十分重要,前者用于快速定位,后者用于比较替换

hashcode标准写法

    override fun hashCode(): Int {
    //用31的好处：VM会自动把它转化成位运算（cpu支持的运算）,运行速度快
        var result = sourceKey.hashCode()
        result += 31 * result + signature.hashCode()
        result += 31 * result + width
        result += 31 * result + height
        appliedTransformation?.let {
            result += 31 * result + it.hashCode()
        }
        result += 31 * result + decodedResourceClass.hashCode()
        result += 31 * result + options.hashCode()
        return result
    }
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equals 标准写法

override fun equals(o: Any?): Boolean {
        if (o is ResourceCacheKey) {
            return o.let {
            //kotlin== 相当于java equals
                return sourceKey == it.sourceKey &&
                        signature == it.signature &&
                        width == it.width &&
                        height == it.height &&
                        appliedTransformation == it.appliedTransformation &&
                        decodedResourceClass == it.decodedResourceClass &&
                        options == it.options
            }
        }
        return false
    }
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put方法分析

大致流程

通过hasdcode 计算出在hash表中存放的位置
如果该位置上没有节点,直接新建节点存放value
该位置上有节点,通过equals方法处理冲突

public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

//onlyIfAbsent =false
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        //表为空则通过resize方法建表
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
            
            //通过(n - 1) & hash 获取在表中index
        if ((p = tab[i = (n - 1) & hash]) == null)
          //当前位置没有值,则新建node存放value
            tab[i] = newNode(hash, key, value, null);
        else {
          //当前位置有值则通过equals处理冲突
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                //hash是通过hashcode计算而来,所以hashcode相等并且equals相等
                //则将e指向p,用于待会将value替换
                e = p;
            else if (p instanceof TreeNode)
                //hashcode相等,equals不相等情况
                //存到红黑树中
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                //hashcode相等,equals不相等情况
                //遍历单链表
                for (int binCount = 0; ; ++binCount) {
                    //遍历到单链表尾部
                    if ((e = p.next) == null) {
                        //在单链表尾部新建节点存放Node
                        p.next = newNode(hash, key, value, null);
                        //TREEIFY_THRESHOLD 默认为 8;
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            //超过8个,将节点树化
                            treeifyBin(tab, hash);
                        break;
                    }
                    //equals相等,则直接break ,待会替换新的value
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            //替换新的value 
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                //onlyIfAbsent =false ,将value替换成新的
                    e.value = value;
                afterNodeAccess(e);
                //把oldValue返回
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
        //size 大于阀值则扩容
            resize();
        afterNodeInsertion(evict);
        return null;
    }





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get 方法分析

大致流程

通过hashcode计算在表中的位置index
遍历单链表或者红黑树,如果equals相等则返回value

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) {
           
           //通过(n - 1) & hash 算出在表中的index , 如果不为null,则判断equals
            if (first.hash == hash && 
                ((k = first.key) == key || (key != null && key.equals(k))))
                //first 节点的equals 相等则返回first
                return first;
          
          //处理冲突的情况
          if ((e = first.next) != null) {
                //遍历红黑树找出equals相等的元素并返回
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
               //遍历单链表找出equals相等的元素并返回
               do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }


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