Quietly tell you

没有什么技术是一篇文档解决不了的,如果有,那就两篇。

ArrarList&HashMap, 这些基础你还记得吗?

ArrarList

创建:

new ArrayList();
下面这是一个ArrayList默认构造方法的源代码,他只进行了一次赋值操作,这里的this.elementData则是Array List中的数据存储表,也就是一个Object[],其中,DEFAULTCAPACITY_EMPTY_ELEMENTDATA是一个静态的公共的Object[],从这段代码可以证明在创建ArrayList的时候,当前实例会默认得到一个固定长度的数组。

    /**
     * Constructs an empty list with an initial capacity of ten.
     */
    public ArrayList() {
        this.elementData = DEFAULTCAPACITY_EMPTY_ELEMENTDATA;
    }

添加元素:

ArrayList.add();
这个ArrayList在创建并得到初始的数组容器后,当添加的内容超过这个数组容器本身的长度后会如何?
我的回答是当超过Array List内部数组本身长度后,他会new一个新的、更长的数组,同时,吧旧数组的内容再重新赋值给新的数组中去,并将this.elementData指针指向新的数组。
且看下面JDK源码,这段代码中执行了这么一句话ensureCapacityInternal(size + 1):

    /**
     * Appends the specified element to the end of this list.
     *
     * @param e element to be appended to this list
     * @return <tt>true</tt> (as specified by {@link Collection#add})
     */
    public boolean add(E e) {
        ensureCapacityInternal(size + 1);  // Increments modCount!!
        elementData[size++] = e;
        return true;
    }

那么这个ensureCapacityInternal()方法就是得到下一个坐标,当下一个坐标超过数组界限是,就会触发new新数组的过程了,如下:

private void ensureCapacityInternal(int minCapacity) {
    if (elementData == DEFAULTCAPACITY_EMPTY_ELEMENTDATA) {
        minCapacity = Math.max(DEFAULT_CAPACITY, minCapacity);
    }

    ensureExplicitCapacity(minCapacity);
}

private void ensureExplicitCapacity(int minCapacity) {
    modCount++;

    // overflow-conscious code
    if (minCapacity - elementData.length > 0)
        grow(minCapacity);
}

从上面代码会发现,当if (minCapacity – elementData.length > 0)时,也就是数组不够时,会触发grow()方法(创建新数组并将旧数组的值复制到新的数组容器中去),JDK代码如下:

    /**
     * Increases the capacity to ensure that it can hold at least the
     * number of elements specified by the minimum capacity argument.
     *
     * @param minCapacity the desired minimum capacity
     */
    private void grow(int minCapacity) {
        // overflow-conscious code
        int oldCapacity = elementData.length;
        int newCapacity = oldCapacity + (oldCapacity >> 1);
        if (newCapacity - minCapacity < 0)
            newCapacity = minCapacity;
        if (newCapacity - MAX_ARRAY_SIZE > 0)
            newCapacity = hugeCapacity(minCapacity);
        // minCapacity is usually close to size, so this is a win:
        elementData = Arrays.copyOf(elementData, newCapacity);
    }

证明:

以上代码证明,我的理解是正确的,并且从这整个流程中可以看出,ArrayList本身也是一个线程不安全的集合,他缺失了原子性。那么如何让他线程安全呢?我使用synchronized修饰add方法和get方法可以使ArrayList变得线程安全。
另外,问道我除了ArrayList之外,用其他的集合方式能否实现线程安全,我只到有那么一个集合,和Array List非常相似,但当时一直没有想起是什么来,因为我几乎没有用到这个类,其实他叫Vector,这里我给出的解释是:
因为ArrayList本身是基于数组的一个List,所以他的查询性能不需质疑,但对于存储性能来说,因为他在存储的时候可能会重新计算索引,所以在存储性能上其实是比较浪费资源的,尤其是在并发场景下,所以一般情况下我用ArrayList都是在单线程中去使用,更多的是用它来承载列表数据。也没有遇到过硬性需求使我必须在多线程的场景下使用ArrayList,如果有这个需求,我会选择前者(修饰add和get方法使其同步)因为在Vector中,也是这么做的。并且Java官方也并不是很鼓励大量使用Vector。

HashMap

HashMap是一个基于哈希表的集合

创建:

new HashMap();
再JVM中,默认存在一个HashMap的初始因子,他的值时0.75f

    /**
     * The load factor used when none specified in constructor.
     */
    static final float DEFAULT_LOAD_FACTOR = 0.75f;
  • 1
  • 2
  • 3
  • 4

当调用HashMap的构造方法时,他执行了下面代码,这段代码就是给当前实例的哈希表设置一个初始因子(这也就是所谓的哈希表了,而这也是他的核心,下面的存取操作也是基于这里的哈希表):

    /**
     * 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
    }

赋值:
Map.put(key, value);
源代码:

    /**
     * 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);
    }

其中,hash(key)这个函数的作用是,将用户所传入的key计算成一个值,这个值我叫他哈希值,int类型,他对应的这个值的内存的对应的数据,也就是value,所以,当put一个重复的key时,他的哈希值计算结果时相同的,那么效果就是,key不变,value被覆盖。

证明:

这是hash函数的源代码,他得到了key的hashCode,同时为了避免在内存中不同实例所在内存的key的冲突,又进行了二次计算

/** 
 * Computes key.hashCode() and spreads (XORs) higher bits of hash
 * to lower.  Because the table uses power-of-two masking, sets of
 * hashes that vary only in bits above the current mask will
 * always collide. (Among known examples are sets of Float keys
 * holding consecutive whole numbers in small tables.)  So we
 * apply a transform that spreads the impact of higher bits
 * downward. There is a tradeoff between speed, utility, and
 * quality of bit-spreading. Because many common sets of hashes
 * are already reasonably distributed (so don't benefit from
 * spreading), and because we use trees to handle large sets of
 * collisions in bins, we just XOR some shifted bits in the
 * cheapest possible way to reduce systematic lossage, as well as
 * to incorporate impact of the highest bits that would otherwise
 * never be used in index calculations because of table bounds.
 */
static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

得到之前的key的哈希值后,直接找出他的value存储位置,并创建Node(key、Value),这个时候,赋值操作就完成了,当然在其中还会有很多场景下的健壮性处理。
这是put函数的源代码,这也是HashMap允许存储null值以及可能get出null的原因:

/**
   * 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;
  }

取值:
下面两个函数,也证明了以上观点,他在取值时,会根据传入的key计算出hash值,并拿到这个Node所在的位置,然后获得这个node的value。

/**
 * 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;
}

以上证明,我的回答是完全整却的!其实当时我也不是很确定我的回答是否正确,因为一直到现在我都在想办法学习新的东西例如分布式这是真的,一些基础的东西可能记忆有些模糊了,直到我今天又翻了源码。

在此贴出文章,在面试的过程中,如果有面试官再问到你这两个集合的问题时,咱们可以大声地喊出来,不要犹豫,对的就是对的。错的对不了,对的错不了。我就是因为当时不确定而丧失了一次机会。。。

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