Map及其子类源码简单分析以及性能比较

1.HashMap

构造：key-value键值对，key采用hash函数来排列，加快查询速度

 

  /**
     * 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) {
        if (table == EMPTY_TABLE) {
            inflateTable(threshold);
        }
        if (key == null)
            return putForNullKey(value);
        int hash = hash(key);
        int i = indexFor(hash, table.length);
        for (Entry<K,V> e = table[i]; e != null; e = e.next) {
            Object k;
            if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
                V oldValue = e.value;
                e.value = value;
                e.recordAccess(this);
                return oldValue;
            }
        }

        modCount++;
        addEntry(hash, key, value, i);
        return null;
    }

HashMap中key支持为null

 

<span style="white-space:pre">	</span>@Test
	public void Test(){
		Map<String, String> map = new HashMap<String, String>();
		map.put(null, "tom1");
		map.put("key2", "tom2");
		map.entrySet();
		for(Entry<String, String> set:map.entrySet()){
			System.out.println(set.getKey()+"<>"+map.get(set.getKey()));
		}
		
	}

 

2.TreeMap

 

构造：键值对，增加一个comparator参数对key进行默认排序

 

  /**
     * Constructs a new, empty tree map, ordered according to the given
     * comparator.  All keys inserted into the map must be <em>mutually
     * comparable</em> by the given comparator: {@code comparator.compare(k1,
     * k2)} must not throw a {@code ClassCastException} for any keys
     * {@code k1} and {@code k2} in the map.  If the user attempts to put
     * a key into the map that violates this constraint, the {@code put(Object
     * key, Object value)} call will throw a
     * {@code ClassCastException}.
     *
     * @param comparator the comparator that will be used to order this map.
     *        If {@code null}, the {@linkplain Comparable natural
     *        ordering} of the keys will be used.
     */
    public TreeMap(Comparator<? super K> comparator) {
        this.comparator = comparator;
    }

put()时是即对key做排序，效率不会太高，所以只适用特定的环境

 

 public V put(K key, V value) {
        Entry<K,V> t = root;
        if (t == null) {
            compare(key, key); // type (and possibly null) check

            root = new Entry<>(key, value, null);
            size = 1;
            modCount++;
            return null;
        }
        int cmp;
        Entry<K,V> parent;
        // split comparator and comparable paths
        Comparator<? super K> cpr = comparator;
        if (cpr != null) {
            do {
                parent = t;
                cmp = cpr.compare(key, t.key);
                if (cmp < 0)
                    t = t.left;
                else if (cmp > 0)
                    t = t.right;
                else
                    return t.setValue(value);
            } while (t != null);
        }
        else {
            if (key == null)
                throw new NullPointerException();
            Comparable<? super K> k = (Comparable<? super K>) key;
            do {
                parent = t;
                cmp = k.compareTo(t.key);
                if (cmp < 0)
                    t = t.left;
                else if (cmp > 0)
                    t = t.right;
                else
                    return t.setValue(value);
            } while (t != null);
        }
        Entry<K,V> e = new Entry<>(key, value, parent);
        if (cmp < 0)
            parent.left = e;
        else
            parent.right = e;
        fixAfterInsertion(e);
        size++;
        modCount++;
        return null;
    }

3.ConcurrentHashMap

并发的HashMap，默认并发级别为16线程

public ConcurrentHashMap(int initialCapacity) {
        this(initialCapacity, DEFAULT_LOAD_FACTOR, DEFAULT_CONCURRENCY_LEVEL);
    }

put写操作是线程安全的，且是并发写到segments中

/**
     * Maps the specified key to the specified value in this table.
     * Neither the key nor the value can be null.
     *
     * <p> The value can be retrieved by calling the <tt>get</tt> method
     * with a key that is equal to the original key.
     *
     * @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>
     * @throws NullPointerException if the specified key or value is null
     */
    @SuppressWarnings("unchecked")
    public V put(K key, V value) {
        Segment<K,V> s;
        if (value == null)
            throw new NullPointerException();
        int hash = hash(key);
        int j = (hash >>> segmentShift) & segmentMask;
        if ((s = (Segment<K,V>)UNSAFE.getObject          // nonvolatile; recheck
             (segments, (j << SSHIFT) + SBASE)) == null) //  in ensureSegment
            s = ensureSegment(j); //确定哪个segment
        return s.put(key, hash, value, false);
    }

size()锁定所有segments，然后统计所有segments中的kv个数

 /**
     * Returns the number of key-value mappings in this map.  If the
     * map contains more than <tt>Integer.MAX_VALUE</tt> elements, returns
     * <tt>Integer.MAX_VALUE</tt>.
     *
     * @return the number of key-value mappings in this map
     */
    public int size() {
        // Try a few times to get accurate count. On failure due to
        // continuous async changes in table, resort to locking.
        final Segment<K,V>[] segments = this.segments;
        int size;
        boolean overflow; // true if size overflows 32 bits
        long sum;         // sum of modCounts
        long last = 0L;   // previous sum
        int retries = -1; // first iteration isn't retry
        try {
            for (;;) {
                if (retries++ == RETRIES_BEFORE_LOCK) {
                    for (int j = 0; j < segments.length; ++j)
                        ensureSegment(j).lock(); // force creation 循环锁定所有segments
                }
                sum = 0L;
                size = 0;
                overflow = false;
                for (int j = 0; j < segments.length; ++j) {
                    Segment<K,V> seg = segmentAt(segments, j);
                    if (seg != null) {
                        sum += seg.modCount;
                        int c = seg.count;
                        if (c < 0 || (size += c) < 0)
                            overflow = true;
                    }
                }
                if (sum == last)
                    break;
                last = sum;
            }
        } finally {
            if (retries > RETRIES_BEFORE_LOCK) {
                for (int j = 0; j < segments.length; ++j)
                    segmentAt(segments, j).unlock();
            }
        }
        return overflow ? Integer.MAX_VALUE : size;
    }

 

4.后记

我的博客：http://blog.csdn.net/caicongyang

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