LRUCache原理分析

一.注释

LRUCache的原理，基本都在注释里面描述清楚了。

/**
 * A cache that holds strong references to a limited number of values. Each time
 * a value is accessed, it is moved to the head of a queue. When a value is
 * added to a full cache, the value at the end of that queue is evicted and may
 * become eligible for garbage collection.
 *
 * <p>If your cached values hold resources that need to be explicitly released,
 * override {@link #entryRemoved}.
 *
 * <p>If a cache miss should be computed on demand for the corresponding keys,
 * override {@link #create}. This simplifies the calling code, allowing it to
 * assume a value will always be returned, even when there's a cache miss.
 *
 * <p>By default, the cache size is measured in the number of entries. Override
 * {@link #sizeOf} to size the cache in different units. For example, this cache
 * is limited to 4MiB of bitmaps:
 * <pre>   {@code
 *   int cacheSize = 4 * 1024 * 1024; // 4MiB
 *   LruCache<String, Bitmap> bitmapCache = new LruCache<String, Bitmap>(cacheSize) {
 *       protected int sizeOf(String key, Bitmap value) {
 *           return value.getByteCount();
 *       }
 *   }}</pre>
 *
 * <p>This class is thread-safe. Perform multiple cache operations atomically by
 * synchronizing on the cache: <pre>   {@code
 *   synchronized (cache) {
 *     if (cache.get(key) == null) {
 *         cache.put(key, value);
 *     }
 *   }}</pre>
 *
 * <p>This class does not allow null to be used as a key or value. A return
 * value of null from {@link #get}, {@link #put} or {@link #remove} is
 * unambiguous: the key was not in the cache.
 *
 * <p>This class appeared in Android 3.1 (Honeycomb MR1); it's available as part
 * of <a href="http://developer.android.com/sdk/compatibility-library.html">Android's
 * Support Package</a> for earlier releases.
 */

1.每次一个元素被访问，它会被move到队列的head位置。当某个元素加入到已经full的池里面，最久未使用的一个元素会被delete。

2.如果元素需要做特殊的释放操作，请重载entryRemoved

3.如果某个key值，没有对应的初始值，可以通过create方法提供默认值。

4.创建一个LRUCache需要复写sizeof方法。

5.这个类是线程安全的。

二：源代码分析：

1.变量

private final LinkedHashMap<K, V> map;
 
    /** Size of this cache in units. Not necessarily the number of elements. */
    private int size;      //count
    private int maxSize;  //容量
 
    private int putCount;
    private int createCount;
    private int evictionCount;  //delete count
    private int hitCount; 　　　　//命中 count
    private int missCount;　　　　//未命中 count　　

2.构造函数

/**
     * @param maxSize for caches that do not override {@link #sizeOf}, this is
     *     the maximum number of entries in the cache. For all other caches,
     *     this is the maximum sum of the sizes of the entries in this cache.
     */
    public LruCache(int maxSize) {
        if (maxSize <= 0) {
            throw new IllegalArgumentException("maxSize <= 0");
        }
        this.maxSize = maxSize;
        this.map = new LinkedHashMap<K, V>(0, 0.75f, true);
    }

这段代码最关键的是，new了一个LinkedHashMap，这个hashmap是可以根据访问先后次序，来控制顺序的。

LinkedHashMap：第一个参数表示初始数据是0个。

0.75f 是经典的加载因子的数值。

第三个参数true表示，以访问的先后顺序来排序。也就是做到了，容量满了以后，删除最久未访问的数据。

3.resize

    /**
     * Sets the size of the cache.
     *
     * @param maxSize The new maximum size.
     */
    public void resize(int maxSize) {
        if (maxSize <= 0) {
            throw new IllegalArgumentException("maxSize <= 0");
        }
 
        synchronized (this) {
            this.maxSize = maxSize;
        }
        trimToSize(maxSize);
    }

maxsize就是容量，也就是说，队列的最大长度，当full的时候，最后一个会被delete。

resize 就是调用trimToSize

4.trimToSize

/**
     * Remove the eldest entries until the total of remaining entries is at or
     * below the requested size.
     *
     * @param maxSize the maximum size of the cache before returning. May be -1
     *            to evict even 0-sized elements.
     */
    public void trimToSize(int maxSize) {
        while (true) {
            K key;
            V value;
            synchronized (this) {
                if (size < 0 || (map.isEmpty() && size != 0)) {
                    throw new IllegalStateException(getClass().getName()
                            + ".sizeOf() is reporting inconsistent results!");
                }
 
                if (size <= maxSize) {
                    break;
                }
 
                Map.Entry<K, V> toEvict = map.eldest();
                if (toEvict == null) {
                    break;
                }
 
                key = toEvict.getKey();
                value = toEvict.getValue();
                map.remove(key);
                size -= safeSizeOf(key, value);
                evictionCount++;
            }
 
            entryRemoved(true, key, value, null);
        }
    }

最外面的是while true 死循环。 然后开始判读size的值。1）获取最后的一个entry，拿到后，getkey & value。

2）map remove掉它。3）然后是获取这个entry占用的大小（不一定是1）。size 减去这个值。 4）evictionCount++

5）entryRemoved 之前说过，就是可以

5.get

/**
     * Returns the value for {@code key} if it exists in the cache or can be
     * created by {@code #create}. If a value was returned, it is moved to the
     * head of the queue. This returns null if a value is not cached and cannot
     * be created.
     */
    public final V get(K key) {
        if (key == null) {
            throw new NullPointerException("key == null");
        }
 
        V mapValue;
        synchronized (this) {
            mapValue = map.get(key);
            if (mapValue != null) {
                hitCount++;
                return mapValue;
            }
            missCount++;
        }
 
        /*
         * Attempt to create a value. This may take a long time, and the map
         * may be different when create() returns. If a conflicting value was
         * added to the map while create() was working, we leave that value in
         * the map and release the created value.
         */
 
        V createdValue = create(key);
        if (createdValue == null) {
            return null;
        }
 
        synchronized (this) {
            createCount++;
            mapValue = map.put(key, createdValue);
 
            if (mapValue != null) {
                // There was a conflict so undo that last put
                /**
     * Returns the value for {@code key} if it exists in the cache or can be
     * created by {@code #create}. If a value was returned, it is moved to the
     * head of the queue. This returns null if a value is not cached and cannot
     * be created.
     */
    public final V get(K key) {
        if (key == null) {
            throw new NullPointerException("key == null");
        }
 
        V mapValue;
        synchronized (this) {
            mapValue = map.get(key);
            if (mapValue != null) {
                hitCount++;
                return mapValue;
            }
            missCount++;
        }
 
        /*
         * Attempt to create a value. This may take a long time, and the map
         * may be different when create() returns. If a conflicting value was
         * added to the map while create() was working, we leave that value in
         * the map and release the created value.
         */
 
        V createdValue = create(key);
        if (createdValue == null) {
            return null;
        }
 
        synchronized (this) {
            createCount++;
            mapValue = map.put(key, createdValue);
 
            if (mapValue != null) {
                // There was a conflict so undo that last put
                map.put(key, mapValue);
            } else {
                size += safeSizeOf(key, createdValue);
            }
        }
 
        if (mapValue != null) {
            entryRemoved(false, key, createdValue, mapValue);
            return mapValue;
        } else {
            trimToSize(maxSize);
            return createdValue;
        }
    }map.put(key, mapValue);
            } else {
                size += safeSizeOf(key, createdValue);
            }
        }
 
        if (mapValue != null) {
            entryRemoved(false, key, createdValue, mapValue);
            return mapValue;
        } else {
            trimToSize(maxSize);
            return createdValue;
        }
    }

1）从map中获取元素，2）没有的话，就创建一个 3）由于存在多线程问题，可能已经好了一个。所以要删除最新创建的这个，或者增加size的值

4）如果create的值不需要，就释放create的值，or trimToSize

6.put

/**
     * Caches {@code value} for {@code key}. The value is moved to the head of
     * the queue.
     *
     * @return the previous value mapped by {@code key}.
     */
    public final V put(K key, V value) {
        if (key == null || value == null) {
            throw new NullPointerException("key == null || value == null");
        }
 
        V previous;
        synchronized (this) {
            putCount++;
            size += safeSizeOf(key, value);
            previous = map.put(key, value);
            if (previous != null) {
                size -= safeSizeOf(key, previous);
            }
        }
 
        if (previous != null) {
            entryRemoved(false, key, previous, value);
        }
 
        trimToSize(maxSize);
        return previous;
    }

map 存在一个多线程 create，put的问题。所以在put的时候，可能由其他线程已经存在了oldvalue值。所以根据

previous = map.put(key, value);

这个特性，判断previous的值，来确认是否是重复put的问题。这里关键是牵涉到size的大小问题。

其他方法，注释已经写的很清楚，不难理解。

总结：

• LruCache 封装了 LinkedHashMap，提供了 LRU 缓存的功能；
• LruCache 通过 trimToSize 方法自动删除最近最少访问的键值对；
• LruCache 不允许空键值；
• LruCache 线程安全；
• LruCache 的源码在不同版本中不一样，需要区分
• 继承 LruCache 时，必须要复写 sizeOf 方法，用于计算每个条目的大小。