标签:
LRU(Least Recently Used)最近最少使用算法
缓存保存了一个强引用(Android 2.3开始,垃圾回收器更倾向于回收弱引用和软引用,软引用和弱引用变得不可靠,Android 3.0中,图片的数据会存储在本地的内存当中,因而无法用一种可预见的方式将其释放)限制值的数量. 每当值被访问的时候,它会被移动到队列的头部. 当缓存已满的时候加入新的值时,队列中最后的值会出队,可能被回收
LRUCache内部维护主要是通过LinkedHashMap实现
这是一个安全的线程,多线程缓存通过同步实现?
默认情况下,缓存的大小是由值的数量决定,重写sizeOf计算不同的值
如果你缓存值需要明确释放,重写entryRemoved()
int maxMemory = (int) Runtime.getRuntime().maxMemory();
int mCacheSize = maxMemory / 8;
//给LruCache分配1/8 4M
mMemoryCache = new LruCache<String, Bitmap>(mCacheSize){
//必须重写此方法,来测量Bitmap的大小
@Override
protected int sizeOf(String key, Bitmap value) {
return value.getRowBytes() * value.getHeight();
}
};
mMemoryCache.put(key, bitmap)
mMemoryCache.get(key)
这个类不允许有空的键值. get,put,remove 返回空值,key对应的值不在缓存中
构造函数,初始化了最大容量和LinkedHashMap
/**
* @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);
}
这里将LinkedHashMap最后一个参数(accessOrder)设置为true,将accessOrder设置为true时,可以使遍历顺序和访问顺序一致,其内部双向链表将会按照近期最少访问到近期最多访问的顺序排列Entry对象
put方法,首先不允许键值为空,然后是线程安全,put的次数加一,size增加,以键值对的形式存入LinkedHashMap,如果之前已经存在了这个键值对,size减少成原来的大小,如果容量超过maxsize,将会删除最近很少访问的entry
/** * 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; }
put方法有一个很关键的地方超过最大值是会删除最近最少访问的
trimToSize首先线程安全,检查当前大小是否大于最大值,如果大于最大值,从LinkedHashMap中去除最近最少(循环删除链表首部元素)被访问的元素,获得键值,删除
/**
* 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);
}
}
get方法,首先key不能为空,线程安全,根据key,从LinkedHashMap中获得value,不为空的话返回,为空的话,创建一个key,创建失败返回null,创建成功,在LinkedHashMap中创建键值对,存在就覆盖,不存在size增加,返回value值
/** * 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; } }
核心代码分析完毕,想知道LinkedHashMap,请听下回哔哔
注:本文源码来自api 23
标签:
原文地址:http://blog.csdn.net/elinavampire/article/details/51478088