W-TinyLFU源码分析(Caffeine)

2021-07-11  本文已影响0人  山里小龙

Caffeine使用一个ConcurrencyHashMap来保存所有数据,那它的过期淘汰策略采用的什么的方式与数据结构呢?其中写过期是使用writeOrderDeque,这个比较简单无需多说,而读过期相对复杂很多,使用W-TinyLFU的结构与算法。

网络上有很多文章介绍W-TinyLFU结构的,大家可以去查一下,这里主要是从源码来分析,总的来说它使用了三个双端队列:accessOrderEdenDeque,accessOrderProbationDeque,accessOrderProtectedDeque,使用双端队列的原因是支持LRU算法比较方便。

accessOrderEdenDeque属于eden区,缓存1%的数据,其余的99%缓存在main区。

accessOrderProbationDeque属于main区,缓存main内数据的20%,这部分是属于冷数据,即将补淘汰。

accessOrderProtectedDeque属于main区,缓存main内数据的80%,这部分是属于热数据,是整个缓存的主存区。

我们先看一下淘汰方法入口:

  void evictEntries() {
    if (!evicts()) {
      return;
    }
    //先从edn区淘汰
    int candidates = evictFromEden();
    //eden淘汰后的数据进入main区,然后再从main区淘汰
    evictFromMain(candidates);
  }

accessOrderEdenDeque对应W-TinyLFU的W(window),这里保存的是最新写入数据的引用,它使用LRU淘汰,这里面的数据主要是应对突发流量的问题,淘汰后的数据进入
accessOrderProbationDeque.代码如下:

int evictFromEden() {
    int candidates = 0;
    Node<K, V> node = accessOrderEdenDeque().peek();
    while (edenWeightedSize() > edenMaximum()) {
      // The pending operations will adjust the size to reflect the correct weight
      if (node == null) {
        break;
      }

      Node<K, V> next = node.getNextInAccessOrder();
      if (node.getWeight() != 0) {
        node.makeMainProbation();
        //先从eden区移除
        accessOrderEdenDeque().remove(node);
        //移除的数据加入到main区的probation队列
        accessOrderProbationDeque().add(node);
        candidates++;

        lazySetEdenWeightedSize(edenWeightedSize() - node.getPolicyWeight());
      }
      node = next;
    }

    return candidates;
  }

数据进入probation队列后,继续执行以下代码:

 void evictFromMain(int candidates) {
    int victimQueue = PROBATION;
    Node<K, V> victim = accessOrderProbationDeque().peekFirst();
    Node<K, V> candidate = accessOrderProbationDeque().peekLast();
    while (weightedSize() > maximum()) {
      // Stop trying to evict candidates and always prefer the victim
      if (candidates == 0) {
        candidate = null;
      }

      // Try evicting from the protected and eden queues
      if ((candidate == null) && (victim == null)) {
        if (victimQueue == PROBATION) {
          victim = accessOrderProtectedDeque().peekFirst();
          victimQueue = PROTECTED;
          continue;
        } else if (victimQueue == PROTECTED) {
          victim = accessOrderEdenDeque().peekFirst();
          victimQueue = EDEN;
          continue;
        }

        // The pending operations will adjust the size to reflect the correct weight
        break;
      }

      // Skip over entries with zero weight
      if ((victim != null) && (victim.getPolicyWeight() == 0)) {
        victim = victim.getNextInAccessOrder();
        continue;
      } else if ((candidate != null) && (candidate.getPolicyWeight() == 0)) {
        candidate = candidate.getPreviousInAccessOrder();
        candidates--;
        continue;
      }

      // Evict immediately if only one of the entries is present
      if (victim == null) {
        candidates--;
        Node<K, V> evict = candidate;
        candidate = candidate.getPreviousInAccessOrder();
        evictEntry(evict, RemovalCause.SIZE, 0L);
        continue;
      } else if (candidate == null) {
        Node<K, V> evict = victim;
        victim = victim.getNextInAccessOrder();
        evictEntry(evict, RemovalCause.SIZE, 0L);
        continue;
      }

      // Evict immediately if an entry was collected
      K victimKey = victim.getKey();
      K candidateKey = candidate.getKey();
      if (victimKey == null) {
        Node<K, V> evict = victim;
        victim = victim.getNextInAccessOrder();
        evictEntry(evict, RemovalCause.COLLECTED, 0L);
        continue;
      } else if (candidateKey == null) {
        candidates--;
        Node<K, V> evict = candidate;
        candidate = candidate.getPreviousInAccessOrder();
        evictEntry(evict, RemovalCause.COLLECTED, 0L);
        continue;
      }

      // Evict immediately if the candidate's weight exceeds the maximum
      if (candidate.getPolicyWeight() > maximum()) {
        candidates--;
        Node<K, V> evict = candidate;
        candidate = candidate.getPreviousInAccessOrder();
        evictEntry(evict, RemovalCause.SIZE, 0L);
        continue;
      }

      // Evict the entry with the lowest frequency
      candidates--;
      //最核心算法在这里:从probation的头尾取出两个node进行比较频率,频率更小者将被remove
      if (admit(candidateKey, victimKey)) {
        Node<K, V> evict = victim;
        victim = victim.getNextInAccessOrder();
        evictEntry(evict, RemovalCause.SIZE, 0L);
        candidate = candidate.getPreviousInAccessOrder();
      } else {
        Node<K, V> evict = candidate;
        candidate = candidate.getPreviousInAccessOrder();
        evictEntry(evict, RemovalCause.SIZE, 0L);
      }
    }
  }

上面的代码逻辑是从probation的头尾取出两个node进行比较频率,频率更小者将被remove,其中尾部元素就是上一部分从eden中淘汰出来的元素,如果将两步逻辑合并起来讲是这样的: 在eden队列通过lru淘汰出来的”候选者“与probation队列通过lru淘汰出来的“被驱逐者“进行频率比较,失败者将被从cache中真正移除。下面看一下它的比较逻辑admit:

  boolean admit(K candidateKey, K victimKey) {
    int victimFreq = frequencySketch().frequency(victimKey);
    int candidateFreq = frequencySketch().frequency(candidateKey);
    //如果候选者的频率高就淘汰被驱逐者
    if (candidateFreq > victimFreq) {
      return true;
      //如果被驱逐者比候选者的频率高,并且候选者频率小于等于5则淘汰者
    } else if (candidateFreq <= 5) {
      // The maximum frequency is 15 and halved to 7 after a reset to age the history. An attack
      // exploits that a hot candidate is rejected in favor of a hot victim. The threshold of a warm
      // candidate reduces the number of random acceptances to minimize the impact on the hit rate.
      return false;
    }
    //随机淘汰
    int random = ThreadLocalRandom.current().nextInt();
    return ((random & 127) == 0);
  }

从frequencySketch取出候选者与被驱逐者的频率,如果候选者的频率高就淘汰被驱逐者,如果被驱逐者比候选者的频率高,并且候选者频率小于等于5则淘汰者,如果前面两个条件都不满足则随机淘汰。

整个过程中你是不是发现protectedDeque并没有什么作用,那它是怎么作为主存区来保存大部分数据的呢?

//onAccess方法触发该方法 
void reorderProbation(Node<K, V> node) {
    if (!accessOrderProbationDeque().contains(node)) {
      // Ignore stale accesses for an entry that is no longer present
      return;
    } else if (node.getPolicyWeight() > mainProtectedMaximum()) {
      return;
    }

    long mainProtectedWeightedSize = mainProtectedWeightedSize() + node.getPolicyWeight();
   //先从probation中移除
   accessOrderProbationDeque().remove(node);
  //加入到protected中
    accessOrderProtectedDeque().add(node);
    node.makeMainProtected();

    long mainProtectedMaximum = mainProtectedMaximum();
  //从protected中移除
    while (mainProtectedWeightedSize > mainProtectedMaximum) {
      Node<K, V> demoted = accessOrderProtectedDeque().pollFirst();
      if (demoted == null) {
        break;
      }
      demoted.makeMainProbation();
      //加入到probation中
      accessOrderProbationDeque().add(demoted);
      mainProtectedWeightedSize -= node.getPolicyWeight();
    }

    lazySetMainProtectedWeightedSize(mainProtectedWeightedSize);
  }

当数据被访问时并且该数据在probation中,这个数据就会移动到protected中去,同时通过lru从protected中淘汰一个数据进入到probation中。

这样数据流转的逻辑全部通了:新数据都会进入到eden中,通过lru淘汰到probation,并与probation中通过lru淘汰的数据进行使用频率pk,如果胜利了就继续留在probation中,如果失败了就会被直接淘汰,当这条数据被访问了,则移动到protected。当其它数据被访问了,则它可能会从protected中通过lru淘汰到probation中。

TinyLFU

传统LFU一般使用key-value形式来记录每个key的频率,优点是数据结构非常简单,并且能跟缓存本身的数据结构复用,增加一个属性记录频率就行了,它的缺点也比较明显就是频率这个属性会占用很大的空间,但如果改用压缩方式存储频率呢? 频率占用空间肯定可以减少,但会引出另外一个问题:怎么从压缩后的数据里获得对应key的频率呢?

TinyLFU的解决方案是类似位图的方法,将key取hash值获得它的位下标,然后用这个下标来找频率,但位图只有0、1两个值,那频率明显可能会非常大,这要怎么处理呢? 另外使用位图需要预占非常大的空间,这个问题怎么解决呢?

TinyLFU根据最大数据量设置生成一个long数组,然后将频率值保存在其中的四个long的4个bit位中(4个bit位不会大于15),取频率值时则取四个中的最小一个。

image

Caffeine认为频率大于15已经很高了,是属于热数据,所以它只需要4个bit位来保存,long有8个字节64位,这样可以保存16个频率。取hash值的后左移两位,然后加上hash四次,这样可以利用到16个中的13个,利用率挺高的,或许有更好的算法能将16个都利用到。

  public void increment(@Nonnull E e) {
    if (isNotInitialized()) {
      return;
    }

    int hash = spread(e.hashCode());
    int start = (hash & 3) << 2;

    // Loop unrolling improves throughput by 5m ops/s
    int index0 = indexOf(hash, 0); //indexOf也是一种hash方法,不过会通过tableMask来限制范围
    int index1 = indexOf(hash, 1);
    int index2 = indexOf(hash, 2);
    int index3 = indexOf(hash, 3);

    boolean added = incrementAt(index0, start);
    added |= incrementAt(index1, start + 1);
    added |= incrementAt(index2, start + 2);
    added |= incrementAt(index3, start + 3);

    //当数据写入次数达到数据长度时就重置
    if (added && (++size == sampleSize)) {
      reset();
    }
  }

给对应位置的bit位四位的Int值加1:

  boolean incrementAt(int i, int j) {
    int offset = j << 2;
    long mask = (0xfL << offset);
    //当已达到15时,次数不再增加
    if ((table[i] & mask) != mask) {
      table[i] += (1L << offset);
      return true;
    }
    return false;
  }

获得值的方法也是通过四次hash来获得,然后取最小值:

  public int frequency(@Nonnull E e) {
    if (isNotInitialized()) {
      return 0;
    }

    int hash = spread(e.hashCode());
    int start = (hash & 3) << 2;
    int frequency = Integer.MAX_VALUE;
    //四次hash
    for (int i = 0; i < 4; i++) {
      int index = indexOf(hash, i);
      //获得bit位四位的Int值
      int count = (int) ((table[index] >>> ((start + i) << 2)) & 0xfL);
      //取最小值
      frequency = Math.min(frequency, count);
    }
    return frequency;
  }

当数据写入次数达到数据长度时就会将次数减半,一些冷数据在这个过程中将归0,这样会使hash冲突降低:

  void reset() {
    int count = 0;
    for (int i = 0; i < table.length; i++) {
      count += Long.bitCount(table[i] & ONE_MASK);
      table[i] = (table[i] >>> 1) & RESET_MASK;
    }
    size = (size >>> 1) - (count >>> 2);
  }

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