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我正在尝试使用 spymemcached 2.6 与更新同步，我发现了以下两种使用方式：

  • 使用 CASMutation 定义一个 CASMutator，这是一种非常侵入性的实现方式，让我们看一个例子：

    public List<Item> addAnItem(final Item newItem) throws Exception {

    // This is how we modify a list when we find one in the cache.
    CASMutation<List<Item>> mutation = new CASMutation<List<Item>>() {

        // This is only invoked when a value actually exists.
        public List<Item> getNewValue(List<Item> current) {
            // Not strictly necessary if you specify the storage as
            // LinkedList (our initial value isn't), but I like to keep
            // things functional anyway, so I'm going to copy this list
            // first.
            LinkedList<Item> ll = new LinkedList<Item>(current);

            // If the list is already "full", pop one off the end.
            if(ll.size() > 10) {
                ll.removeLast();
            }
            // Add mine first.
            ll.addFirst(newItem);

            return ll;
        }
    };

    // The initial value -- only used when there's no list stored under
    // the key.
    List<Item> initialValue=Collections.singletonList(newItem);

    // The mutator who'll do all the low-level stuff.
    CASMutator<List<Item>> mutator = new CASMutator<List<Item>>(client, transcoder);

    // This returns whatever value was successfully stored within the
    // cache -- either the initial list as above, or a mutated existing
    // one
    return mutator.cas("myKey", initialValue, 0, mutation);
}

或通过使用cas方法

cas(String key, long casId, Object value)

完成后：

gets(String key, Transcoder<T> tc)

第二个真的更简单，我明白我为什么要使用 CASMutation ......我真的很高兴能得到一些关于使用这个 couchbase 客户端的反馈。

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1 回答 1

3

CASMutator/CASMutation 捕获最佳实践和工作流程，以便为您完成正确的事情。

你的反例看起来更简单，因为你没有说你实际上用这些方法做什么。您在上面发布的示例显示了一个从 memcached 中拉出的列表，一个新项目添加到其中，有条件地从中删除一些项目，然后将其放回原处。您发布的文本中至少有一半仍然需要编写。

如果你不使用CASMutator，你最终会重新发明它，而这并不是那么简单。这就是它今天为您所做的：

public T cas(final String key, final T initial, int initialExp,
        final CASMutation<T> m) throws Exception {
    T rv=initial;

    boolean done=false;
    for(int i=0; !done && i<max; i++) {
        CASValue<T> casval=client.gets(key, transcoder);
        T current=null;
        // If there were a CAS value, check to see if it's compatible.
        if(casval != null) {
            T tmp = casval.getValue();
            current=tmp;
        }
        // If we have anything mutate and CAS, else add.
        if(current != null) {
            // Declaring this impossible since the only way current can
            // be non-null is if casval was set.
            assert casval != null : "casval was null with a current value";

            rv=m.getNewValue(current);
            // There are three possibilities here:
            //  1) It worked and we're done.
            //  2) It collided and we need to reload and try again.
            //  3) It disappeared between our fetch and our cas.
            // We're ignoring #3 because it's *extremely* unlikely and the
            // behavior will be fine in this code -- we'll do another gets
            // and follow it up with either an add or another cas depending
            // on whether it exists the next time.
            if(client.cas(key, casval.getCas(), rv, transcoder)
                    == CASResponse.OK) {
                done=true;
            }
        } else {
            // No value found, try an add.
            if(initial == null) {
                done = true;
                rv = null;
            } else if(client.add(key, initialExp, initial, transcoder).get()) {
                done=true;
                rv=initial;
            }
        }
    }
    if(!done) {
        throw new RuntimeException("Couldn't get a CAS in " + max
            + " attempts");
    }

    return rv;
}
于 2011-06-15T19:11:43.027 回答