blocks|key|1492462|text|也许使用RMI？是否有一个实例用作服务器，其余实例用作客户端？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1492463|我认为这将比从磁盘重新加载复杂得多。|1492464|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Using RMI perhaps? Have one instance working as server and the rest as clients? 

I think it would be much more complicated than reloading from disk.

blocks|key|1277703|text|将图表保存到磁盘，然后使用MappedByteBuffer将其映射到内存中。这两个进程应该使用相同的内存，该内存将与页面缓存共享。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1277704|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/nio/MappedByteBuffer.html^0|D|G|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Save your graph to disk, then map it into memory with <a href="http://docs.oracle.com/javase/7/docs/api/java/nio/MappedByteBuffer.html" rel="nofollow noreferrer">MappedByteBuffer</a>. Both processes should use the same memory, which will be shared with the page cache.

blocks|key|1492536|text|如果构建图形的开销很大，也许你可以序列化对象。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1492537|ByteArrayOutputStream+bos+=+new+ByteArrayOutputStream();
++++++++ObjectOutputStream+out+=+new+ObjectOutputStream(bos);
++++++++out.writeObject(graph);
++++++++out.flush();
++++++++byte+b[]+=+bos.toByteArray();
//you+can+use+FileOutputStream+instead+of+a+ByteArrayOutputStream|code-block|syntax|javascript|1492538|然后，您可以从该文件构建对象|1492539|ByteArrayInputStream+inputBuffer+=+new+ByteArrayInputStream(b);
++++++++ObjectInputStream+inputStream+=+new+ObjectInputStream(inputBuffer);
++++++++try+{
++++++++++++Graph+graph+=+(Graph)+inputStream.readObject();

++++++++}+finally+{
++++++++++++if+(inputStream+!=+null)+{
++++++++++++++++inputStream.close();
++++++++++++}
++++++++}|1492540|只需将ByteArrayInputStream替换为FileInputStream即可|1492541|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If is expensive to build your graph maybe you can serialize the object.

<pre><code>ByteArrayOutputStream bos = new ByteArrayOutputStream();
 ObjectOutputStream out = new ObjectOutputStream(bos);
 out.writeObject(graph);
 out.flush();
 byte b[] = bos.toByteArray();
//you can use FileOutputStream instead of a ByteArrayOutputStream
</code></pre>

Then you can build your object from the file

<pre><code>ByteArrayInputStream inputBuffer = new ByteArrayInputStream(b);
 ObjectInputStream inputStream = new ObjectInputStream(inputBuffer);
 try {
 Graph graph = (Graph) inputStream.readObject();

 } finally {
 if (inputStream != null) {
 inputStream.close();
 }
 }
</code></pre>

Just replace the ByteArrayInputStream with a FileInputStream

blocks|key|1277819|text|两个JVM听起来比实际情况更复杂。您是否考虑过进行一种“热部署”设置，即主程序加载图形，显示UI，然后请求(或自动查找)一个jar/类文件来加载包含实际算法代码的文件？这样，您的算法代码将在与图相同的jvm中运行，但您不必仅仅为了重新加载新的算法实现而重新加载图。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1277820|更新以在评论中解决OP的问题：|1277821|下面是你如何组织你的代码，使你的算法是可交换的。不同的算法做什么并不重要，只要它们对相同的输入数据进行操作即可。只需定义一个如下所示的接口，并让您的图形算法实现它。|1277822|public+interface+GraphAlgorithm+{
++public+void+doStuff(Map<whatever>+myBigGraph)
}|code-block|syntax|javascript|1277823|如果您的算法要将结果显示到某种小部件上，您也可以将其传入，或者让doStuff()返回某种results对象。|1277824|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Two JVMs sounds more complicated than it needs to be. Have you considered doing a kind of "hot deploy" setup, where your main program loads up the graph, displays the UI, and then asks for (or automatically looks for) a jar/class file to load that contains your actual algorithm code? That way your algorithm code would be running in the same jvm as your graph, but you wouldn't have to reload the graph just to reload a new algorithm implementation.

UPDATE to address OP's question in comment: 

Here's how you could structure your code so that your algorithms would be swappable. It doesn't matter what the various algorithms do, so long as they are operating on the same input data. Just define an interface like the following, and have your graph algorithms implement it.

<pre><code>public interface GraphAlgorithm {
 public void doStuff(Map&lt;whatever&gt; myBigGraph)
}
</code></pre>

If your algorithms are displaying results to some kind of widget, you could pass that in as well, or have doStuff() return some kind of results object.

blocks|key|1277851|text|如果问题只是在没有名称冲突的情况下动态加载和运行代码，那么定制的类加载器就足够了。对于新的运行，只需在新的类加载器中缓存所有类文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1277852|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

if the problem is just to dynamicly load and run your code without name clashes a custom class loader could be enough.
for a new run just cache all class files in a new classloader.

blocks|key|1277874|text|你是否考虑过简单地使用少量的样本数据来测试你的算法？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1277875|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Have you considered simply using a smaller amount of sample data for testing your algorithms?

blocks|key|1492734|text|你有没有考虑过OSGi平台？它驻留在单个JVM中，但将允许您使用算法升级捆绑包，而无需重启平台。因此，您可能会有一个长期运行的捆绑包，其中包含巨大的数据结构和访问数据的短期算法捆绑包。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1492735|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Did you consider OSGi platform? It lives in a single JVM, but will allow you to upgrade bundles with algorithms without platform restart. Thus you may have a long-term running bundle with your huge data structures and short-term algorithm bundles taking access to the data.

blocks|key|1492751|text|Terracotta在多个JVM实例之间共享内存，因此您可以轻松地将集群应用到您的系统。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1492752|entityMap|0|LINK|mutability|MUTABLE|url|http://www.terracotta.org/^0|0|A|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

<a href="http://www.terracotta.org/" rel="nofollow noreferrer">Terracotta</a> shares memory between many JVM instances so you can easily apply cluster to your system.

blocks|key|1277976|text|哇！派对迟到了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1277977|如果它在本地机器上，类似于映射的字节缓冲区，则有apache直接内存。http://directmemory.apache.org/|offset|length|1277978|如果你想把它分发出去，那就试试http://hazelcast.org/。它被许多大型项目所使用。当然，您的对象必须是可序列化的。|1277979|entityMap|0|LINK|mutability|MUTABLE|url|http://directmemory.apache.org/|1|http://hazelcast.org/^0|0|Z|V|0|0|F|L|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@$D|X|E|Y|1|Z]]|A|$]]|$1|H|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]|P|$5|K|L|M|A|$N|Q]]]]

Woo! late to the party. 

If its on a local machine, similar to mapped byte buffers, there is apache direct memory. <a href="http://directmemory.apache.org/" rel="nofollow">http://directmemory.apache.org/</a>

If you want it distributed, give <a href="http://hazelcast.org/" rel="nofollow">http://hazelcast.org/</a> a try. Its used by a lot of large projects. Of course, your objects must be serializable.

I build a huge graph in JVM (Scala) which I want to use repeatedly, tweaking algorithms. I'd rather not reload it each time from disk. Is there a way to have it sit in one JVM while connecting from another, where the algorithms are being developed?

How can I share memory between two JVM instances?

数据结构

服务器

我在JVM (Scala)中构建了一个巨大的图，我想反复使用它来调整算法。我不想每次都从磁盘重新加载它。有没有办法让它驻留在一个JVM中，同时从另一个正在开发算法的JVM进行连接？

问如何在两个JVM实例之间共享内存？
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