akka 的集群访问方式

akka  中采用startProxy分区代理 访问 ，跟使用shardRegion 来访问的区别

 

这两种访问方式是不是重了呢。

 而另外这是一个单例代理

private fun startUniverseCwarManager() {
  val settings = ClusterSingletonManagerSettings.create(actorSystem).withRole(ClusterRole.universe_cwar.name)
  actorSystem.actorOf(
    ClusterSingletonManager.props(UniverseCwarManager.props(), Handoff, settings),
    UNIVERSE_CWAR_MANAGER
  )
}

以上创建的集群单例，通过以下方式进行访问
protected fun startUniverseProxy(universeRole: UniverseRole) {

  val proxySettings = ClusterSingletonProxySettings.create(actorSystem).withRole(universeRole.clusterRole.name)
  val actorRef: ActorRef = actorSystem.actorOf(ClusterSingletonProxy.props(universeRole.proxyPath, proxySettings))
  universeProxies[universeRole] = actorRef
}



worldActor 中 又 创建了一个访问对象

/**
 * actor分布在universe_cwar节点上
 */
private fun startUniverseCwarShardRegion() {
  val settings = ClusterShardingSettings.create(actorSystem).withRole(ClusterRole.universe_cwar.name)
  val region = ClusterSharding.get(actorSystem).start(
    GameWorldShard.universe_cwar.name,
    UniverseCwarWorld.props(),
    settings,
    UCWorldMessageExtractor(),
    ShardCoordinator.LeastShardAllocationStrategy(5, 1),
    Handoff
  )
  logger.info("SharedRegion $region started.")

  ClusterClientReceptionist.get(actorSystem).registerService(region)
}

通过这种方式创建的集群分片 ， 通过以下 代理进行访问

/**
 * Retrieve the actor reference of the [[ShardRegion]] actor responsible for the named entity type.
 * The entity type must be registered with the [[#start]] or [[#startProxy]] method before it
 * can be used here. Messages to the entity is always sent via the `ShardRegion`.
 */

就是只要  

protected fun startUniverseCwarShardProxy() {
  ClusterSharding.get(actorSystem).startProxy(
    GameWorldShard.universe_cwar.name, Optional.of(ClusterRole.universe_cwar.name),
    UCWorldMessageExtractor()
  )
    .let { logger.info("UniverseCwar shard proxy $it started.") }
} 

开始的方式， 就能以以下的方式获取到

ClusterSharding.get(context.system()).let {

ucWorldShardProxy = it.shardRegion(GameWorldShard.universe_cwar.name)
}