Question

我仅在没有任何shellcode的情况下从代理在UnetStack中创建路由。我要从路由表中删除路由吗？

以下是我的代码：

我想做的是每10秒在两条路线之间切换。因此，从我的源节点（节点5）到目标节点（节点1），路由从5变为3到4到1或从5变为2到4到1

模拟

platform = RealTimePlatform
channel.model = ProtocolChannelModel

channel.soundSpeed = 1500.mps          // c
channel.communicationRange = 100.m     // Rc

// run the simulation infinately
simulate  {
    // Destination node
    node '1', remote: 1101, address: 1, location: [ 0.m, 0.m, 0.m], shell: 5101, stack: { container ->
        container.add 'new_routing_agent', new new_routing_agent();
        container.add 'routing', new Router();
        container.add 'rdp', new RouteDiscoveryProtocol();
    }

    node '2', remote: 1102, address: 2, location: [ 0.m, 0.m, -75.m], shell: 5102, stack: { container ->
        container.add 'new_routing_agent', new new_routing_agent();
        container.add 'routing', new Router();
        container.add 'rdp', new RouteDiscoveryProtocol();
    }

    // neighbor node for node 5, and will be a next node for node 5 during routing
    node '3', remote: 1103, address: 3, location: [0.m, 0.m, -90.m], shell: 5103, stack: { container ->
        container.add 'new_routing_agent', new new_routing_agent();
        container.add 'routing', new Router();
        container.add 'rdp', new RouteDiscoveryProtocol();
    }

//Neighbor node for node 5 ,but not a next node for node 5
    node '4', remote: 1104, address: 4, location: [0.m, 0.m, -150.m], shell: 5104, stack: {container ->
        container.add 'new_routing_agent', new new_routing_agent();
        container.add 'routing', new Router();
        container.add 'rdp', new RouteDiscoveryProtocol();
    }

// Source node
    node '5', remote: 1105, address: 5, location: [0.m, 0.m, -160.m], shell: true, stack: {container ->
        container.add 'new_routing_agent', new new_routing_agent();
        container.add 'routing', new Router();
        container.add 'rdp', new RouteDiscoveryProtocol();
    }
}

代理商

class new_routing_agent extends UnetAgent {
def router
int addr
def flag

void addroute(int to, int via) {
    router.send new RouteDiscoveryNtf(to: to, nextHop: via)
}

void deleteRoutes(){
//        router.send new RouteDiscoveryNtf().setRoute()

}

void routeDynamically(){
    while(1){
        deleteRoutes()
        if(flag){
            addroute 1, 2
        }
        else{
            addroute 1,3
        }
        flag = !flag
        sleep(10000)
    }
}

void startup() {
    flag = false
    def phy = agentForService Services.PHYSICAL
    subscribe topic(phy)

    router = agentForService Services.ROUTING
    subscribe topic(router)

    def nodeInfo = agentForService Services.NODE_INFO
    addr = nodeInfo.address

    switch (addr){
        case 1:
            addroute 1, 1
            break
        case 2:
            addroute 1, 4
            break
        case 3:
            addroute 1, 4
            break
        case 4:
            addroute 1, 1
            break
        case 5:
            routeDynamically()
            break
        default:
            addroute 1, 1
            break
    }
}

void processMessage(Message msg) {

    }
}

Answer 1

作为ROUTING服务的一部分，UnetStack当前未将删除路由的消息标准化。但是，您可以使用与Shell删除路由相同的机制，请记住，将来的UnetStack版本可能会更改该机制。有关如何执行此操作的详细信息，请参见：How to delete route entries from routing table using an agent in UnetStack。

从Unetstack中的路由表中删除路由

1 个答案: