fix DeleteTestGossipConnection

This is a one-sided operation, so we shouldn't be touching the
remote's Peers.

Also fix the associated test... When both 1 and 2 have lost their
connection to 3, they no longer have a route to 3 and hence gc it.

And because they no longer have a route, gossip from them won't get
through to 3.
This commit is contained in:
Matthias Radestock committed 2015-12-16 13:18:05 +00:00
1 parent 5401037743
commit 81df06b3f4
1 file changed
+3 -15
+3 -15
View File
@@ -65,16 +65,9 @@ func (router *Router) NewTestGossipConnection(r *Router) *mockGossipConnection {
}
func (router *Router) DeleteTestGossipConnection(r *Router) {
fromName := router.Ourself.Peer.Name
toName := r.Ourself.Peer.Name
fromPeer := r.Peers.Fetch(fromName)
toPeer := router.Peers.Fetch(toName)
r.Peers.Dereference(fromPeer)
router.Peers.Dereference(toPeer)
conn, _ := router.Ourself.ConnectionTo(toName)
router.Peers.Dereference(conn.Remote())
router.Ourself.handleDeleteConnection(conn)
}
@@ -136,15 +129,10 @@ func TestGossipTopology(t *testing.T) {
// Drop the connection from 1 to 3
r1.DeleteTestGossipConnection(r3)
sendPendingGossip(r1, r2, r3)
checkTopology(t, r1, r1.tp(r2), r2.tp(r1), r3.tp(r1, r2))
checkTopology(t, r2, r1.tp(r2), r2.tp(r1), r3.tp(r1, r2))
checkTopology(t, r1, r1.tp(r2), r2.tp(r1))
checkTopology(t, r2, r1.tp(r2), r2.tp(r1))
// r3 still thinks r1 has a connection to it
checkTopology(t, r3, r1.tp(r2, r3), r2.tp(r1), r3.tp(r1, r2))
// On a timer, r3 will gossip to r1
r3.SendAllGossip()
sendPendingGossip(r1, r2, r3)
checkTopology(t, r1, r1.tp(r2), r2.tp(r1), r3.tp(r1, r2))
}
func TestGossipSurrogate(t *testing.T) {