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[cluster-bus] Send a MEET packet to a node if there is no inbound link #1307
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In some cases, when meeting a new node, if the handshake times out, we can end up with an inconsistent view of the cluster where the new node knows about all the nodes in the cluster, but the cluster does not know about this new node (or vice versa). To detect this inconsistency, we now check if a node has an outbound link but no inbound link, in this case it probably means this node does not know us. In this case we (re-)send a MEET packet to this node to do a new handshake with it. Signed-off-by: Pierre Turin <[email protected]>
Signed-off-by: Pierre Turin <[email protected]>
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Codecov ReportAll modified and coverable lines are covered by tests ✅
Additional details and impacted files@@ Coverage Diff @@
## unstable #1307 +/- ##
============================================
- Coverage 70.69% 70.67% -0.02%
============================================
Files 115 115
Lines 63153 63163 +10
============================================
Hits 44643 44643
- Misses 18510 18520 +10
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What if we disconnect the outbound link if inbound link is not available? I think it will lead to the same reconnection flow. Would it help with having simpler code and one unified flow. I'm not sure if it will perform the MEET
operation though.
} | ||
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/* If this is a MEET packet from an unknown node, we still process | ||
* the gossip section here since we have to trust the sender because | ||
* of the message type. */ | ||
if (!sender && type == CLUSTERMSG_TYPE_MEET) clusterProcessGossipSection(hdr, link); | ||
/* If this is a MEET packet from an unknown node, we still process | ||
* the gossip section here since we have to trust the sender because | ||
* of the message type. */ | ||
clusterProcessGossipSection(hdr, link); | ||
} |
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We don't need this change.
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True, but this double if with the same condition was driving me crazy.
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I don't mind this. But in general we avoid changes to the lines of code not related to the PR.
src/cluster_legacy.h
Outdated
@@ -60,6 +60,7 @@ typedef struct clusterLink { | |||
#define nodeIsPrimary(n) ((n)->flags & CLUSTER_NODE_PRIMARY) | |||
#define nodeIsReplica(n) ((n)->flags & CLUSTER_NODE_REPLICA) | |||
#define nodeInHandshake(n) ((n)->flags & CLUSTER_NODE_HANDSHAKE) | |||
#define nodeIsMeeting(n) ((n)->flags & CLUSTER_NODE_MEET) |
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How about nodeInMeetProcess
/ nodeInMeetState
?
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Ah yes, I do prefer nodeInMeetState()
.
src/cluster_legacy.c
Outdated
clusterDelNode(node); | ||
return 1; | ||
} | ||
if (node->link != NULL && node->inbound_link == NULL && | ||
!nodeInHandshake(node) && !nodeIsMeeting(node) && !nodeTimedOut(node) && |
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Should we create a macro for this node state check? Not readable at this point.
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nodeInNormalState()
?
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nodeInHealthyState()
?
[llength [R 0 CLUSTER NODES]] == 26 && | ||
[llength [R 1 CLUSTER NODES]] == 26 && | ||
[llength [R 2 CLUSTER NODES]] == 26 |
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Could we match certain value in the string output? I don't like this magic number comparison which can change in the near future.
In this case we would just re-open an outbound connection, which the other node will accept, but it won't force the other node to recognize us as being part of the cluster if it doesn't trust us yet. The only way to force the other node to add us to its cluster view is for us to send a MEET packet. |
Update test to check node IDs instead of relying on number of words. Rename nodeIsMeeting() to nodeInMeetState(). Introduce nodeInNormalState() macro. Signed-off-by: Pierre Turin <[email protected]>
CLUSTER MEET is an admin operation but I guess we are fine with the case of reinitiating it if the operation wasn't successful in first place and retry it. |
proc cluster_3_nodes_all_know_each_other {} { | ||
set node0_id [dict get [get_myself 0] id] | ||
set node1_id [dict get [get_myself 1] id] | ||
set node2_id [dict get [get_myself 2] id] | ||
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if { | ||
[cluster_get_node_by_id 0 $node0_id] != {} && | ||
[cluster_get_node_by_id 0 $node1_id] != {} && | ||
[cluster_get_node_by_id 0 $node2_id] != {} && | ||
[cluster_get_node_by_id 1 $node0_id] != {} && | ||
[cluster_get_node_by_id 1 $node1_id] != {} && | ||
[cluster_get_node_by_id 1 $node2_id] != {} && | ||
[cluster_get_node_by_id 2 $node0_id] != {} && | ||
[cluster_get_node_by_id 2 $node1_id] != {} && | ||
[cluster_get_node_by_id 2 $node2_id] != {} && | ||
[llength [R 0 CLUSTER LINKS]] == 4 && | ||
[llength [R 1 CLUSTER LINKS]] == 4 && | ||
[llength [R 2 CLUSTER LINKS]] == 4 | ||
} { | ||
return 1 | ||
} else { | ||
return 0 | ||
} | ||
} |
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From ChatGPT:
proc cluster_3_nodes_all_know_each_other {} { | |
set node0_id [dict get [get_myself 0] id] | |
set node1_id [dict get [get_myself 1] id] | |
set node2_id [dict get [get_myself 2] id] | |
if { | |
[cluster_get_node_by_id 0 $node0_id] != {} && | |
[cluster_get_node_by_id 0 $node1_id] != {} && | |
[cluster_get_node_by_id 0 $node2_id] != {} && | |
[cluster_get_node_by_id 1 $node0_id] != {} && | |
[cluster_get_node_by_id 1 $node1_id] != {} && | |
[cluster_get_node_by_id 1 $node2_id] != {} && | |
[cluster_get_node_by_id 2 $node0_id] != {} && | |
[cluster_get_node_by_id 2 $node1_id] != {} && | |
[cluster_get_node_by_id 2 $node2_id] != {} && | |
[llength [R 0 CLUSTER LINKS]] == 4 && | |
[llength [R 1 CLUSTER LINKS]] == 4 && | |
[llength [R 2 CLUSTER LINKS]] == 4 | |
} { | |
return 1 | |
} else { | |
return 0 | |
} | |
} | |
proc cluster_nodes_all_know_each_other {num_nodes} { | |
# Collect node IDs dynamically | |
set node_ids {} | |
for {set i 0} {$i < $num_nodes} {incr i} { | |
lappend node_ids [dict get [get_myself $i] id] | |
} | |
# Check if all nodes know each other | |
foreach node_id $node_ids { | |
foreach check_node_id $node_ids { | |
for {set node_index 0} {$node_index < $num_nodes} {incr node_index} { | |
if {[cluster_get_node_by_id $node_index $check_node_id] == {}} { | |
return 0 | |
} | |
} | |
} | |
} | |
# Verify cluster link counts for each node | |
set expected_links [expr {2 * ($num_nodes - 1)}] | |
for {set i 0} {$i < $num_nodes} {incr i} { | |
if {[llength [R $i CLUSTER LINKS]] != $expected_links} { | |
return 0 | |
} | |
} | |
return 1 | |
} |
In some cases, when meeting a new node, if the handshake times out, we can end up with an inconsistent view of the cluster where the new node knows about all the nodes in the cluster, but the cluster does not know about this new node (or vice versa).
To detect this inconsistency, we now check if a node has an outbound link but no inbound link, in this case it probably means this node does not know us. In this case we (re-)send a MEET packet to this node to do a new handshake with it.
This fixes the bug described in #1251.