mirror of https://github.com/status-im/nim-eth.git
discv4: fix ping pong time removal algorithm
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@ -64,6 +64,8 @@ const
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ID_SIZE = 256
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BOND_EXPIRATION = initDuration(hours = 12)
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proc len(r: RoutingTable): int
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proc toNodeId*(pk: PublicKey): NodeId =
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readUintBE[256](keccak256.digest(pk.toRaw()).data)
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@ -125,24 +127,46 @@ proc cmp(x, y: (TimeKey, int64)): int =
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proc removeTooOldPingPongTime(k: KademliaProtocol) =
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const
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MaxEntries = 128
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MaxRC = 5
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MinEntries = 128
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MaxRC = MinEntries div 8
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if k.pingPongTime.len < MaxEntries:
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# instead of using fixed limit, we use dynamic limit
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# with minimum entries = 128.
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# remove 25% of too old entries if we need more space.
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# the reason maxEntries is twice routing table because we
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# store ping and pong time.
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let
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maxEntries = max(k.routing.len * 2, MinEntries)
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maxRemove = maxEntries div 4
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if k.pingPongTime.len < maxEntries:
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return
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k.pingPongTime.sort(cmp, order = SortOrder.Descending)
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# it is safe to remove this table sort?
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# because we already using ordered table to store time from
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# older value to newer value
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when false:
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k.pingPongTime.sort(cmp, order = SortOrder.Descending)
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var
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num = 0
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rc: array[MaxRC, TimeKey]
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rci = 0
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numRemoved = 0
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rc: array[MaxRC, TimeKey] # 784 bytes(MinEntries/8*sizeof(TimeKey))
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for v in keys(k.pingPongTime):
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if num < MaxRC:
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rc[num] = v
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inc num
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# using fixed size temp on stack possibly
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# requires multiple iteration to remove
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# old entries
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while numRemoved < maxRemove:
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for v in keys(k.pingPongTime):
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rc[rci] = v
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inc rci
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inc numRemoved
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if rci >= MaxRC or numRemoved >= maxRemove: break
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for i in 0..<num:
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k.pingPongTime.del(rc[i])
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for i in 0..<rci:
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k.pingPongTime.del(rc[i])
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rci = 0
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proc updateLastPingReceived(k: KademliaProtocol, n: Node, t: Time) =
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k.removeTooOldPingPongTime()
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