docs: public rendezvous procedure #1203
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AlejandroCabeza
approved these changes
Sep 27, 2024
libp2p/protocols/rendezvous.nim
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| await rdv.advertise(ns, ttl, rdv.peers) | ||
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| proc requestLocally*(rdv: RendezVous, ns: string): seq[PeerRecord] = | ||
| ## This procedure returns all the peers already registered on the | ||
| ## given namespace. This function is synchronous |
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This function is synchronous
Not sure this is necessary.
libp2p/protocols/rendezvous.nim
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| @@ -555,6 +560,10 @@ proc requestLocally*(rdv: RendezVous, ns: string): seq[PeerRecord] = | |||
| proc request*( | |||
| rdv: RendezVous, ns: string, l: int = DiscoverLimit.int, peers: seq[PeerId] | |||
| ): Future[seq[PeerRecord]] {.async.} = | |||
| ## This async procedure request discovers and returns peers for | |||
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request discovers
Is it a typo?
libp2p/protocols/rendezvous.nim
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| @@ -555,6 +560,10 @@ proc requestLocally*(rdv: RendezVous, ns: string): seq[PeerRecord] = | |||
| proc request*( | |||
| rdv: RendezVous, ns: string, l: int = DiscoverLimit.int, peers: seq[PeerId] | |||
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it's better to use a more descriptive name for the parameter than l.
| @@ -646,6 +655,10 @@ proc unsubscribeLocally*(rdv: RendezVous, ns: string) = | |||
| return | |||
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| proc unsubscribe*(rdv: RendezVous, ns: string, peerIds: seq[PeerId]) {.async.} = | |||
| ## The async unsubscribe procedure removes peers from a namespace by | |||
| ## sending an "Unregister" message to each connected peer. The operation | |||
| ## is bounded by a timeout for unsubscribing from all peers. | |||
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Where is this timeout?
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Three changes on this PR:
advertisefrom method to proc.Related to these comments: #1183 (comment), #1183 (comment) and #1183 (comment)