Performance optimization for determining friends in WHO <mask>.

Construct a hash set of the user's channels and check that rather than
querying channel membership, to reduce the number of locks that need to be
acquired.
This commit is contained in:
Alex Jaspersen 2020-05-08 01:55:47 +00:00
parent b3cfcc1289
commit a6e4a26cbb
2 changed files with 21 additions and 16 deletions

@ -1022,21 +1022,6 @@ func (client *Client) ModeString() (str string) {
return "+" + client.modes.String()
}
// IsFriend() returns true if the given otherClient shares a channel with this client, or if they are the same user.
func (client *Client) IsFriend(otherClient *Client) bool {
if client == otherClient {
return true
}
for _, channel := range client.Channels() {
if channel.hasClient(otherClient) {
return true
}
}
return false
}
// Friends refers to clients that share a channel with this client.
func (client *Client) Friends(capabs ...caps.Capability) (result map[*Session]bool) {
result = make(map[*Session]bool)

@ -2636,8 +2636,28 @@ func whoHandler(server *Server, client *Client, msg ircmsg.IrcMessage, rb *Respo
}
}
} else {
// Construct set of channels the client is in.
userChannels := make(map[*Channel]bool)
for _, channel := range client.Channels() {
userChannels[channel] = true
}
// Another client is a friend if they share at least one channel, or they are the same client.
isFriend := func(otherClient *Client) bool {
if client == otherClient {
return true
}
for _, channel := range otherClient.Channels() {
if userChannels[channel] {
return true
}
}
return false
}
for mclient := range server.clients.FindAll(mask) {
if isOper || !mclient.HasMode(modes.Invisible) || mclient.IsFriend(client) {
if isOper || !mclient.HasMode(modes.Invisible) || isFriend(mclient) {
client.rplWhoReply(nil, mclient, rb)
}
}