26
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
this post was submitted on 04 Nov 2023
26 points (100.0% liked)
Guild Wars 2
1504 readers
1 users here now
Welcome to c/guildwars2 on the Lemmyverse.
SotM: Love Spooky Season (@avater)
Links
Official
Builds
- Snow Crows (raids)
- Discretize (fractals)
- Gw2Mists (WvW)
- Metabattle
- Hardstuck
- GuildJen
Tools
Addons
Related Communities
- r/guildwars2 (Reddit)
- c/mmorpg
founded 1 year ago
MODERATORS
I don't get from where information about map full comes from It doesnt make sense to use subgroup to determine it at all. Map is much bigger than 50 people, checking squads for map capability would always be wrong because there are people without squads. They must be checking number of players on the entire instance, and count all the players. Codding a search for commander instance IP in squad function to only work in subgroup one doesn't make any sense. I have joined many squads by queuing where commander was in the subgroup without any problems
Re-posting from my
lemmy.wtf
account, since outgoing federation fromlemmy.sdf.org
appears to be broken at the moment:@necropola@lemmy.sdf.org said:
I have even seen squads to explicitly use subgroup 1 as a lobby/queue for players not in the commander's map (yet) which would be really bad, if LFG actually uses the first player of subgroup 1 to check whether the squad's map is full.
If there is no commander my guess would be it won't show map is full ever. Because game has no idea which instance it should check.
Of course I don't know for a fact that LFG actually checks subgroup 1 to figure out whether a squad's map is full, but it seemed pretty plausible.
Also, I didn't say that map full detection is in any way related to the squad size limit. The question is how does LFG determine whether a squad's map is full? Does it actually check the map of the squad's commander (squad might not have one) or does it check the map of the first player in subgroup 1 (which is what I think).