I’ll explain with an example - there are 3 Lemmy instances A, B, C

A federates with B but not C

B Federates with both A and C

Does A indirectly get the content from C due to B’s federation with it?

edit: Formatting for clarity

  • 0x4E4F
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    1 year ago

    No, A doesn’t get the content from C, but if C is not defederated from A, yes, C gets the posts from A (feed). It’s called asymmetric defederation. For example, that’s the case with sh.itjust.works and beehaw.org. Those of us that subscribed to communities on beehaw.org before it defederated from us, we still get posts from beehaw in our feed, but beehaw doesn’t get any from us. And if we actually comment on the post, that comment will be visible only to members of sh.itjust.works, no one else in the federation, because the sync feed from beehaw takes presence over ours, so the instances that can talk to beehaw, get the sync feed from beehaw first (or if it arrives second, ours is discarded, because beehaw is the original instance on which the post was made).

  • Izzy@lemmy.world
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    1 year ago

    I’m going to guess not because then we would have infinite loops of federated content. It probably checks the source instance before syncing content.

    • WontonSoup@lemmy.worldOP
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      1 year ago

      That makes sense if it were to become circular. The way I was thinking of it is like

      B gets C’s content making C’s part of B’s. so because C’s is a part of B then A gets it.

      But I guess C’s content isn’t really B’s its only there through federation? Trying to wrap my head around all this, appreciate the reply.

  • Myrhial@discuss.online
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    1 year ago

    If I have understood it correctly then A should only get content from B because they are directly linked. Content from C will display on B (via ‘all’) but will not be sent over from C. Else A would never be able to block C.

  • zumi@lemmy.sdf.org
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    1 year ago

    Just also want to add:

    When A federates with B, it does not get B’s content. A only gets content from B that A’s users subscribe to.