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While going through ROS-1 wiki pages, at several places I read that the message between a publisher and subscriber does not route via Master. Rather the message travels directly from Publisher to Subscriber. But in the following Wikipedia page (https://en.wikipedia.org/wiki/Publish%E2%80%93subscribe_pattern), under heading TOPOLOGIES, it is mentioned that in a broker-based pub-sub system, the messages are routed via broker (i.e. master). So, I request someone who is aware of the latest ROS-1 communication design to help in clarifying this doubt please. Thanks.

Originally posted by Aakashp on ROS Answers with karma: 41 on 2020-12-23

Post score: 1


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it is mentioned that in a broker-based pub-sub system, the messages are routed via broker [..]

ROS 1 does not use broker-based pub-sub, so that would seem to answer your question (ie: the master is not directly involved in message exchange, as you've already read in multiple other places).

(i.e. master)

the ROS Master != a broker.

But in the following Wikipedia page, [..]

did you intend to include a link?

Originally posted by gvdhoorn with karma: 86574 on 2020-12-23

This answer was ACCEPTED on the original site

Post score: 3

Original comments

Comment by Aakashp on 2020-12-23:
@gvdhoorn : Yes, sorry for forgetting to include the link. Here it is: (also updating in question): https://en.wikipedia.org/wiki/Publish%E2%80%93subscribe_pattern (in Topologies section)

Comment by Aakashp on 2020-12-23:
@gvdhoorn : Thanks a lot for your prompt response. But please do let me know in case you wish to comment anything after going through the Wikipedia link mentioned in above comment, and also in question (heading Topologies)

Comment by gvdhoorn on 2020-12-23:
No, my answer would be the same.

It's true that in a broker based pub-sub system, the broker can be involved in message exchange.

But ROS 1 does not implement such a pub-sub system, so there is no broker, and consequently, no message passing through/via the master.

Comment by Aakashp on 2020-12-24:
@gvdhoorn : Thanks again. I have marked the question as 'answered'.


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