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by anitil 1878 days ago
Does clean room in this case mean you didn't look at the Redis source? Is there some licensing condition where this is required?
1 comments

No, I don't believe that's what it means in this case. I took the usage to mean that since Rust has memory model which is different enough from C to require a redesign.

I looked at many implementations of Redis and read many KV papers. My redesign reason was similar to a "clean room Rust" reason, I desired a memory model that used shared memory that was independent of the protocol (Redis RESP in this case), allowing multiple processes with different types of protocol services to attach to it.

"Rewrite", then.
Yeah. "Refactor" may be more accurate.