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Date: | 2006-03-08 (19:36) |
From: | William Lovas <wlovas@s...> |
Subject: | Re: [Caml-list] STM support in OCaml |
On Wed, Mar 08, 2006 at 11:52:05AM +1100, skaller wrote: > On Tue, 2006-03-07 at 19:05 +0000, Asfand Yar Qazi wrote: > > As for claim 1. "Lock-free" doesn't mean what you think it does. > > I know what STM does, thank you: I intend to implement it > myself in my own programming language. Maybe you should > read more carefully. > > I said "protected by a mutex under the hood." which means > sure, the programmer is not writing locks, but they're used > in the implementation and the associated costs are still paid. As the quoted paper said, though, some running transaction can always commit -- so not all of the associated costs are still paid. In particular, the cost of potential non-termination due to deadlock or livelock is not paid. It doesn't matter that there's a mutex under the hood, since it's used only in such a way as to preserve the property that some transaction can always complete. > I really hate it when people try to throw papers against > simple logic. I said what the tradeoffs where: In this case, i think the paper was right :) Its statement of "lock-free"ness was with regards to semantics, not performance. You may be correct that there is some other property, regarding performance, that the STM Haskell implementation does not have, but it is not what their paper calls "lock-free". William