Hello Christophe,
On Fri, Mar 12, 2010 at 08:27:15PM +0100, Christophe Grand wrote:
> See my memoize5: the call isn't computed inside the swap!s
That doesn't mean, that it is not computed several times!
user=> (defn f
[x]
(println "Got" x "from" (Thread/currentThread))
(Thread/sleep 5000)
(case x
3 (f 2)
2 (f 1)
1 :done))
#'user/f
user=> (def f (memoize5 f))
#'user/f
user=> (-> #(do (f 3) (println "Done for" (Thread/currentThread))) Thread.
.start)
(Thread/sleep 2500)
(-> #(do (f 3) (println "Done for" (Thread/currentThread))) Thread.
.start)
Got 3 from #<Thread Thread[Thread-2,5,main]>
Got 3 from #<Thread Thread[Thread-3,5,main]>
Got 2 from #<Thread Thread[Thread-2,5,main]>
Got 2 from #<Thread Thread[Thread-3,5,main]>
Got 1 from #<Thread Thread[Thread-2,5,main]>
Got 1 from #<Thread Thread[Thread-3,5,main]>
Done for #<Thread Thread[Thread-2,5,main]>
Done for #<Thread Thread[Thread-3,5,main]>
Hmm? Wasn't f supposed to memoized? The problem is, that the call to f
is not guarded.
> Since you use a lock I think some clever combination of memoized functions
> can create a deadlock.
No. In the protected area we simply create a promise and fire off a
thread, which does the computation. Then we return immediatelly. So no
deadlock possibility here. However we trade one lock for the „other“
(the promise). What is the possibility of a deadlock here? Well, the
computation of f never completes. But this is not a problem of memoize.
The only way memoize could cause a problem here is that the computation
of f somehow calls f again with the same arguments. Then it would
deadlock on the promise, but without memoize we would also have a
infinite loop here...
Sincerely
Meikel
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