#2252: Extreme performance degradation on minor code change
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Reporter: simona | Owner:
Type: bug | Status: new
Priority: normal | Milestone:
Component: Compiler | Version: 6.8.2
Severity: normal | Resolution:
Keywords: | Difficulty: Unknown
Testcase: | Architecture: x86_64 (amd64)
Os: Linux |
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Comment (by simonpj):
You made a mistake in the new `GoodPerform.hs`; you wrote
{{{
replicateM_ 1000 $ do
--res1 <- fixpoint initial bottom bottom vn 1
--unless (res1 `subset` res1) $ putStrLn "something's wrong"
res2 <- fixpoint initial bottom bottom vn 1
unless (res2 `subset` res2) $ putStrLn "something's wrong"
}}}
but you meant to use two calls to `replicateM_`, didn't you?
Anyway I know what's going on. Here is the crucial fragment of the two
forms, after desguaring:
{{{
Bad:
replicateM_ 1000 (f 10 >>= (\res1 -> f 20 >>= (\res2 -> return ())))
Good:
replicateM_ 1000 (f 10) >> replicateM_ 1000 (f 20)
}}}
The `f` is the `fixpoint` function; the `10` and `20` are the constant
args to those two calls. Just look at it! The key point is that in both
cases, the subexpression `(f 10)` is outside any lambdas, and hence is
executed just once. It's just as I said, although obscured by clutter:
the computation is unaffected by IO state, so there is no need for it to
be repeated each time.
Furthermore, in the `good` case, the call `(f 20)` is '''also#'' outside
any lambdas, and ''hence is computed only once''. But in the `bad` case,
the call `(f 20)` is inside the `(\ res1 -> ...)` abstraction, and so is
computed once for each call of the (\res1...); that is 1000 times.
So that's why there's the big difference. It is a little puzzling, but it
becomes much clearer when you desugar the do-notation.
The moral, as often, is that if you put a constant expression inside a
loop, GHC will often compute it just once. But this occurs much more
often in benchmarks than in real programs
Simon
--
Ticket URL: <http://hackage.haskell.org/trac/ghc/ticket/2252#comment:7>
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