When is memoization automatic in GHC Haskell? -


i can't figure out why m1 apparently memoized while m2 not in following:

m1      = ((filter odd [1..]) !!)  m2 n    = ((filter odd [1..]) !! n)

m1 10000000 takes 1.5 seconds on first call, , fraction of on subsequent calls (presumably caches list), whereas m2 10000000 takes same amount of time (rebuilding list each call). idea what's going on? there rules of thumb if , when ghc memoize function? thanks.

ghc not memoize functions.

it does, however, compute given expression in code @ once per time surrounding lambda-expression entered, or @ once ever if @ top level. determining lambda-expressions can little tricky when use syntactic sugar in example, let's convert these equivalent desugared syntax:

m1' = (!!) (filter odd [1..])              -- nb: see below! m2' = \n -> (!!) (filter odd [1..]) n

(note: haskell 98 report describes left operator section (a %) equivalent \b -> (%) b, ghc desugars (%) a. these technically different because can distinguished seq. think might have submitted ghc trac ticket this.)

given this, can see in m1', expression filter odd [1..] not contained in lambda-expression, computed once per run of program, while in m2', filter odd [1..] computed each time lambda-expression entered, i.e., on each call of m2'. explains difference in timing seeing.

actually, versions of ghc, optimization options, share more values above description indicates. can problematic in situations. example, consider function

f = \x -> let y = [1..30000000] in foldl' (+) 0 (y ++ [x])

ghc might notice y not depend on x , rewrite function to

f = let y = [1..30000000] in \x -> foldl' (+) 0 (y ++ [x])

in case, new version less efficient because have read 1 gb memory y stored, while original version run in constant space , fit in processor's cache. in fact, under ghc 6.12.1, function f twice fast when compiled without optimizations compiled -o2.


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