GHC Haskell 什么时候自动记忆？

Question

I can't figure out why m1 is apparently memoized while m2 is not in the following:

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

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

m1 10000000 takes about 1.5 seconds on the first call, and a fraction of that on subsequent calls (presumably it caches the list), whereas m2 10000000 always takes the same amount of time (rebuilding the list with each call). Any idea what's going on? Are there any rules of thumb as to if and when GHC will memoize a function? Thanks.

Answer 1

GHC does not memoize functions.

It does, however, compute any given expression in the code at most once per time that its surrounding lambda-expression is entered, or at most once ever if it is at top level. Determining where the lambda-expressions are can be a little tricky when you use syntactic sugar like in your example, so let's convert these to equivalent desugared syntax:

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

(Note: The Haskell 98 report actually describes a left operator section like (a %) as equivalent to \\b -> (%) ab , but GHC desugars it to (%) a . These are technically different because they can be distinguished by seq . I think I might have submitted a GHC Trac ticket about this.)

Given this, you can see that in m1' , the expression filter odd [1..] is not contained in any lambda-expression, so it will only be computed once per run of your program, while in m2' , filter odd [1..] will be computed each time the lambda-expression is entered, ie, on each call of m2' . That explains the difference in timing you are seeing.

---

Actually, some versions of GHC, with certain optimization options, will share more values than the above description indicates. This can be problematic in some situations. For example, consider the function

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

GHC might notice that y does not depend on x and rewrite the function to

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

In this case, the new version is much less efficient because it will have to read about 1 GB from memory where y is stored, while the original version would run in constant space and fit in the processor's cache. In fact, under GHC 6.12.1, the function f is almost twice as fast when compiled without optimizations than it is compiled with -O2 .

Answer 2

m1 is computed only once because it is a Constant Applicative Form, while m2 is not a CAF, and so is computed for each evaluation.

See the GHC wiki on CAFs: http://www.haskell.org/haskellwiki/Constant_applicative_form

Answer 3

There is a crucial difference between the two forms: the monomorphism restriction applies to m1 but not m2, because m2 has explicitly given arguments. So m2's type is general but m1's is specific. The types they are assigned are:

m1 :: Int -> Integer
m2 :: (Integral a) => Int -> a

Most Haskell compilers and interpreters (all of them that I know of actually) do not memoize polymorphic structures, so m2's internal list is recreated every time it's called, where m1's is not.

Answer 4

I'm not sure, because I'm quite new to Haskell myself, but it appears that it's beacuse the second function is parametrized and the first one is not. The nature of the function is that, it's result depends on input value and in functional paradigm especailly it depends ONLY on the input. Obvious implication is that a function with no parameters returns always the same value over and over, no matter what.

Aparently there's an optimizing mechanizm in GHC compiler that exploits this fact to compute the value of such a function only once for whole program runtime. It does it lazily, to be sure, but does it nonetheless. I noticed it myself, when I wrote the following function:

primes = filter isPrime [2..]
    where isPrime n = null [factor | factor <- [2..n-1], factor `divides` n]
        where f `divides` n = (n `mod` f) == 0

Then to test it, I entered GHCI and wrote: primes !! 1000 primes !! 1000 . It took a few seconds, but finally I got the answer: 7927 . Then I called primes !! 1001 primes !! 1001 and got the answer instantly. Similarly in an instant I got the result for take 1000 primes , because Haskell had to compute the whole thousand-element list to return 1001st element before.

Thus if you can write your function such that it takes no parameters, you probably want it. ;)