loop optimization by removing constants

[Azmo]

the current version of Data.List.foldl' is:

foldl'           :: (a -> b -> a) -> a -> [b] -> a
foldl' f a []     = a
foldl' f a (x:xs) = let a' = f a x in a' `seq` foldl' f a' xs

which for some reason is faster if removing the first parameter, e.g:

foldl2'           :: (a -> b -> a) -> a -> [b] -> a
foldl2' f = loop
  where loop a []     = a
        loop a (x:xs) = let a' = f a x in a' `seq` loop a' xs

---

Testing speed difference:

module Main (main) where

import System (getArgs)

sumFoldl' :: Int -> Int
sumFoldl' n = foldl' (+) 0 [1..n]

sumFoldl2' :: Int -> Int
sumFoldl2' n = foldl2' (+) 0 [1..n]

-- copy of Data.List.foldl'
foldl'           :: (a -> b -> a) -> a -> [b] -> a
foldl' f a []     = a
foldl' f a (x:xs) = let a' = f a x in a' `seq` foldl' f a' xs

-- altered version of foldl'
foldl2'           :: (a -> b -> a) -> a -> [b] -> a
foldl2' f = loop
  where loop a []     = a
        loop a (x:xs) = let a' = f a x in a' `seq` loop a' xs

main = do
  [v,n] <- getArgs
  case v of
    "1" -> print (sumFoldl' (read n))
    "2" -> print (sumFoldl2' (read n))

---

I have no idea how the optimization of loops is done, but the request is that this kind of loop rewriting should be done by the compiler, not by the user (to get a faster loop). It seems to just be a matter of finding and removing constants (identifiers just passed along).

[kirsten]

The optimization you suggest is called the "static argument transformation". There's already a ticket to implement it at:  http://hackage.haskell.org/trac/ghc/ticket/888

and if you're interested, there's an explanation there of why this isn't easy.

[simonpj]

Indeed. I've copied the example over to #888 (always good to have more examples), and am closing this one.

Simon