Re: [GHC Users] Dictionary sharing
Jonas Almström Duregård wrote:
Thank you for your response Edward,
You write that it is usually only evaluated once, do you know the
circumstances under which it is evaluated more than once? I have some
examples of this but they are all very large.
Only the dictionaries for type class instances that do not depend on
other instances will be CAFs and evaluated at most once. When an
instance has such dependencies, as for example (from your initial
mail in this thread),
instance List a = List [a] where
list = permutations list
then dictionaries will be created on demand (causing re-evaluation of
'list' in this particular case). More precisely, when the compiler
finds that a function needs a List [a] instance where only a List a
instance is available, it will create a fresh dictionary for List [a]
using the above implementation.
I am not aware of GHC providing any caching or memoisation mechanism
for this, so I think that your solution of building your own using
Typeable is appropriate.
Best regards,
Bertram
-- Example program showing addresses of various Ord dictionaries.
-- Contents may be hazardous if swallowed! Keep away from children!
{-# LANGUAGE MagicHash, Rank2Types #-}
module Main where
import GHC.Exts
import GHC.Int
newtype GetDict = GetDict { unGetDict :: forall a . Ord a = a - Int }
-- Evil hack for extracting the address of a dictionary from a function
-- call. Note that these addresses may change during GC!
getDict :: Ord a = a - Int
getDict = unGetDict (unsafeCoerce# getDict') where
getDict' :: Addr# - Addr# - Int
getDict' d _ = I# (addr2Int# d)
{-# NOINLINE bar #-}
-- newListDict is designed to force the creation of a new Ord [a]
-- dictionary given an Ord a dictionary, and return the new dictionary's
-- address.
getListDict :: Ord a = a - Int
getListDict x = unGetDict (GetDict (\x - getDict [x])) x
main = do
print $ getDict (1 :: Int) -- using a CAF dictionary
print $ getDict (2 :: Int) -- same as previous
print $ getDict (2 :: Word) -- a different CAF dictionary
print $ getDict ([1] :: [Int]) -- also a CAF!
print $ getDict ([2] :: [Int]) -- same as previous
print $ getListDict (1 :: Int) -- a dynamically created dictionary
print $ getListDict (2 :: Int) -- different from previous
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[GHC Users] Dictionary sharing
Hi, Is there a way to ensure that functions in a class instance are treated as top level definitions and not re-evaluated? For instance if I have this: class C a where list :: [a] instance List a = List [a] where list = permutations list How can I ensure that list :: [[X]] is evaluated at most once for any type X (throughout my program)? I assume this is potentially harmful, since list can never be garbage collected and there may exist an unbounded number of X's. I currently have a solution that uses Typeable to memoise the result of the function based on its type. Is there an easier way? Regards, Jonas ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: [GHC Users] Dictionary sharing
Hello Jonas, Like other top-level definitions, these instances are considered CAFs (constant applicative forms), so these instances will in fact usually be evaluated only once per type X. import System.IO.Unsafe class C a where dflt :: a instance C Int where dflt = unsafePerformIO (putStrLn bang return 2) main = do print (dflt :: Int) print (dflt :: Int) print (dflt :: Int) ezyang@javelin:~/Dev/haskell$ ./caf bang 2 2 2 Cheers, Edward Excerpts from Jonas Almström Duregård's message of Fri Jun 29 07:25:42 -0400 2012: Hi, Is there a way to ensure that functions in a class instance are treated as top level definitions and not re-evaluated? For instance if I have this: class C a where list :: [a] instance List a = List [a] where list = permutations list How can I ensure that list :: [[X]] is evaluated at most once for any type X (throughout my program)? I assume this is potentially harmful, since list can never be garbage collected and there may exist an unbounded number of X's. I currently have a solution that uses Typeable to memoise the result of the function based on its type. Is there an easier way? Regards, Jonas ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: [GHC Users] Dictionary sharing
Thank you for your response Edward, You write that it is usually only evaluated once, do you know the circumstances under which it is evaluated more than once? I have some examples of this but they are all very large. The real issue I was having was actually not with a list but with a memoised function i.e. something like: class C a where memoised :: Int - a Perhaps functions are treated differently? Regards, Jonas On 29 June 2012 15:55, Edward Z. Yang ezy...@mit.edu wrote: Hello Jonas, Like other top-level definitions, these instances are considered CAFs (constant applicative forms), so these instances will in fact usually be evaluated only once per type X. import System.IO.Unsafe class C a where dflt :: a instance C Int where dflt = unsafePerformIO (putStrLn bang return 2) main = do print (dflt :: Int) print (dflt :: Int) print (dflt :: Int) ezyang@javelin:~/Dev/haskell$ ./caf bang 2 2 2 Cheers, Edward Excerpts from Jonas Almström Duregård's message of Fri Jun 29 07:25:42 -0400 2012: Hi, Is there a way to ensure that functions in a class instance are treated as top level definitions and not re-evaluated? For instance if I have this: class C a where list :: [a] instance List a = List [a] where list = permutations list How can I ensure that list :: [[X]] is evaluated at most once for any type X (throughout my program)? I assume this is potentially harmful, since list can never be garbage collected and there may exist an unbounded number of X's. I currently have a solution that uses Typeable to memoise the result of the function based on its type. Is there an easier way? Regards, Jonas ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: [GHC Users] Dictionary sharing
I say usually because while I believe this to be true for the current implementation of GHC, I don't think we necessary give this operational guarantee. But yes, your real problem is that there is a world of difference between functions and non-functions. You will need to use one of the usual tricks for memoising functions, or forgo using a function altogether and lean on laziness. Edward Excerpts from Jonas Almström Duregård's message of Fri Jun 29 11:21:46 -0400 2012: Thank you for your response Edward, You write that it is usually only evaluated once, do you know the circumstances under which it is evaluated more than once? I have some examples of this but they are all very large. The real issue I was having was actually not with a list but with a memoised function i.e. something like: class C a where memoised :: Int - a Perhaps functions are treated differently? Regards, Jonas On 29 June 2012 15:55, Edward Z. Yang ezy...@mit.edu wrote: Hello Jonas, Like other top-level definitions, these instances are considered CAFs (constant applicative forms), so these instances will in fact usually be evaluated only once per type X. import System.IO.Unsafe class C a where dflt :: a instance C Int where dflt = unsafePerformIO (putStrLn bang return 2) main = do print (dflt :: Int) print (dflt :: Int) print (dflt :: Int) ezyang@javelin:~/Dev/haskell$ ./caf bang 2 2 2 Cheers, Edward Excerpts from Jonas Almström Duregård's message of Fri Jun 29 07:25:42 -0400 2012: Hi, Is there a way to ensure that functions in a class instance are treated as top level definitions and not re-evaluated? For instance if I have this: class C a where list :: [a] instance List a = List [a] where list = permutations list How can I ensure that list :: [[X]] is evaluated at most once for any type X (throughout my program)? I assume this is potentially harmful, since list can never be garbage collected and there may exist an unbounded number of X's. I currently have a solution that uses Typeable to memoise the result of the function based on its type. Is there an easier way? Regards, Jonas ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
