Question

为了避免CAF（资源共享），我尝试转换为功能使用伪参数，但没有成功（noCafB）。我看过How to make a CAF not a CAF in Haskell?，因此尝试了noCafC和noCafD。编译时 -O0，然后每次都会评估带有伪参数的函数。但是，对于-O2，似乎GHC将这些函数转换为 CAF。这是预期的行为（GHC的优化）吗？

module Main where
import Debug.Trace

cafA :: [Integer]
cafA = trace "hi" (map (+1) $ [1..])
noCafB :: a -> [Integer]
noCafB _ = trace "hi" (map (+1) $ [1..])
noCafC :: a -> [Integer]
noCafC _ = trace "hi" (map (+1) $ [1..])
{-# NOINLINE noCafC #-}
noCafD :: a -> [Integer]
noCafD _ = trace "hi" (map (+1) $ myEnumFrom 0 1)
{-# NOINLINE noCafD #-}
myEnumFrom :: a -> Integer -> [Integer]
myEnumFrom _ n =  enumFrom n
{-# NOINLINE myEnumFrom #-}

main :: IO ()
main = do
  putStrLn "cafA"
  print $ (cafA !! 1 + cafA !! 2)
  putStrLn "noCafB"
  print $ (noCafB 0 !! 1 + noCafB 0 !! 2)
  putStrLn "noCafC"
  print $ (noCafC 0 !! 1 + noCafC 0 !! 2)
  putStrLn "noCafD"
  print $ (noCafD 0 !! 1 + noCafD 0 !! 2)

-O2

的结果

$ stack ghc -- --version
The Glorious Glasgow Haskell Compilation System, version 7.10.3

$ stack ghc -- -O2 cafTest.hs
[1 of 1] Compiling Main             ( cafTest.hs, cafTest.o )
Linking cafTest ...
$ ./cafTest
cafA
hi
7
noCafB
7
noCafC
7
noCafD
hi
7

-O0

的结果

$ stack ghc -- -O0 cafTest.hs
[1 of 1] Compiling Main             ( cafTest.hs, cafTest.o )
Linking cafTest ...
$ ./cafTest
cafA
hi
7
noCafB
hi
hi
7
noCafC
hi
hi
7
noCafD
hi
hi
7

我也试过没有trace，但结果是一样的。在-O2下，我发现incInt函数的结果是通过检查分析输出来共享的。为什么会出现这种情况？

incIntOrg :: [Integer]
incInt = map (+1) [1..]

incInt :: a -> [Integer]  -- results IS shared. should it be?
incInt _ = map (+1) $ myEnum 0 1
{-# NOINLINE incInt #-}
myEnum :: a -> Integer -> [Integer]
myEnum _ n =  enumFrom n
{-# NOINLINE myEnum #-}
main :: IO ()
main = do
  print (incInt 0 !!  9999999)
  print (incInt 0 !!  9999999)
  print (incInt 0 !!  9999999)

任何评论都将深深感激。感谢。

避免在Haskell中使用CAF

0 个答案: