haskell - 列表理解中的表达式是否被冗余评估？

Question

let strings = ["foo", "bar", "baz"]
in  [filter (== char) (concat strings) | char <- "oa"]

GHC 是否在 char == 'o' 时评估 concat strings，然后在 char == 'a' 时再次评估？或者它是否记得 concat strings == "foobarbaz" 供以后使用？

我意识到我可以通过重构这段代码来避免这种不确定性，但我对代码的编译方式很感兴趣。

Answer 1

GHC 对其进行评估的次数可能远少于 9 次。事实上它确实如此，我们可以证明使用 Debug.Trace.trace

module Main (main) where 
import Debug.Trace

x = let strings = ["foo", "bar", "baz"]
    in  [filter (== char) (trace "\nconcat strings\n" (concat strings)) | char <- "oaxyz"]

main = do
    print x

这是对“oaxyz”进行 5 次 -O0 计算，一次 -O1 和 -O2 计算:

! 529)-> touch LC.hs ; ghc -O0 -o lc0 LC.hs 
[1 of 1] Compiling Main             ( LC.hs, LC.o )
Linking lc0 ...

(! 530)-> touch LC.hs ; ghc -O1 -o lc1 LC.hs 
[1 of 1] Compiling Main             ( LC.hs, LC.o )
Linking lc1 ...

(! 531)-> ./lc0; ./lc1; ./lc2

concat strings


concat strings


concat strings


concat strings


concat strings

["oo","aa","","","z"]

concat strings

["oo","aa","","","z"]

concat strings

["oo","aa","","","z"]