Re: [Haskell-cafe] Hints for Euler Problem 11

17 Aug 2007

      On Thu, 2007-08-16 at 12:50 -0700, Kim-Ee Yeoh wrote:
...
Aaron Denney wrote:
...
On 2007-08-15, Pekka Karjalainen  wrote:
...
A little style issue here on the side, if I may. You don't need to use
(++) to join multiline string literals.
text = "If you want to have multiline string literals \
       \in your source code, you can break them up with \
       \backslashes. Any whitespace characters between \
       \two backslashes will be ignored."
I find the first far more readable.  The compiler should be able to
assemble it all at compile time, right?
'Course not. The (++) function like all Haskell functions is only a
/promise/ to do its job. What does "assembling at compile time"
mean here:
s = "I will not write infinite loops " ++ s
Let's check, shall we?  I've never used core before, but there's a first
time for everything:

  % cat C.hs

  module Test where

  x = "Foo" ++ "Bar"
  y = "Zot" ++ y

  % ghc -ddump-simpl C.hs

  ==================== Tidy Core ====================
  Test.x :: [GHC.Base.Char]
  [GlobalId]
  []
  Test.x =
    GHC.Base.++
      @ GHC.Base.Char (GHC.Base.unpackCString# "Foo") (GHC.Base.unpackCString# "Bar")

  Rec {
  Test.y :: [GHC.Base.Char]
  [GlobalId]
  []
  Test.y = GHC.Base.++ @ GHC.Base.Char (GHC.Base.unpackCString# "Zot") Test.y
  end Rec }

If I interpret it correctly, the compiler does approximately nothing -
reasonably enough, since we didn't ask for optimization.  With -O:

  % ghc -ddump-simpl C.hs -O

  ==================== Tidy Core ====================
  Rec {
  Test.y :: [GHC.Base.Char]
  [GlobalId]
  [Str: DmdType]
  Test.y = GHC.Base.unpackAppendCString# "Zot" Test.y
  end Rec }

  Test.x :: [GHC.Base.Char]
  [GlobalId]
  [Str: DmdType]
  Test.x = GHC.Base.unpackCString# "FooBar"

y gets turned into an unpackAppendCString#, which I can only presume is
a sensible way to represent a cyclic list, while x gets concatenated
compile-time.

-k

Re: [Haskell-cafe] Hints for Euler Problem 11

Ketil Malde