[Haskell-beginners] Are these solution the Haskell way ?

Frerich Raabe raabe at froglogic.com
Wed May 13 07:17:46 UTC 2015


Roelof,

since these functions are all pure (i.e. solely defined through their 
arguments) you can apply equational reasoning, which means: everytime the 
'plusplus' function is called you can replace that call with the definition 
of the function without changing the behaviour of the program.

That allows you to just evaluate a call manually, expanding each function 
call by hand. I added some extra parentheses to emphasize the functions being 
evaluate:

   plusplus [1,2,3] [4,5,6]                    | apply 'plusplus', second 
pattern for 'plusplus' matches
   == 1 : (plusplus [2,3] [4,5,6])             | "
   == 1 : (2 : (plusplus [3] [4,5,6]))         | "
   == 1 : (2 : (3 : (plusplus [] [4,5,6])))    | apply 'plusplus', first 
pattern for 'plusplus' matches
   == 1 : (2 : (3 : [4,5,6])                   | apply (:) function
   == 1 : (2 : [3,4,5,6])                      | "
   == 1 : [2,3,4,5,6]                          | "
   == [1,2,3,4,5,6]                            | "


On 2015-05-13 08:55, Roelof Wobben wrote:
> Thanks,
> 
>  So the answer is x and the rest of xs and ys.
>  How do x then get added to ys.
> 
>  Roelof
> 
>  Daniel P. Wright schreef op 13-5-2015 om 8:52:
> 
>> Ah, that's just a small syntactic issue -- in Haskell, operators are infix 
>> (go between the arguments) by default, but named functions are
>> not. So you would have to write it:
>> 
>> plusplus [] xs = xs
>> plusplus (x:xs) ys = x : plusplus xs ys
>> 
>> 2015-05-13 15:39 GMT+09:00 Roelof Wobben <r.wobben at home.nl>:
>> 
>> Thanks,
>> 
>> If I re-implement it like this :
>> 
>> plusplus :: [a] -> [a] -> [a]
>> plusplus [] (xs) = xs
>> plusplus (x:xs) yx = x : xs plusplus yx
>> 
>> main = print $ ["a","b"] plusplus ["c","d"]
>> 
>> I see this error appear :
>> 
>> src/Main.hs at 3:26-3:40
>> Couldn't match expected type ‘([a0] -> [a0] -> [a0]) -> [a] -> [a]’ with 
>> actual type
>> [a] Relevant bindings include yx :: [a] (bound at 
>> /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:17) 
>> xs ::
>> [a] (bound at 
>> /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:13) 
>> x :: a (bound at
>> /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:11) 
>> plusplus :: [a] -> [a] -> [a] (bound at
>> /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:2:1) 
>> The function
>> xs is applied to two arguments, but its type
>> [a] has none …
>> 
>> So for me not a aha moment. I was hoping I would get it
>> 
>> Roelof
>> 
>> Daniel P. Wright schreef op 13-5-2015 om 8:27:
>> 
>> Hi Roelof,
>> 
>> If you don't consider it cheating (and I suggest you shouldn't, having had 
>> a stab at the answers), you will find great enlightenment looking
>> at how these functions are *actually* implemented in the wild. Did you know 
>> that there is a "source" link for each function on Hackage? By
>> clicking on that, for your first example, you can compare the actual 
>> implementation of (++) with your "plusplus" function:
>> 
>> http://hackage.haskell.org/package/base-4.8.0.0/docs/src/GHC-Base.html#%2B%2B 
>> [2]
>> 
>> (By the way, it's that function in particular that gave me one of my first 
>> "Aha!" moments in Haskell... it's quite beautiful in its
>> simplicity).
>> 
>> One thing with viewing the source on Hackage is that sometimes it can be a 
>> little more confusing than it needs to be for the sake of
>> efficiency. A really good source for good, readable examples of Prelude 
>> functions in Haskell is the Haskell Report:
>> 
>> https://www.haskell.org/onlinereport/standard-prelude.html [3]
>> 
>> (In this case, though, the implementation is the same).
>> 
>> Having a shot at defining these library functions yourself, as you have 
>> done, and then comparing your version with the "official" version in
>> the prelude is a great way to learn good style!
>> 
>> -Dani.
>> 
>> 2015-05-13 15:10 GMT+09:00 Roelof Wobben <r.wobben at home.nl>:
>> Hello,
>> 
>> For practising pattern matching and recursion I did recreate some commands 
>> of Data,list.
>> 
>> My re-implementation of ++ :
>> 
>> plusplus :: [a] -> [a] -> [a]
>> plusplus [] [] = [] ;
>> plusplus [] (xs) = xs
>> plusplus (xs) [] = xs
>> plusplus (xs) yx = plusplus' (reverse xs) yx
>> 
>> plusplus' :: [a] -> [a] -> [a]
>> plusplus' [] (xs) = xs
>> plusplus' (x:xs) yx = plusplus' xs (x:yx)
>> 
>> main = print $ plusplus ["a","b"] ["c","d"]
>> 
>> my re-implementation of init :
>> 
>> import Data.Maybe
>> 
>> -- | The main entry point.
>> init' :: [a] -> Maybe [a]
>> init' [] = Nothing
>> init' [x] = Just []
>> init' (x:xs) = Just (x:fromMaybe xs (init' xs))
>> 
>> main = print . init' $ [1,3]
>> 
>> my re-implementation of last :
>> 
>> -- | The main entry point.
>> last' :: [a] -> Maybe a
>> last' [] = Nothing
>> last' [x] = Just x
>> last' (_:xs) = last' xs
>> 
>> main = print . last' $ []
>> 
>> Now I wonder if these solutions are the haskell way ? if not so, how can I 
>> improve them ,
>> 
>> Roelof
>> 
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-- 
Frerich Raabe - raabe at froglogic.com
www.froglogic.com - Multi-Platform GUI Testing


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