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

Wed May 13 11:40:50 UTC 2015

The evaluation would go the way Frerich shows it. In particular you should
know that these two notations are equivalent.

[1,2,3] == 1 : 2 : 3 : []

The cons (:) operator (with type a -> [a] -> [a]) adds an element to the
start of a list. Also, as it is right associative

    1 : 2 : 3 : []
==  1 : (2 : (3 : []))   { right associativity }
==  1 : (2 : [3])        { add element to empty list }
==  1 : [2, 3]           { add element }
==  [1, 2, 3]            { add element }

On 13 May 2015 at 12:50, Roelof Wobben <r.wobben at home.nl> wrote:

>  Hello,
>
> I see it.
>
> plusplus [1,2] [3,4]
> plusplus  [1] : plusplus [2] [3,4]
> plusplus [1] :  [2] : plusplus [] [3,4]
> plusplus [1] : [2] : [3,4]
> [1:2:3:4]
>
> I think I need some more practice on this.
> Does anyone know a good exercise ?
>
> Roelof
>
>
> Daniel P. Wright schreef op 13-5-2015 om 9:08:
>
> Think about what the possible values of "xs" might be, and trace through
> the next call to "plusplus xs ys".
>
>  It would help if I didn't name the variables stupidly... here is a
> slightly better version:
>
>  plusplus [] ys = ys
> plusplus (x:xs) ys = x : plusplus xs ys
>
>  If it helps, try tracing through the steps required to evaluate
> "plusplus [1, 2] [3, 4]" manually (using a pen and paper, not on the
> computer)
>
> 2015-05-13 15:55 GMT+09:00 Roelof Wobben <r.wobben at home.nl>:
>
>>  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
>>>
>>>  (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
>>>
>>>  (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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>>>
>>>
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-- 
Regards

Sumit Sahrawat
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