Haskell: parse error on input ‘=’ in where











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3
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I am new in Haskell and practicing Algorithm in Haskell follow the book "Pearls of Functional Algorithm Design"



This is the algo to find the smallest natural number not in a given finite set X of natural numbers



import Data.List
import Data.Array

minfree xs = if null ([0..b-1] \ us)
then head ([b..] \ vs)
else head ([0..b-1] \ us)
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2


Pff the compiler error like this



Prelude Data.Array Data.List> :load 01_the_smallest_free_number.hs 
[1 of 1] Compiling Main ( 01_the_smallest_free_number.hs, interpreted )

01_the_smallest_free_number.hs:11:29: error:
parse error on input ‘=’
Perhaps you need a 'let' in a 'do' block?
e.g. 'let x = 5' instead of 'x = 5'
|
11 | b = div (length xs) 2
| ^
Failed, no modules loaded.


Well obveriously add let before b is not the right answer I've tried.



Then I replaced all b to div (length xs) 2 that works, so seems here is the problem to be, but I don't get it










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  • 2




    Bad identation causes the problem I guess.
    – Sereja Bogolubov
    Nov 11 at 8:35















up vote
3
down vote

favorite












I am new in Haskell and practicing Algorithm in Haskell follow the book "Pearls of Functional Algorithm Design"



This is the algo to find the smallest natural number not in a given finite set X of natural numbers



import Data.List
import Data.Array

minfree xs = if null ([0..b-1] \ us)
then head ([b..] \ vs)
else head ([0..b-1] \ us)
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2


Pff the compiler error like this



Prelude Data.Array Data.List> :load 01_the_smallest_free_number.hs 
[1 of 1] Compiling Main ( 01_the_smallest_free_number.hs, interpreted )

01_the_smallest_free_number.hs:11:29: error:
parse error on input ‘=’
Perhaps you need a 'let' in a 'do' block?
e.g. 'let x = 5' instead of 'x = 5'
|
11 | b = div (length xs) 2
| ^
Failed, no modules loaded.


Well obveriously add let before b is not the right answer I've tried.



Then I replaced all b to div (length xs) 2 that works, so seems here is the problem to be, but I don't get it










share|improve this question


















  • 2




    Bad identation causes the problem I guess.
    – Sereja Bogolubov
    Nov 11 at 8:35













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I am new in Haskell and practicing Algorithm in Haskell follow the book "Pearls of Functional Algorithm Design"



This is the algo to find the smallest natural number not in a given finite set X of natural numbers



import Data.List
import Data.Array

minfree xs = if null ([0..b-1] \ us)
then head ([b..] \ vs)
else head ([0..b-1] \ us)
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2


Pff the compiler error like this



Prelude Data.Array Data.List> :load 01_the_smallest_free_number.hs 
[1 of 1] Compiling Main ( 01_the_smallest_free_number.hs, interpreted )

01_the_smallest_free_number.hs:11:29: error:
parse error on input ‘=’
Perhaps you need a 'let' in a 'do' block?
e.g. 'let x = 5' instead of 'x = 5'
|
11 | b = div (length xs) 2
| ^
Failed, no modules loaded.


Well obveriously add let before b is not the right answer I've tried.



Then I replaced all b to div (length xs) 2 that works, so seems here is the problem to be, but I don't get it










share|improve this question













I am new in Haskell and practicing Algorithm in Haskell follow the book "Pearls of Functional Algorithm Design"



This is the algo to find the smallest natural number not in a given finite set X of natural numbers



import Data.List
import Data.Array

minfree xs = if null ([0..b-1] \ us)
then head ([b..] \ vs)
else head ([0..b-1] \ us)
where (us, vs) = (partition (<b) xs)
b = div (length xs) 2


Pff the compiler error like this



Prelude Data.Array Data.List> :load 01_the_smallest_free_number.hs 
[1 of 1] Compiling Main ( 01_the_smallest_free_number.hs, interpreted )

01_the_smallest_free_number.hs:11:29: error:
parse error on input ‘=’
Perhaps you need a 'let' in a 'do' block?
e.g. 'let x = 5' instead of 'x = 5'
|
11 | b = div (length xs) 2
| ^
Failed, no modules loaded.


Well obveriously add let before b is not the right answer I've tried.



Then I replaced all b to div (length xs) 2 that works, so seems here is the problem to be, but I don't get it







haskell






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asked Nov 11 at 7:47









Aries_is_there

1518




1518








  • 2




    Bad identation causes the problem I guess.
    – Sereja Bogolubov
    Nov 11 at 8:35














  • 2




    Bad identation causes the problem I guess.
    – Sereja Bogolubov
    Nov 11 at 8:35








2




2




Bad identation causes the problem I guess.
– Sereja Bogolubov
Nov 11 at 8:35




Bad identation causes the problem I guess.
– Sereja Bogolubov
Nov 11 at 8:35












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7
down vote



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This is bad indentation:



     where (us, vs) = (partition (<b) xs)
b = div (length xs) 2


Since the second equality is indented more than the first, it continues the first one, as if we wrote



     where (us, vs) = (partition (<b) xs) b = div (length xs) 2


triggering the error.



You instead want:



     where (us, vs) = (partition (<b) xs)
b = div (length xs) 2


so that both equations are indented in the same way.






share|improve this answer





















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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    7
    down vote



    accepted










    This is bad indentation:



         where (us, vs) = (partition (<b) xs)
    b = div (length xs) 2


    Since the second equality is indented more than the first, it continues the first one, as if we wrote



         where (us, vs) = (partition (<b) xs) b = div (length xs) 2


    triggering the error.



    You instead want:



         where (us, vs) = (partition (<b) xs)
    b = div (length xs) 2


    so that both equations are indented in the same way.






    share|improve this answer

























      up vote
      7
      down vote



      accepted










      This is bad indentation:



           where (us, vs) = (partition (<b) xs)
      b = div (length xs) 2


      Since the second equality is indented more than the first, it continues the first one, as if we wrote



           where (us, vs) = (partition (<b) xs) b = div (length xs) 2


      triggering the error.



      You instead want:



           where (us, vs) = (partition (<b) xs)
      b = div (length xs) 2


      so that both equations are indented in the same way.






      share|improve this answer























        up vote
        7
        down vote



        accepted







        up vote
        7
        down vote



        accepted






        This is bad indentation:



             where (us, vs) = (partition (<b) xs)
        b = div (length xs) 2


        Since the second equality is indented more than the first, it continues the first one, as if we wrote



             where (us, vs) = (partition (<b) xs) b = div (length xs) 2


        triggering the error.



        You instead want:



             where (us, vs) = (partition (<b) xs)
        b = div (length xs) 2


        so that both equations are indented in the same way.






        share|improve this answer












        This is bad indentation:



             where (us, vs) = (partition (<b) xs)
        b = div (length xs) 2


        Since the second equality is indented more than the first, it continues the first one, as if we wrote



             where (us, vs) = (partition (<b) xs) b = div (length xs) 2


        triggering the error.



        You instead want:



             where (us, vs) = (partition (<b) xs)
        b = div (length xs) 2


        so that both equations are indented in the same way.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 11 at 8:04









        chi

        72.1k279133




        72.1k279133






























             

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