Haskell: parse error on input ‘=’ in where











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












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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










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












1 Answer
1






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





















    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














     

    draft saved


    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53246799%2fhaskell-parse-error-on-input-in-where%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    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






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53246799%2fhaskell-parse-error-on-input-in-where%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Bressuire

            Vorschmack

            Quarantine