How does Haskell evaluate this signature?












0















ggt_euklid :: Nat1 -> (Nat1 -> Nat1)



I am trying to learn partial application, I know that in this case, if the parentheses would be left out, I would get the same result, but I do not know how this signature should be evaluated.



As far as I have understood, parentheses signify that it is a function? Would that not imply that ggt_euklid takes a value Nat1 and returns a function?



Below is the complete function:



ggt_euklid x y

| x == y = x

|x>y =ggt_euklid(x-y) y 

|x<y =ggt_euklid x (y-x)





haskell signature currying partial-application







share|improve this question


















  • 3





    "parentheses signify that it is a function?" No, the -> signifies that it's a function.

    – sepp2k
    Nov 21 '18 at 23:36











  • Possible duplicate of Example of deep understanding of currying

    – jberryman
    Nov 21 '18 at 23:54
















0















ggt_euklid :: Nat1 -> (Nat1 -> Nat1)



I am trying to learn partial application, I know that in this case, if the parentheses would be left out, I would get the same result, but I do not know how this signature should be evaluated.



As far as I have understood, parentheses signify that it is a function? Would that not imply that ggt_euklid takes a value Nat1 and returns a function?



Below is the complete function:



ggt_euklid x y

| x == y = x

|x>y =ggt_euklid(x-y) y 

|x<y =ggt_euklid x (y-x)





haskell signature currying partial-application







share|improve this question


















  • 3





    "parentheses signify that it is a function?" No, the -> signifies that it's a function.

    – sepp2k
    Nov 21 '18 at 23:36











  • Possible duplicate of Example of deep understanding of currying

    – jberryman
    Nov 21 '18 at 23:54














0












0








0








ggt_euklid :: Nat1 -> (Nat1 -> Nat1)



I am trying to learn partial application, I know that in this case, if the parentheses would be left out, I would get the same result, but I do not know how this signature should be evaluated.



As far as I have understood, parentheses signify that it is a function? Would that not imply that ggt_euklid takes a value Nat1 and returns a function?



Below is the complete function:



ggt_euklid x y

| x == y = x

|x>y =ggt_euklid(x-y) y 

|x<y =ggt_euklid x (y-x)





haskell signature currying partial-application







share|improve this question














ggt_euklid :: Nat1 -> (Nat1 -> Nat1)



I am trying to learn partial application, I know that in this case, if the parentheses would be left out, I would get the same result, but I do not know how this signature should be evaluated.



As far as I have understood, parentheses signify that it is a function? Would that not imply that ggt_euklid takes a value Nat1 and returns a function?



Below is the complete function:



ggt_euklid x y

| x == y = x

|x>y =ggt_euklid(x-y) y 

|x<y =ggt_euklid x (y-x)





haskell signature currying partial-application




haskell signature currying partial-application






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 21 '18 at 23:17









MinimaxMinimax

53312




53312








  • 3





    "parentheses signify that it is a function?" No, the -> signifies that it's a function.

    – sepp2k
    Nov 21 '18 at 23:36











  • Possible duplicate of Example of deep understanding of currying

    – jberryman
    Nov 21 '18 at 23:54














  • 3





    "parentheses signify that it is a function?" No, the -> signifies that it's a function.

    – sepp2k
    Nov 21 '18 at 23:36











  • Possible duplicate of Example of deep understanding of currying

    – jberryman
    Nov 21 '18 at 23:54








3




3





"parentheses signify that it is a function?" No, the -> signifies that it's a function.

– sepp2k
Nov 21 '18 at 23:36





"parentheses signify that it is a function?" No, the -> signifies that it's a function.

– sepp2k
Nov 21 '18 at 23:36













Possible duplicate of Example of deep understanding of currying

– jberryman
Nov 21 '18 at 23:54





Possible duplicate of Example of deep understanding of currying

– jberryman
Nov 21 '18 at 23:54












2 Answers
2






active

oldest

votes


















2















Would that not imply that ggt_euklid takes a value Nat1 and returns a
function?




No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.



The Arrow -> always be right-associativity (when no parentheses), i.e.:



Nat1 -> Nat1 -> Nat1


is equivalent to



Nat1 -> (Nat1 -> Nat1)


which is corresponding to the function application always be left-associativity. (when no parentheses) for example:



ggt_euklid 1 2


is equivalent to 



(ggt_euklid 1) 2


Here 



(ggt_euklid 1) ~ Nat1 -> Nat1


and 



(ggt_euklid 1) 2 ~ Nat1


So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.






share|improve this answer

































    4














    You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:



    ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
    
ggt_euklid x = y -> result
    
  where result | x == y = x
    
               | x > y = ggt_euklid (x-y) y 
    
               | x < y = ggt_euklid x (y-x)


    Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.






    share|improve this answer























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2















      Would that not imply that ggt_euklid takes a value Nat1 and returns a
      function?




      No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.



      The Arrow -> always be right-associativity (when no parentheses), i.e.:



      Nat1 -> Nat1 -> Nat1


      is equivalent to



      Nat1 -> (Nat1 -> Nat1)


      which is corresponding to the function application always be left-associativity. (when no parentheses) for example:



      ggt_euklid 1 2


      is equivalent to 



      (ggt_euklid 1) 2


      Here 



      (ggt_euklid 1) ~ Nat1 -> Nat1


      and 



      (ggt_euklid 1) 2 ~ Nat1


      So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.






      share|improve this answer






























        2















        Would that not imply that ggt_euklid takes a value Nat1 and returns a
        function?




        No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.



        The Arrow -> always be right-associativity (when no parentheses), i.e.:



        Nat1 -> Nat1 -> Nat1


        is equivalent to



        Nat1 -> (Nat1 -> Nat1)


        which is corresponding to the function application always be left-associativity. (when no parentheses) for example:



        ggt_euklid 1 2


        is equivalent to 



        (ggt_euklid 1) 2


        Here 



        (ggt_euklid 1) ~ Nat1 -> Nat1


        and 



        (ggt_euklid 1) 2 ~ Nat1


        So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.






        share|improve this answer




























          2












          2








          2








          Would that not imply that ggt_euklid takes a value Nat1 and returns a
          function?




          No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.



          The Arrow -> always be right-associativity (when no parentheses), i.e.:



          Nat1 -> Nat1 -> Nat1


          is equivalent to



          Nat1 -> (Nat1 -> Nat1)


          which is corresponding to the function application always be left-associativity. (when no parentheses) for example:



          ggt_euklid 1 2


          is equivalent to 



          (ggt_euklid 1) 2


          Here 



          (ggt_euklid 1) ~ Nat1 -> Nat1


          and 



          (ggt_euklid 1) 2 ~ Nat1


          So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.






          share|improve this answer
















          Would that not imply that ggt_euklid takes a value Nat1 and returns a
          function?




          No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.



          The Arrow -> always be right-associativity (when no parentheses), i.e.:



          Nat1 -> Nat1 -> Nat1


          is equivalent to



          Nat1 -> (Nat1 -> Nat1)


          which is corresponding to the function application always be left-associativity. (when no parentheses) for example:



          ggt_euklid 1 2


          is equivalent to 



          (ggt_euklid 1) 2


          Here 



          (ggt_euklid 1) ~ Nat1 -> Nat1


          and 



          (ggt_euklid 1) 2 ~ Nat1


          So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 24 '18 at 15:29

























          answered Nov 22 '18 at 10:26









          assembly.jcassembly.jc

          2,0891214




          2,0891214

























              4














              You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:



              ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
              
ggt_euklid x = y -> result
              
  where result | x == y = x
              
               | x > y = ggt_euklid (x-y) y 
              
               | x < y = ggt_euklid x (y-x)


              Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.






              share|improve this answer




























                4














                You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:



                ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
                
ggt_euklid x = y -> result
                
  where result | x == y = x
                
               | x > y = ggt_euklid (x-y) y 
                
               | x < y = ggt_euklid x (y-x)


                Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.






                share|improve this answer


























                  4












                  4








                  4







                  You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:



                  ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
                  
ggt_euklid x = y -> result
                  
  where result | x == y = x
                  
               | x > y = ggt_euklid (x-y) y 
                  
               | x < y = ggt_euklid x (y-x)


                  Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.






                  share|improve this answer













                  You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:



                  ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
                  
ggt_euklid x = y -> result
                  
  where result | x == y = x
                  
               | x > y = ggt_euklid (x-y) y 
                  
               | x < y = ggt_euklid x (y-x)


                  Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 21 '18 at 23:24









                  amalloyamalloy

                  59.7k5105156




                  59.7k5105156






























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