CODEWITHSUNDEEP
haskelltypesparameterspattern-matching
Alexander Robertson
Alexander Robertson

Reputation: 13

How do I pattern match on two parameters in haskell

I am quite new to Haskell, so apologies if there is any incorrect or confusing syntax. I have massively simplified what I am trying to do to make it more understandable.

To start, I have two user defined types:

data Foo = A String | B Int | C

type Bar = (Int, String, [Int])

I am trying to write a function such that:

myfunc :: Foo -> Bar -> Bar


--if Foo is A s,
--  increment intA
--  append s to stringA
--  return new Bar

myfunc (A s) intA stringA listA = (intA + 1) stringA++s listA 


--if Foo is B i, 
--  if listA[1]<listA[0]
--    increment intA by i
--    increment intA
--    return new Bar
--  else
--    increment intA
--    return new Bar

myfunc (B i) intA stringA (x:y:xs) = if y<x then ((intA+i)+1 stringA xs) else ((intA+1) stringA xs)



--if Foo is C,
--  increment intA
--  add listA[0], listA[1]
--  prepend to listA
--  return new Bar

myfunc (C) intA stringA (top:second:xs) = (intA + 1) stringA top+second:xs

So, there is a different definition of myfunc for each possible value of Foo.

I then want to access the values in the second parameter, Bar, in order to return an 'updated' Bar, updated in different ways depending on the Foo used.

I am currently encountering an error on the myfunc (B i) version of myfunc:

Couldn't match type ‘(Int, String, [Int])’ with ‘[Bar]’
      Expected type: [Bar]
        Actual type: Bar

Which I interpret as the compiler expecting a list of Bar, which I don't understand.

Upvotes: 1

Views: 466

Answers (1)

chepner
chepner

Reputation: 531345

A Bar value is a single tuple, not 3 separate values. Match an existing value, and create a new value for return, just as you would any other 3-tuple.

myfunc (A s) (intA, stringA, listA) = ((intA + 1), stringA++s, listA)
-- etc

Upvotes: 2

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