CODEWITHSUNDEEP
haskelloverloading
Overt_Agent
Overt_Agent

Reputation: 519

Overloading a function with a data type in Haskell

If I've declared a data type thus:

data ExampleType = TypeA (Int, Int) | TypeB (Int, Int, Int)

How can I declare a function that takes either TypeA or TypeB and performs different operations on that basis? My current attempt is:

exampleFunction :: ExampleType -> Int
exampleFunction (TypeA(firstInt, secondInt)) = --Fn body
exampleFunction (TypeB(firstInt, secondInt, thirdInt)) = --Fn body

But I'm getting a Duplicate type signatures error, so I'm clearly missing something.

Upvotes: 0

Views: 180

Answers (2)

Julia Path
Julia Path

Reputation: 2366

Your code should not cause such an error. However there some things wrong with your question. First the use of a tuple is unsusual for a Product type (TypeA (Int, Int)). Instead you would just declare TypeA as a data constructor that takes two arguments of type Int instead of one of type (Int, Int). Furthermore TypeA and TypeB are not two different types but two different data constructors of the same sum type ExampleType. To reflect that I renamed them to DataA and DataB in the below code.

data ExampleType = DataA Int Int | DataB Int Int Int
exampleFunction :: ExampleType -> Int
exampleFunction (DataA x y) = x + y
exampleFunction (DataB x y z) = x + y + z

Upvotes: 2

Landei
Landei

Reputation: 54584

Works for me: 

data ExampleType = TypeA (Int, Int) | TypeB (Int, Int, Int)

exampleFunction :: ExampleType -> Int
exampleFunction (TypeA (a,b)) = a + b
exampleFunction (TypeB (a,b,c)) = a + c + c

main = print $ exampleFunction (TypeA (2,3))

http://ideone.com/jsIBVF

Note that you typically wouldn't use tuples as components of your type, as this makes it quite hard to get to the data. If you have not a very good reason, simply use data ExampleType = TypeA Int Int | TypeB Int Int Int

Upvotes: 4

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