CODEWITHSUNDEEP
haskellfunctional-programming
jiipeezz
jiipeezz

Reputation: 245

Showing a number with preceding '+' sign

I'm new to functional programming and Haskell, and having really difficult time understanding making of your own data types. As a learning source I have been using http://learnyouahaskell.com/, but I'm still unable to grasp some ideas.

My specific problem is, that I'm trying to create a show Instance for my custom data type, which always shows '+' symbol in front of whatever is showed (in my case input 555 would be "+555"). This is how I've been trying to solve it.

data CountryCode = CountryCode Integer deriving Eq

instance Show CountryCode where
    show _ = "+" : _

And this is what I get when I try to load it.

[1 of 1] Compiling Main             ( phonetest.hs, interpreted )

phonetest.hs:6:14: error:
    • Couldn't match type ‘[Char]’ with ‘Char’
      Expected type: String
        Actual type: [[Char]]
    • In the expression: "+" : _
      In an equation for ‘show’: show _ = "+" : _
      In the instance declaration for ‘Show CountryCode’

phonetest.hs:6:20: error:
    • Found hole: _ :: [[Char]]
    • In the second argument of ‘(:)’, namely ‘_’
      In the expression: "+" : _
      In an equation for ‘show’: show _ = "+" : _
    • Relevant bindings include
        show :: CountryCode -> String (bound at phonetest.hs:6:5)
Failed, modules loaded: none.

I can get something out of the error message, but not enough to make it work correctly. I also tried show (CountryCode _) = "+" : _ but haskell still complains. To me it seems pretty logical, but clearly there's some basic haskell knowledge I'm lacking.

Upvotes: 1

Views: 100

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476503

You should unpack the data constructor, and thus obtain the parameter x. We can then use guards to check if the value is positive or negative. In case it is positive, we can prepend 'x' to the result of show x:

instance Show CountryCode where
    show (CountryCode x) | x >= 0 = '+' : show x
                         | otherwise = show x

or we can omit branching in a positive and negative case like @chepner suggests with:

instance Show CountryCode where
    show (CountryCode x) = '+' : show x

Upvotes: 4

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