Haskell: Deriving Show for custom type

Question

I have this type definition:

data Operace = Op (Int->Int->Int) String (Int->Int->Int) deriving Show

I want to print this type into the interactive shell (GHCi). All that should be printed is the String field.

I tried this:

instance Show Operace where
    show (Op op str inv) = show str

But I still keep getting

No instance for (Show (Int -> Int -> Int))
  arising from the 'deriving' clause of a data type declaration
Possible fix:
  add an instance declaration for (Show (Int -> Int -> Int))
  or use a standalone 'deriving instance' declaration,
       so you can specify the instance context yourself
When deriving the instance for (Show Operace)

I don't want to add Show for (Int->Int->Int), all I want to print is the string.

Thanks for help!

EDIT:

For future reference, the fixed version is:

data Operace = Op (Int->Int->Int) String (Int->Int->Int)

instance Show Operace where
    show (Op _ str _) = str

score 35 · Accepted Answer · answered May 21 '11 at 13:43

35

The instance declaration you made is the correct way to go. It seems you forgot to remove that faulty deriving clause from the original data declaration.

data Operace = Op (Int->Int->Int) String (Int->Int->Int)

instance Show Operace where
   show (Op op str inv) = show str

answered May 21 '11 at 13:43

hugomg

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1

Yes, that was it, thanks. I expected it would work like in C-like languages, where you declare list of parent types in class definition. – Matěj Zábský May 21 '11 at 13:50
1

@mzabsky: This is a powerful mechanism, as you can define additional type class instances for an existing type without touching its definition. – hammar May 21 '11 at 19:34

score 30 · Answer 2 · answered May 21 '11 at 14:22

30

You can derive Show, just import Text.Show.Functions first.

answered May 21 '11 at 14:22

Thomas M. DuBuisson

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Haskell: Deriving Show for custom type

2 Answers2

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