CODEWITHSUNDEEP
haskell
Jordan
Jordan

Reputation: 1161

Why do I get a No instance for (Show t0) arising from a use of show error?

I'm stuck on this problem and I'm very new to Haskell, I was trying to complete the first Euler problem with the following code:

main = putStrLn . show . sum $ [3,6..1000]:[5,10..1000]

And here's the first part of the error:

euler/1.hs:1:19:
    No instance for (Show t0) arising from a use of `show'
    The type variable `t0' is ambiguous
    Possible fix: add a type signature that fixes these type variable(s)
    Note: there are several potential instances:
      instance Show Double -- Defined in `GHC.Float'
      instance Show Float -- Defined in `GHC.Float'
      instance (Integral a, Show a) => Show (GHC.Real.Ratio a)
        -- Defined in `GHC.Real'
      ...plus 23 others
    In the first argument of `(.)', namely `show'
    In the second argument of `(.)', namely `show . sum'
    In the expression: putStrLn . show . sum

Upvotes: 3

Views: 1339

Answers (2)

HaskellElephant
HaskellElephant

Reputation: 9891

Unfortunately the error message you get is missing the problem. Loading it in ghci 7.4.2 gives much more helpful error messages:

No instance for (Num [t0])
  arising from a use of `sum'
Possible fix: add an instance declaration for (Num [t0])
In the second argument of `(.)', namely `sum'
In the second argument of `(.)', namely `show . sum'
In the expression: putStrLn . show . sum

No instance for (Enum [t0])
  arising from the arithmetic sequence `5, 10 .. 1000'
Possible fix: add an instance declaration for (Enum [t0])
In the second argument of `(:)', namely `[5, 10 .. 1000]'
In the second argument of `($)', namely
  `[3, 6 .. 1000] : [5, 10 .. 1000]'
In the expression:
  putStrLn . show . sum $ [3, 6 .. 1000] : [5, 10 .. 1000]

This seems to hint that the problem is in the expression [3, 6 .. 1000] : [5, 10 .. 1000] and indeed it is because the type of : is a -> [a] -> [a] and x:xs simply adds the element x to the beginning of the list xs. In [3, 6 .. 1000] : [5, 10 .. 1000] you are trying to add [3, 6 .. 1000] as an element to [5, 10 .. 1000], but lists in haskell must contain elements of the same type (they are homogeneous) so that will not work. This is because [3, 6 .. 1000] has the type [Int] and the elements of [5, 10 .. 1000] has the type Int.

I think that what you are looking for is ++ which has the type [a] -> [a] -> [a] and xs ++ ys is the concatenation of the list xs and the list ys:

ghci> putStrLn . show . sum $ [3,6..1000] ++ [5,10..1000]
267333

Upvotes: 4

Karolis Juodelė
Karolis Juodelė

Reputation: 3770

What do you expect [3,6..1000]:[5,10..1000] to do? x : xs preppeds an object to a list of objects. Here both arguments are lists of integers. Did you want ++ instead (concatenation).

It needs to be said that your approach is not right, no matter what you expected : to do. I'll not expand on this in case you want to figure it out yourself.

Upvotes: 6

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