CODEWITHSUNDEEP
haskell
hNewb
hNewb

Reputation: 393

Couldn't Match Expected Type Against Inferred Type, Rigid Type Variable Error

What is wrong with this function ?

test :: Show s => s
test = "asdasd"

String is an instance of the Show class, so it seems correct.

The error is 

src\Main.hs:224:7:
    Couldn't match expected type `s' against inferred type `[Char]'
      `s' is a rigid type variable bound by
          the type signature for `test' at src\Main.hs:223:13
    In the expression: "asdasd"
    In the definition of `test': test = "asdasd"

Upvotes: 32

Views: 11895

Answers (2)

sepp2k
sepp2k

Reputation: 370162

test :: Foo a => a means "for any type which is an instance of Foo, test is a value of that type". So in any place where you can use a value of type X where X is an instance Foo, you can use a value of type Foo a => a.

Something like test :: Num a => a; test = 42 works because 42 can be a value of type Int or Integer or Float or anything else that is an instance of Num.

However "asdasd" can't be an Int or anything else that is an instance of Show - it can only ever be a String. As a consequence it does not match the type Show s => s.

Upvotes: 39

user395760
user395760

Reputation:

Yes, String is an instance of Show. But that doesn't allow using a string as an abritary Show value. 1 can be Num a => a because there's an 1 :: Integer, an 1 :: Double, an 1 :: Word16, etc. If "asdasd" could be of type Show a => a, there would be "asdasd" :: Bool, "asdasd" :: String, "asdasd" :: Int, etc. There isn't. Therfore, "asdasd" can't be of type Show a => a. The type of a string constant doesn't get much more general than String.

Upvotes: 8

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