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

Question

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"

Answer 1

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 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 .

Answer 2

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 .