Not in scope: data constructor in Haskell

Question

Please tell me what is the problem?

data Stack' v = Stack' [v] Int deriving (Show)
...
type StackInt = Stack' Int 

main = print(StackInt [1,2,3] 4)

The error i am getting is

Not in scope: data constructor `Stackint'

What is wrong?

Answer 1

It looks to me like you are confusing the concepts of types and constructors, this is a common problem as they live in separate namespaces and are often given the same name. In the Haskell expression

data SomeType = SomeType Int

say, you are actually defining the type SomeType and a constructor SomeType. The type is not a function in the normal sense, but the constructor is. If you asked ghci for the type of SomeType you would get this:

:t SomeType
SomeType :: Int -> SomeType

Now, a type declaration is just a shorthand for a longer type definition, in your case making StackInt a synonym of Stack' Int. But in order to construct a value of this type you still need to use the constructor Stack' (which has type [v] -> Int -> Stack' v). So your code should be

data Stack' v = Stack' [v] Int deriving (Show)

main = print(Stack' [1,2,3] 4)

If you wanted to be sure that the type was Stack' Int then you could add a function

data Stack' v = Stack' [v] Int deriving (Show)

stackInt :: [Int] -> Int -> Stack' Int
stackInt list i = Stack' list i

main = print(stackInt [1,2,3] 4)

EDIT: Not also that I've written stackInt list i = Stack' list i for transparency here, but you can write it more elegantly just as stackInt = Stack'. It is the type constraint that makes sure that you get the right type here.

You could also have both the new function and the type synonym if you wanted, ie

data Stack' v = Stack' [v] Int deriving (Show)
type StackInt = Stack' Int

stackInt :: [Int] -> Int -> StackInt
stackInt list i = Stack' list i

main = print(stackInt [1,2,3] 4)

Answer 2

The name of the constructor is Stack', not StackInt. Creating a type alias using type does not create an alias for the constructors (which wouldn't make sense since there may be many constructors for the type and their names don't have to be related the type name at all).