Question

I am trying to create a backward list using Haskell’s recursive types

data RevList a = Snoc a (RevList a) | Lin
    deriving Show 

mkrevlst [] = Lin
mkrevlst (x:xs) = mkrevlst xs Snoc x 

When I do > mkrevlst [1,2,3] ,the output I am expecting is : ((Lin Snoc 3) Snoc 2) Snoc 1

When I run this I get an error. I am new to Haskell & I am not able to make out where is mistake is.
Where am I going wrong?

Thank you.

Answer 1

I’m not sure what this line was supposed to be, but it doesn’t make sense as is:

mkrevlst (x:xs) = mkrevlst xs Snoc x 

The expression mkrevlist xs presumably has type RevList a, since the base case above returns Lin. Applying this to two more arguments will indeed result in a type error.

It looks like you’re expecting Snoc to be used infix, is that correct? In Haskell, identifiers made of alphanumeric characters are prefix, unless surrounded by backticks, e.g. mkrevlist xs `Snoc` x. Identifiers made of symbols are infix, unless surrounded in parentheses, and infix data constructors specifically must start with a colon. So you could also define your data type like this:

data RevList a = a :| (RevList a) | Lin
    deriving Show 

Also, note that even if you do use Snoc infix, the order of its arguments are still backwards from how you’re using it in mkrevlist.