In my textbook, there is an example of how to take out the first and third character from a string using parsers. I have tried to write it down as it stands in the book.
type Parser a = String -> [(a,String)]
item = \inp -> case inp of
[]-> []
(x:xs) -> [(x,xs)]
p = do
x <- item
item
y <- item
return (x,y)
However I get the message:
dataTest.hs:47:11:
parse error on input ‘<-’
Perhaps this statement should be within a 'do' block?
As I was not able to figure out what went wrong, I wrote down this for a less abstract version of the code:
q = item >>= \x1 ->
item >>= \x2 ->
item >>= \x3 ->
return (x1, x3)
But as shown below:
*Main> q "abc"
([('a',"bc")],[('a',"bc")],[('a',"bc")])
I get the wrong output.
So I have two questions:
It's just a formatting thing. Start a newline:
p = do
x <- item
item
y <- item
return (x,y)
As for q, remember that do {x <- item; f x} is equivalent to item >>= \x -> f x, and the second one is redundant, so:
q = do
x1 <- item
item
x3 <- item
return (x1, x3)
...which is equivalent to p.