CODEWITHSUNDEEP
haskell
nullromo
nullromo

Reputation: 2647

Extracting Information from Haskell Object

I'm new to Haskell and I'm confused on how to get values out of function results. In my particular case, I am trying to parse Haskell files and see which AST nodes appear on which lines. This is the code I have so far:

import Language.Haskell.Parser
import Language.Haskell.Syntax

getTree :: String -> IO (ParseResult HsModule)
getTree path = do
               file <- readFile path
               let tree = parseModuleWithMode (ParseMode path) file
               return tree

main :: IO ()
main = do
       tree <- getTree "ex.hs"
       -- <do something with the tree other than print it>
       print tree

So on the line where I have the comment, I have a syntax tree as tree. It appears to have type ParseResult HsModule. What I want is just HsModule. I guess what I'm looking for is a function as follows:

extract :: ParseResult a -> a

Or better yet, a general Haskell function

extract :: AnyType a -> a

Maybe I'm missing a major concept about Haskell here?

p.s. I understand that thinking of these things as "Objects" and trying to access "Fields" from them is wrong, but I'd like an explanation of how to deal with this type of thing in general.

Upvotes: 2

Views: 553

Answers (3)

amalloy
amalloy

Reputation: 92012

Looking for a general function of type

extract :: AnyType a -> a

does indeed show a big misunderstanding about Haskell. Consider the many things AnyType might be, and how you might extract exactly one object from it. What about Maybe Int? You can easily enough convert Just 5 to 5, but what number should you return for Nothing?

Or what if AnyType is [], so that you have [String]? What should be the result of 

extract ["help", "i'm", "trapped"]

or of 

extract []

?

ParseResult has a similar "problem", in that it uses ParseOk to contain results indicating that everything was fine, and ParseFailed to indicate an error. Your incomplete pattern match successfully gets the result if the parse succeeded, but will crash your program if in fact the parse failed. By using ParseResult, Haskell is encouraging you to consider what you should do if the code you are analyzing did not parse correctly, rather than to just blithely assume it will come out fine.

Upvotes: 7

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476813

The definition of ParseResult is:

data ParseResult a = ParseOk a | ParseFailed SrcLoc String

(obtained from source code)

So there are two possibilities: either the parsing succeeded, and it will return a ParseOk instance, or something went wrong during the parsing in which case you get the location of the error, and an error message with a ParseFailed constructor.

So you can define a function:

getData :: ParseResult a -> a
getData (ParseOk x) = x
getData (ParseFailed _ s) = error s

It is better to then throw an error as well, since it is always possible that your compiler/interpreter/analyzer/... parses a Haskell program containing syntax errors.

Upvotes: 3

nullromo
nullromo

Reputation: 2647

I just figured out how to do this. It seems that when I was trying to define

extract :: ParseResult a -> a
extract (ParseResult a) = a

I actually needed to use

extract :: ParseResult a -> a
extract (ParseOk a) = a

instead. I'm not 100% sure why this is.

Upvotes: 1

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