How do I parse Haskell source into an AST in Haskell?

Question

I am trying to parse a Haskell source file into an AST. I'm new to parsers so I'm not sure how to make my own. I was just wondering if there is a function that takes a file and returns an AST. I also thought it might work to turn a file into a string using readFile and then parsing the string, but I still don't know where to go from there. Is there some well-known way to do this? If not, how can I do it myself?

What syntax are you supposed to be parsing? Haskell? Something else? — Louis Wasserman
– Louis Wasserman, Commented Jul 29, 2013 at 22:50
Have you looked at the haskell-src-exts package. In particular, the Language.Haskell.Exts.Parser module contains parsers for many kinds of Haskell AST fragments. — J. Abrahamson
– J. Abrahamson, Commented Jul 30, 2013 at 1:03

Dirk Holsopple · Accepted Answer · 2013-07-30 13:37:26Z

3

Try Language.Haskell.Meta.Parse. In particular, parseHsModule will parse an entire module for you. There are also functions for parsing smaller chunks, such as parseExp which parses an expression.

answered Jul 30, 2013 at 13:37

Dirk Holsopple

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