Haskell -- parser combinators keywords

Question

I am working on building a parser in Haskell using parser combinators. I have an issue with parsing keywords such as "while", "true", "if" etc

So the issue I am facing is that after a keyword there is a requirement that there is a separator or whitespace, for example in the statement if cond then stat1 else stat2 fi;x = 1 with this statement all keywords have either a space in front of them or a semi colon. However in different situations there can be different separators.

Currently I have implemented it as follows:

keyword :: String -> Parser String
keyword k = do
  kword <- leadingWS (string k)
  check (== ';') <|> check isSpace <|> check (== ',') <|> check (== ']')
  junk
  return word

however the problem with this keyword parser is that it will allow programs which have statements like if; cond then stat1 else stat2 fi

We tried passing in a (Char -> Bool) to keyword, which would then be passed to check. But this wouldn’t work because where we parse the keyword we don’t know what kind of separator is allowed.

I was wondering if I could have some help with this issue?

Just because a keyword is followed by a semi-colon, that doesn't mean it is being used correctly. Keywords are always part of some larger syntactic structure. Namely, a keyword is just a specific token which your grammar is built upon. The only difference between a keyword and an identifier is that the keyword is recognized by the grammar, rather than treated as a generic token left to later stages of parsing and evaluation. — chepner
– chepner, Commented Sep 8, 2016 at 18:34
The semi-colon itself is another token; you should have another combinator that recognizes correct collections of other tokens. (In this sense, the parser is really doing both lexical analysis--converting a stream of bytes into a stream of tokens--and parsing--converting a stream of tokens into an abstract syntax tree.) — chepner
– chepner, Commented Sep 8, 2016 at 18:43

Steven Shaw · Accepted Answer · 2016-09-09 11:26:01Z

3

Don't try to handle the separators in keyword but you need to ensure that keyword "if" will not be confused with an identifier "iffy" (see comment by sepp2k).

keyword :: String -> Parser String
keyword k = leadingWS $ try (do string k
                                notFollowedBy alphanum)

Handling separators for statements would go like this:

statements = statement `sepBy` semi
statement  = ifStatement <|> assignmentStatement <|> ...

edited Sep 9, 2016 at 11:26

answered Sep 8, 2016 at 23:55

Steven Shaw

6,3293 gold badges37 silver badges44 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

sepp2k Over a year ago

If you do it like this, the identifier "iffy" could be instead parsed as the keyword "if" followed by the identifier "fy". I believe (and I admit I'm totally guessing) the reason why OP decided that "after a keyword there is a requirement that there is a separator or whitespace" is to disallow this.

Steven Shaw Over a year ago

Oh i see, perhaps I missed that point in the question. I'll give it some thought and update my answer if I have anything to add.

Steven Shaw Over a year ago

I pushed a more complete example to github.com/steshaw/haskell-shebang-examples/blob/master/stack/…

Collectives™ on Stack Overflow

Haskell -- parser combinators keywords

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related