[Haskell-cafe] Associative prefix operators in Parsec

[Christian Maeder]

The simplest solution is to parse the prefixes yourself and do not put 
it into the table.

(Doing the infixes "&" and "|" by hand is no big deal, too, and possibly 
easier then figuring out the capabilities of buildExpressionParser)

Cheers C.

Am 07.03.2012 13:08, schrieb Troels Henriksen:
> Consider a simple language of logical expressions:
>
>> import Control.Applicative
>> import Text.Parsec hiding ((<|>), many)
>> import Text.Parsec.String
>> import Text.Parsec.Expr
>>
>> data Expr = Truth
>>            | Falsity
>>            | And Expr Expr
>>            | Or Expr Expr
>>            | Not Expr
>>              deriving (Show, Eq)
>
> I define a simple expression parser using Parsec:
>
>> expr :: Parser Expr
>> expr    = buildExpressionParser table (lexeme term)
>>          <?>  "expression"
>>
>> term :: Parser Expr
>> term    =  between (lexeme (char '(')) (lexeme (char ')')) expr
>>          <|>  bool
>>          <?>  "simple expression"
>>
>> bool :: Parser Expr
>> bool =     lexeme (string "true" *>  pure Truth)
>>         <|>  lexeme (string "false" *>  pure Falsity)
>>
>> lexeme :: Parser a ->  Parser a
>> lexeme p = p<* spaces
>>
>> table   = [ [prefix "!" Not ]
>>            , [binary "&" And AssocLeft ]
>>            , [binary "|" Or AssocLeft ]
>>            ]
>>
>> binary  name fun assoc = Infix (do{ lexeme (string name); return fun }) assoc
>> prefix  name fun       = Prefix (do{ lexeme (string name); return fun })
>
> Now this doesn't work:
>
>> test1 = parseTest expr "!!true"
>
> But this does:
>
>> test2 = parseTest expr "!(!true)"
>
> I have studied the code for buildExpressionParser, and I know why this
> happens (prefix operators are treated as nonassociative), but it seems
> like one would often want right-associative prefix operators (so test1
> would work).  Is there a common workaround or solution for this problem?
> I assume the nonassociativity in Parsec is by design and not a bug.
>