Re: [Haskell-cafe] Why is Megaparsec treating these two operators differently?

Thanks, Brandon! How did you know that? I changed them to "#1" and "#2" and now it works[1]. But before making that change, why would "a # b ## c # d" evaluate, even though "a ## b" would not? [1] https://github.com/JeffreyBenjaminBrown/digraphs-with-text/tree/master/howto... The corrected file is called "experim.hs"; the old one, uncorrected, is called "experim.buggy.hs". On Sun, Oct 23, 2016 at 2:03 PM, Brandon Allbery wrote:

On Sun, Oct 23, 2016 at 4:15 PM, Jeffrey Brown wrote:

...

[ [ InfixN # symbol "#" *> pure (Pair) ] , [ InfixN # symbol "##" *> pure (Pair) ] ]

Combinator parsers can't rearrange themselves to do longest token matching. So the ## operator will take the first case, match against `symbol "#"` and aOperator will succeed; the the next token match will hit the unconsumed "#" and fail. If you place "##" first then it will match "##" but not "#", which would the match the second rule.

-- brandon s allbery kf8nh sine nomine associates allbery.b@gmail.com ballbery@sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net

-- Jeff Brown | Jeffrey Benjamin Brown Website https://msu.edu/~brown202/ | Facebook https://www.facebook.com/mejeff.younotjeff | LinkedIn https://www.linkedin.com/in/jeffreybenjaminbrown(I often miss messages here) | Github https://github.com/jeffreybenjaminbrown