in reply to Re: Parser Performance Question
in thread Parser Performance Question

Please, stop talking in riddles. Do you mean "a\\"b"? 
($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,

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Re^3: Parser Performance Question (updated)
by LanX (Saint) on Oct 05, 2017 at 13:48 UTC
    > Please, stop talking in riddles.

    Riddles are an efficient instrument for motivation in didactics. =)

    For instance a friend of mine spends loads of time in hacker competitions... ;-)

    > Do you mean "a\\"b"?

    What do you think? xD

    Cheers Rolf
    (addicted to the Perl Programming Language and ☆☆☆☆ :)
    Je suis Charlie!

    update

      DB<103> '"a\\"b"' =~  / " (?: [^"] | \\" )* " /x ; print $&        #
+ nonsense
"a\"

  DB<104> '"a\\"b"' =~  / " (?: \\" | [^"] )* " /x ; print $&        #
+ better
"a\"b"

  DB<105> '"a\\"b"' =~  / " (?: \\\\ | \\" | [^"] )* " /x ; print $& #
+ best
"a\"b"

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    Why #nonsense? Because the or-branch will never match, no matter which input.

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[d/l]

      The last two are still incorrect, as q< \\" > will be parsed as an escaped quote. /" (?: [^"\\] | \\. )* "/sx is the bestest solution :-P

      Edit: except when I'm an idiot and confuse (?= for (?:

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[d/l]
        > q< \\" > will be parsed as an escaped quote

        I don't understand, it IS an escaped quote and it works for me:

          DB<175> $re = qr/ " (?: \\\\ | \\" | [^"] )* " /x;

  DB<176> p $a= q<x  "..\\".."  x>
x  "..\".."  x

  DB<177> $a =~ /$re/; print $&
"..\".."

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        update

        maybe you should show us how you'd escape a double-quote.

        Cheers Rolf
        (addicted to the Perl Programming Language and ☆☆☆☆ :)
        Je suis Charlie!

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