in reply to Re^4: Parser Performance Question (updated)
in thread Parser Performance Question
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 $& "..\".."
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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Re^6: Parser Performance Question (updated)
by Eily (Monsignor) on Oct 06, 2017 at 08:02 UTC | |
by LanX (Saint) on Oct 06, 2017 at 08:37 UTC | |
by Eily (Monsignor) on Oct 06, 2017 at 08:45 UTC | |
by LanX (Saint) on Oct 06, 2017 at 11:28 UTC | |
by Eily (Monsignor) on Oct 06, 2017 at 12:37 UTC | |
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by LanX (Saint) on Oct 06, 2017 at 10:48 UTC |