in reply to reverse quotemeta?

You could do... 
$x =~ s/(?<!\\)\\n/\n/g;

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the (?<!\\) is to make sure the \ in \n isn't escaped.

                - Ant

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Re: Re: reverse quotemeta?
by chipmunk (Parson) on Jun 26, 2001 at 21:50 UTC
    Your substitution will correctly leave the \n in \\n alone, but what about \\\n? The backslash preceeding \n is itself preceeded by a backslash, so the \n should be converted in that case. \\\\n, on the other hand, shouldn't get the \n converted.

    What is needed is something to match pairs of backslashes before the \n, like this: s/((?<!\\)(?:\\\\)*)\\n/$1\n/g;

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      If you are going to do that, then you need to unescape the backslashes as well! While you can do that via (something close to one of these): 

      s#(?<!\\)((?:\\\\)*)\\n# '\\'x(length($1)/2) . "\n" #ge;
# or
s#(?<!\\)(\\*)\1\\n#$1\n#g;

      [download]
      I find that having to escape your backslashes only when they preceed a (desired) newline to be confusing so I'd go with japhy's excellent sugestion or perhaps: s#\\([\\n])# $1 eq "n" ? "\n" : $1 #ge

              - tye (but my friends call me "Tye")
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