Regular expression substitution question

rduz has asked for the wisdom of the Perl Monks concerning the following question:

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Re: Regular expression substitution question by duff (Parson) on Mar 31, 2004 at 18:19 UTC
Sure it can be done ... `$output =~ s/\x7E(?!$)/\x7D\x5E/g;` [download] See perlre where it talks about zero-width lookaheads duff	[reply] [d/l]
Re: Re: Regular expression substitution question by Roy Johnson (Monsignor) on Mar 31, 2004 at 19:01 UTC
And for us anal-retentive types who hate to replace something with itself: `s/(?<=\x7E(?!$))/\x5E/g;` [download] Oh, just shoot me now. I didn't notice 7E->7D. I must not be anal-retentive enough. Thanks, Enlil. `s/~(?!$)/}^/g;` [download] The PerlMonk `tr///` Advocate	[reply] [d/l] [select]
Re: Re: Regular expression substitution question by rduz (Initiate) on Apr 01, 2004 at 03:08 UTC
Thanks for the answers. This indeed works quite well.	[reply]
Re: Regular expression substitution question by Enlil (Parson) on Mar 31, 2004 at 18:20 UTC
Try: `$output =~ s/\x7E(?!$)/\x7D\x5E/g;` [download] update:Yeah, what he said (alas, beaten to the punch again) -enlil	[reply] [d/l]
Re: Regular expression substitution question by davido (Cardinal) on Mar 31, 2004 at 18:20 UTC
The idea I've employed here is to use a positive zero-width lookahead assertion to ensure that a match can only occur if there is one more character following the tilde. `$output =~ s/\x7E(?=.)/\x7D\x5E/g;` [download] It could also be expressed without the zero-width assertion like this: `$output =~ s/\x7E\G(.)/\x7D\x5E$1/g;` But the latter will probably run slower because it requires capturing parens. Update: Use the solutions with lookahead (Enlil's, duff's, or mine); they work, whereas the method that I've "crossed out" skips occurrences of the tilde if they're contiguous. Dave	[reply] [d/l] [select]