Re: s/// with an empty pattern uses the previous pattern of a s///

Yes, this is documented in perlop:

If the PATTERN evaluates to the empty string, the last successfully matched regular expression is used instead. In this case, only the g and c flags on the empty pattern is honoured - the other flags are taken from the original pattern. If no match has previously succeeded, this will (silently) act instead as a genuine empty pattern (which will always match).

If you want to $pat = ''; /$pat/ to match only empty strings, then add $pat = qr/^\z/ if not length($pat);
If you want to $pat = ''; /$pat/ to match everything, then add $pat = qr/.*/ if not length($pat);

Comment on Re: s/// with an empty pattern uses the previous pattern of a s/// Select or Download Code

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Re^2: s/// with an empty pattern uses the previous pattern of a s/// by ikegami (Patriarch) on May 17, 2007 at 15:59 UTC
Oh! Another to fix the problem is to add a no-op to the regexp. `$\ = "\n"; $pat = 'b'; $_='abc'; s/$pat/!/g; print; $pat = ''; $_='abc'; s/$pat/@/g; print; $pat = ''; $_='abc'; s/(?:$pat)/#/g; print;` [download] outputs `a!c a@c #a#b#c#` [download]	[reply] [d/l] [select]
Re^3: s/// with an empty pattern uses the previous pattern of a s/// by ckeith100 (Initiate) on May 17, 2007 at 16:25 UTC
Thank you for these tips. I'm not looking to replace the empty scalar, I'm looking to replace whatever is in that scalar with the value of a different scalar. Those values are determined dynamically so the best answer is to use, as you suggested, something like: `$pat && ($msg =~ s/$pat/$replace/mg);` [download] I was doing some more searching and discovered this exact quote in the perlop man page, as you said in reply #1, however in my copy of the man page it is listed in the `m//` operator rather than the substitute operator I wasn't aware that it applied there also. It does make sense that it would do, but I have to note this to try to scrape together a little dignity after having just posted a question on something in the man pages like a newbie coder :) Thanks again for your help! Colin	[reply] [d/l]
Re^4: s/// with an empty pattern uses the previous pattern of a s/// by ikegami (Patriarch) on May 17, 2007 at 17:46 UTC
That won't work if `$pat` is zero. Fix: `length($pat) && ($msg =~ s/$pat/$replace/mg);` [download] I have to note this to try to scrape together a little dignity after having just posted a question on something in the man pages like a newbie coder :) It's not well known, and it's easy to forget, because it's not a situation that's usually encountered. No dignity lost. By the way compiling the pattern is not enough. I ran a test that showed `$pat = qr//; /$pat/` is the same thing as `m//` even though the stringification of `$pat` is not an empty string.	[reply] [d/l] [select]
Re^4: s/// with an empty pattern uses the previous pattern of a s/// by diotalevi (Canon) on May 17, 2007 at 17:50 UTC
I think even many old-timers rarely see this bug. You may want to change your expression to something like `length( $path ) && ($msg =~ s/$pat/$replace/mg);` because the expression 0 is false but non-empty and potentially meaningful to you. ⠤⠤ ⠙⠊⠕⠞⠁⠇⠑⠧⠊	[reply] [d/l]
Re^4: s/// with an empty pattern uses the previous pattern of a s/// by jdporter (Paladin) on May 17, 2007 at 18:16 UTC
I'm looking to replace whatever is in that scalar with the value of a different scalar. Maybe I'm being dense, but why not just `$msg =~ s/./$replace/s;` [download] A word spoken in Mind will reach its own level, in the objective world, by its own weight*	[reply] [d/l]