bug#31816: closed (Re: bug#31816: Saved Sub String Only Saves Last)

[Mark Otto]

Dear Eric,

Thank you for your thorough explanation of the greediness of sed.  If I was
thinking about sed's greediness, I should have thought that it would be
consistent at every point, including being greedy before my back
reference.  The nongreedy perl operators are intuitive, but their matching
process still needs to be thought through.

I found an explanation of the difference between greedy and non-greedy here
<https://stackoverflow.com/questions/3075130/what-is-the-difference-between-and-regular-expressions>
:

Consider the input 101000000000100.  Using 1.*1, * is greedy  It will match
all the way to the end, and then backtrack until it can match a 1, leaving
you with 1010000000001.  .*? is non-greedy. * will match nothing, but then
will try to match extra characters until it matches a 1, eventually
matching 101.  All quantifiers have a non-greedy mode: .*?, .+?, .{2,6}?,
and even .??.

Sed is a UNIX standard, so I could think harder about how it works rather
than jumping to "It's a bug!"

Best wishes,
Mark

On Mon, Jun 18, 2018 at 4:05 PM GNU bug Tracking System <
address@hidden> wrote:

> Your bug report
>
> #31816: Saved Sub String Only Saves Last
>
> which was filed against the sed package, has been closed.
>
> The explanation is attached below, along with your original report.
> If you require more details, please reply to address@hidden
>
> --
> 31816: http://debbugs.gnu.org/cgi/bugreport.cgi?bug=31816
> GNU Bug Tracking System
> Contact address@hidden with problems
>
>
>
> ---------- Forwarded message ----------
> From: Eric Blake <address@hidden>
> To: address@hidden, address@hidden, GNU bug control <
> address@hidden>
> Cc:
> Bcc:
> Date: Mon, 18 Jun 2018 15:04:23 -0500
> Subject: Re: bug#31816: Saved Sub String Only Saves Last
> tag 31816 notabug
> thanks
>
> On 06/13/2018 12:03 PM, Mark Otto wrote:
> > If I use a saved substring it should capture the maximum number of
> > characters that fit the pattern, in this case  [0-9][0-9]*.
>
> Sed already does that (an operator is as greedy as possible, given what
> has already been matched earlier in the line).  However, you are
> misunderstanding how greedy operators work.
>
> >
> > echo "I'm 2254 years old"|sed "s/^..*\([0-9][0-9]*\) /She's \1 /"
> > She's 4 years old"
>
> That is correct output.  Remember, in sed, every pattern is evaluated
> from left to right to find the longest possible substring that will
> match, where patterns on the left use a shorter substring only if
> patterns on the right are not possible with the longest substring.
> Since .* is a greedy pattern, you have matched:
>
> "I" "'m 225" "4"
>   ^.  .*       \([0-9][0-9]*\)
>
> >
> >
> > She should be 2254 years old.
>
> If you want the second pattern to match longer as a higher priority than
> the first .* pattern being greedy, you have to use some other pattern on
> the first use, such as:
>
> echo "I'm 2254 years old" | sed "s/^..*[^0-9]\([0-9][0-9]*\)/She's \1/"
>
> which matches as:
>
> "I" "'m" " "     "2254"
>   ^.  .*   [^0-9]  \([0-9][0-9]*\)
>
> where my explicit match of a non-digit forced the .* to be less greedy.
>
> Or, you can use other languages, like perl, which have the extension of
> non-greedy operators, as in:
>
> echo "I'm 2254 years old" | perl -pe "s/^..*?([0-9]+) /She's \1/"
>
> perl is more like 'sed -E', but has the additional '.*?' non-greedy
> counterpart to '.*' that sed lacks.
>
> >
> > It does search correctly because without the substring it replaces all
> the
> > digits:
> >
> > echo "I'm 2287 years old"|sed "s/^..*[0-9][0-9]*/She's many/"
> > She's many years old"
>
> That output is still correct, but wasn't doing what you claimed it was
> doing.  Again, it was matching:
>
> "I" "'m 228" "7"
>   ^.  .*       [0-9][0-9]*
>
> then replacing that entire match.
>
> As such, I'm marking this as not a bug.  But feel free to comment
> further if you still need help.
>
> --
> Eric Blake, Principal Software Engineer
> Red Hat, Inc.           +1-919-301-3266
> Virtualization:  qemu.org | libvirt.org
>
>
>
>
> ---------- Forwarded message ----------
> From: Mark Otto <address@hidden>
> To: address@hidden
> Cc:
> Bcc:
> Date: Wed, 13 Jun 2018 13:03:16 -0400
> Subject: Saved Sub String Only Saves Last
> If I use a saved substring it should capture the maximum number of
> characters that fit the pattern, in this case  [0-9][0-9]*.
>
> echo "I'm 2254 years old"|sed "s/^..*\([0-9][0-9]*\) /She's \1 /"
> She's 4 years old"
>
>
> She should be 2254 years old.
>
> It does search correctly because without the substring it replaces all the
> digits:
>
> echo "I'm 2287 years old"|sed "s/^..*[0-9][0-9]*/She's many/"
> She's many years old"
>
>
> Here is my version information:
>
> sed --version # On Windows 10
> sed (GNU sed) 4.4
> Copyright (C) 2017 Free Software Foundation, Inc.
> License GPLv3+: GNU GPL version 3 or later <
> http://gnu.org/licenses/gpl.html>.
> This is free software: you are free to change and redistribute it.
> There is NO WARRANTY, to the extent permitted by law.
>
> Written by Jay Fenlason, Tom Lord, Ken Pizzini,
> and Paolo Bonzini.
> GNU sed home page: <http://www.gnu.org/software/sed/>.
> General help using GNU software: <http://www.gnu.org/gethelp/>.
> E-mail bug reports to: <address@hidden>.
>