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Subject: |
Saved Sub String Only Saves Last |
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Wed, 13 Jun 2018 13:03:16 -0400 |
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/"
Here is my version information:
sed --version # On Windows 10
sed (GNU sed) 4.4
Copyright (C) 2017 Free Software Foundation, Inc.
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.
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--- Begin Message ---
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Subject: |
Re: bug#31816: Saved Sub String Only Saves Last |
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Date: |
Mon, 18 Jun 2018 15:04:23 -0500 |
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Mozilla/5.0 (X11; Linux x86_64; rv:52.0) Gecko/20100101 Thunderbird/52.8.0 |
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
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