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Why is regex [0-9]0,2 not greedy in sed?

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3

echo '123980925sriten34=ienat' | sed -e 's/^.*?([1-9][0-9]0,2+)([%=+-]).*/ 1 2 /'

is giving the result:

 4 =

I am expecting:

 34 =

What am I not understanding?

(Oh and I even added the + and ? to make doubly sure, but afaik 0,2 should be greedy without them.)

shell-script sed regular-expression

share|improve this question

edited Jul 20 at 11:09

Jeff Schaller♦

48.4k1111 gold badges7272 silver badges161161 bronze badges

asked Jul 19 at 23:52

runrinrunrin

1622 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  perl -pe 's/^.*?([1-9][0-9]0,2)([%=+-]).*/ $1 $2 /' is less annoying
  
  – user1133275
  Jul 19 at 23:59
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Isn't it more to do with the fact that the preceding .* is greedy?
  
  – steeldriver
  Jul 20 at 0:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  ... perhaps you're thinking that the following ? makes it non-greedy?
  
  – steeldriver
  Jul 20 at 0:49
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  wrt i even added the + and ? to make doubly sure - that will probably make doubly sure the regexp won't work. You can't just throw random characters into a regexp and hope they'll somehow improve it. Also ? and + are GNU sed only so if you aren't running GNU sed then they're going to be treated as literal chars - the POSIX equivalents are 0,1 and 1, respectively.
  
  – Ed Morton
  Jul 20 at 5:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  + and ? aren't BRE, but even if they were, stacking the repetition specifiers (* and ? or n,m and +) isn't defined.
  
  – ilkkachu
  Jul 20 at 9:06
  

  
  
  
  

add a comment | 

3

echo '123980925sriten34=ienat' | sed -e 's/^.*?([1-9][0-9]0,2+)([%=+-]).*/ 1 2 /'

is giving the result:

 4 =

I am expecting:

 34 =

What am I not understanding?

(Oh and I even added the + and ? to make doubly sure, but afaik 0,2 should be greedy without them.)

shell-script sed regular-expression

share|improve this question

edited Jul 20 at 11:09

Jeff Schaller♦

48.4k1111 gold badges7272 silver badges161161 bronze badges

asked Jul 19 at 23:52

runrinrunrin

1622 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  perl -pe 's/^.*?([1-9][0-9]0,2)([%=+-]).*/ $1 $2 /' is less annoying
  
  – user1133275
  Jul 19 at 23:59
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Isn't it more to do with the fact that the preceding .* is greedy?
  
  – steeldriver
  Jul 20 at 0:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  ... perhaps you're thinking that the following ? makes it non-greedy?
  
  – steeldriver
  Jul 20 at 0:49
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  wrt i even added the + and ? to make doubly sure - that will probably make doubly sure the regexp won't work. You can't just throw random characters into a regexp and hope they'll somehow improve it. Also ? and + are GNU sed only so if you aren't running GNU sed then they're going to be treated as literal chars - the POSIX equivalents are 0,1 and 1, respectively.
  
  – Ed Morton
  Jul 20 at 5:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  + and ? aren't BRE, but even if they were, stacking the repetition specifiers (* and ? or n,m and +) isn't defined.
  
  – ilkkachu
  Jul 20 at 9:06
  

  
  
  
  

add a comment | 

echo '123980925sriten34=ienat' | sed -e 's/^.*?([1-9][0-9]0,2+)([%=+-]).*/ 1 2 /'

is giving the result:

 4 =

I am expecting:

 34 =

What am I not understanding?

(Oh and I even added the + and ? to make doubly sure, but afaik 0,2 should be greedy without them.)

shell-script sed regular-expression

share|improve this question

edited Jul 20 at 11:09

Jeff Schaller♦

48.4k1111 gold badges7272 silver badges161161 bronze badges

asked Jul 19 at 23:52

runrinrunrin

1622 bronze badges

echo '123980925sriten34=ienat' | sed -e 's/^.*?([1-9][0-9]0,2+)([%=+-]).*/ 1 2 /'

 4 =

 34 =

share|improve this question

edited Jul 20 at 11:09

Jeff Schaller♦

48.4k1111 gold badges7272 silver badges161161 bronze badges

asked Jul 19 at 23:52

runrinrunrin

1622 bronze badges

share|improve this question

edited Jul 20 at 11:09

Jeff Schaller♦

48.4k1111 gold badges7272 silver badges161161 bronze badges

asked Jul 19 at 23:52

runrinrunrin

1622 bronze badges

edited Jul 20 at 11:09

Jeff Schaller♦

48.4k1111 gold badges7272 silver badges161161 bronze badges

edited Jul 20 at 11:09

Jeff Schaller♦

48.4k1111 gold badges7272 silver badges161161 bronze badges

asked Jul 19 at 23:52

runrinrunrin

1622 bronze badges

asked Jul 19 at 23:52

runrinrunrin

1622 bronze badges

1 Answer
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11

The problem, as steeldriver states,
isn’t that the [0-9]0,2 is non-greedy;
the problem is that the .*? before it is greedy. 
sed supports BRE and ERE, neither of which supports non-greedy matching. 
That’s a feature of PCREs. 
For example, the following commands:

$ echo 'aQbQc' | sed 's/.*?Q/X/'
$ echo 'aQbQc' | sed 's/.*Q/X/'
$ echo 'aQbQc' | sed -r 's/.*?Q/X/'
$ echo 'aQbQc' | sed -r 's/.*Q/X/'

all output

Xc

(I’m not sure why it just ignores the ?.)

See Non-greedy match with SED regex (emulate perl's .*?).

Your description of the function that you want to perform is skimpy,
but I believe that I’ve reverse engineered it. 
You can get the desired effect by not matching the characters
before the number you want to match until after you’ve found the number:

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/! 1 2 /' -e 's/.*!//'
 34 =

replacing the ! with any string known not to appear in the input data. 
If you have no such string, but you’re using GNU sed, you can use newline:

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/n 1 2 /' -e 's/.*n//'
 34 =

which, of course, cannot appear in any line.

share|improve this answer

answered Jul 20 at 1:37

G-ManG-Man

15.2k99 gold badges4343 silver badges8181 bronze badges

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  wrt I’m not sure why it just ignores the ? - because ? after another repetition RE metachar (* in this case) is undefined behavior per POSIX and since in other contexts it means zero-or-1 just ignoring it is as reasonable approach as any. Essentially .*? should be treated as bug in a regexp as it doesn't have any sensible meaning (zero-or one repetitions of zero-or-more repetitions of any character - huh?).
  
  – Ed Morton
  Jul 20 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  On a GNU system, the repetition operators stack, ? after + (using ERE syntax) makes the whole thing optional (same as *), and 2,4? matches 0, 2, 3 or 4 repetitions (try grep -Eoe 'ba2,4?b' against bb, bab, baab). *? is just the same as * since * can already match zero repetitions. The thing where ? makes a pattern non-greedy is a feature of Perl regexes.
  
  – ilkkachu
  Jul 20 at 9:03
  
  

  
  
  
  

add a comment | 

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11

The problem, as steeldriver states,
isn’t that the [0-9]0,2 is non-greedy;
the problem is that the .*? before it is greedy. 
sed supports BRE and ERE, neither of which supports non-greedy matching. 
That’s a feature of PCREs. 
For example, the following commands:

$ echo 'aQbQc' | sed 's/.*?Q/X/'
$ echo 'aQbQc' | sed 's/.*Q/X/'
$ echo 'aQbQc' | sed -r 's/.*?Q/X/'
$ echo 'aQbQc' | sed -r 's/.*Q/X/'

all output

Xc

(I’m not sure why it just ignores the ?.)

See Non-greedy match with SED regex (emulate perl's .*?).

Your description of the function that you want to perform is skimpy,
but I believe that I’ve reverse engineered it. 
You can get the desired effect by not matching the characters
before the number you want to match until after you’ve found the number:

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/! 1 2 /' -e 's/.*!//'
 34 =

replacing the ! with any string known not to appear in the input data. 
If you have no such string, but you’re using GNU sed, you can use newline:

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/n 1 2 /' -e 's/.*n//'
 34 =

which, of course, cannot appear in any line.

share|improve this answer

answered Jul 20 at 1:37

G-ManG-Man

15.2k99 gold badges4343 silver badges8181 bronze badges

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  wrt I’m not sure why it just ignores the ? - because ? after another repetition RE metachar (* in this case) is undefined behavior per POSIX and since in other contexts it means zero-or-1 just ignoring it is as reasonable approach as any. Essentially .*? should be treated as bug in a regexp as it doesn't have any sensible meaning (zero-or one repetitions of zero-or-more repetitions of any character - huh?).
  
  – Ed Morton
  Jul 20 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  On a GNU system, the repetition operators stack, ? after + (using ERE syntax) makes the whole thing optional (same as *), and 2,4? matches 0, 2, 3 or 4 repetitions (try grep -Eoe 'ba2,4?b' against bb, bab, baab). *? is just the same as * since * can already match zero repetitions. The thing where ? makes a pattern non-greedy is a feature of Perl regexes.
  
  – ilkkachu
  Jul 20 at 9:03
  
  

  
  
  
  

add a comment | 

11

The problem, as steeldriver states,
isn’t that the [0-9]0,2 is non-greedy;
the problem is that the .*? before it is greedy. 
sed supports BRE and ERE, neither of which supports non-greedy matching. 
That’s a feature of PCREs. 
For example, the following commands:

$ echo 'aQbQc' | sed 's/.*?Q/X/'
$ echo 'aQbQc' | sed 's/.*Q/X/'
$ echo 'aQbQc' | sed -r 's/.*?Q/X/'
$ echo 'aQbQc' | sed -r 's/.*Q/X/'

all output

Xc

(I’m not sure why it just ignores the ?.)

See Non-greedy match with SED regex (emulate perl's .*?).

Your description of the function that you want to perform is skimpy,
but I believe that I’ve reverse engineered it. 
You can get the desired effect by not matching the characters
before the number you want to match until after you’ve found the number:

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/! 1 2 /' -e 's/.*!//'
 34 =

replacing the ! with any string known not to appear in the input data. 
If you have no such string, but you’re using GNU sed, you can use newline:

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/n 1 2 /' -e 's/.*n//'
 34 =

which, of course, cannot appear in any line.

share|improve this answer

answered Jul 20 at 1:37

G-ManG-Man

15.2k99 gold badges4343 silver badges8181 bronze badges

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  wrt I’m not sure why it just ignores the ? - because ? after another repetition RE metachar (* in this case) is undefined behavior per POSIX and since in other contexts it means zero-or-1 just ignoring it is as reasonable approach as any. Essentially .*? should be treated as bug in a regexp as it doesn't have any sensible meaning (zero-or one repetitions of zero-or-more repetitions of any character - huh?).
  
  – Ed Morton
  Jul 20 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  On a GNU system, the repetition operators stack, ? after + (using ERE syntax) makes the whole thing optional (same as *), and 2,4? matches 0, 2, 3 or 4 repetitions (try grep -Eoe 'ba2,4?b' against bb, bab, baab). *? is just the same as * since * can already match zero repetitions. The thing where ? makes a pattern non-greedy is a feature of Perl regexes.
  
  – ilkkachu
  Jul 20 at 9:03
  
  

  
  
  
  

add a comment | 

The problem, as steeldriver states,
isn’t that the [0-9]0,2 is non-greedy;
the problem is that the .*? before it is greedy. 
sed supports BRE and ERE, neither of which supports non-greedy matching. 
That’s a feature of PCREs. 
For example, the following commands:

$ echo 'aQbQc' | sed 's/.*?Q/X/'
$ echo 'aQbQc' | sed 's/.*Q/X/'
$ echo 'aQbQc' | sed -r 's/.*?Q/X/'
$ echo 'aQbQc' | sed -r 's/.*Q/X/'

all output

Xc

(I’m not sure why it just ignores the ?.)

See Non-greedy match with SED regex (emulate perl's .*?).

Your description of the function that you want to perform is skimpy,
but I believe that I’ve reverse engineered it. 
You can get the desired effect by not matching the characters
before the number you want to match until after you’ve found the number:

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/! 1 2 /' -e 's/.*!//'
 34 =

replacing the ! with any string known not to appear in the input data. 
If you have no such string, but you’re using GNU sed, you can use newline:

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/n 1 2 /' -e 's/.*n//'
 34 =

which, of course, cannot appear in any line.

share|improve this answer

answered Jul 20 at 1:37

G-ManG-Man

15.2k99 gold badges4343 silver badges8181 bronze badges

$ echo 'aQbQc' | sed 's/.*?Q/X/'
$ echo 'aQbQc' | sed 's/.*Q/X/'
$ echo 'aQbQc' | sed -r 's/.*?Q/X/'
$ echo 'aQbQc' | sed -r 's/.*Q/X/'

Xc

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/! 1 2 /' -e 's/.*!//'
 34 =

$ echo '123980925sriten34=ienat' | sed -e 's/([1-9][0-9]0,2+)([%=+-]).*/n 1 2 /' -e 's/.*n//'
 34 =

share|improve this answer

answered Jul 20 at 1:37

G-ManG-Man

15.2k99 gold badges4343 silver badges8181 bronze badges

answered Jul 20 at 1:37

G-ManG-Man

15.2k99 gold badges4343 silver badges8181 bronze badges

answered Jul 20 at 1:37

G-ManG-Man

15.2k99 gold badges4343 silver badges8181 bronze badges