Is there any way to see the string that was matched in grep?

Question

I'm not talking about -o option. Posix says:

The search for a matching sequence starts at the beginning of a string and stops when the first sequence matching the expression is found, where "first" is defined to mean "begins earliest in the string". If the pattern permits a variable number of matching characters and thus there is more than one such sequence starting at that point, the longest such sequence is matched. For example, the BRE "bb*" matches the second to fourth characters of the string "abbbc", and the ERE "(wee|week)(knights|night)" matches all ten characters of the string "weeknights".

And I want to verify what is being said in posix and this tutorial regTutorialSite:

A POSIX-compliant engine will still find the leftmost match. If you apply Set|SetValue to Set or SetValue once, it will match Set.

How to "apply once"? When i run grep -o the result is two strings: Set and SetValue, but not just " one leftmost " . That is, I read about one thing, but in practice I get something else. So, how to see what string was matched by regex?

(Perhaps the question was formulated incorrectly or could have been better)

Answer 1

grep is named after the g/re/p command of the ed editor. It's about printing the lines that match the given regular expression.

What portion of the line matches is not relevant then.

The GNU implementation has added these two extensions over the standard:

• -o that prints all the non-empty matches of the regexp
• --color that highlights all the matches

But both match the regexp differently from how grep without them does as they carry on looking for more matches after the first in a manner similar to that of ed's s/pattern/<&>/g command (g being the key here).

No grep implementation that I know has a way to output the one and only match that grep without -o/--color matches on.

You'd need to use other tools such as sed, awk or perl.

For instance, to see what grep regex matches on, you could do:

sed -n 's/regex/<&>/p'

Which would print the matching lines with <...> around the matched portion. To print only the matched portion:

sed -n '
  /regex/ {
    s//\
&\
/
    s/^.*\n\(.*\)\n.*$/\1/p
  }'

For grep -E regex:

awk 'match($0, "regex") {print substr($0, RSTART, RLENGTH)}'

(awk regexps are similar to those supported by grep with -E; or used the same approach as above with sed -E where supported)

For grep -P regex:

perl -lne 'print $& if /regex/'

(grep -P initially from GNU grep is for perl regular expressions, but via PCRE2 (formerly PCRE) which are not fully equivalent to perl ones, ast-open grep has its own variant of perl-like regular expressions).

With the perl regexps (grep implementations that support both -P and -o such as GNU grep when built with optional PCRE2 support), you can also do:

grep -Po '^.*?\K(?:regex)'

The ^.*? matches as few characters as possible starting with the start of the line which prevents the regex from matching more than once. \K marks the start of what's to be Kept (and output with -o) from that. The (?:...) grouping is in case there's an alternation operator in the regex (as in a|b), avoiding the capturing variant ((...)) in case the regex has some \1 or (?1)... operators.

Beware it doesn't report empty matches.