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I thought that in regular expressions, the "greediness" applies to quantifiers rather than matches as a whole. However, I observe that 

grep -E --color=auto 'a+(ab)?' <(printf "aab")

returns aab rather than aab.

The same applies to sed. On the other hand, in pcregrep and other tools, it is really the quantifier that is greedy. Is this a specific behaviour of grep?

N.B. I checked both grep (BSD grep) 2.5.1-FreeBSD and grep (GNU grep) 3.1

oguz ismail
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Joseph Stack
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  • Related: [Greedy vs. Reluctant vs. Possessive Quantifiers](https://stackoverflow.com/questions/5319840) – kvantour Dec 02 '19 at 11:41
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    ok, I understood the question better now, `echo 'aab' | grep -P 'a+(ab)?'` highlights `aa` whereas `echo 'aab' | grep -E 'a+(ab)?'` highlights `aab` meaning it optional `ab` matched even though it wasn't required.. I think it is because of longest match wins.. for example, `echo 'aab' | grep -E 'a+|a+b'` highlights `aab` because that's the longest match whereas `echo 'aab' | grep -P 'a+|a+b'` highlights `aa` because in PCRE, alternation precedence is left to right for matches starting from same location – Sundeep Dec 02 '19 at 12:02
  • yes, I had read somewhere about longest match too. But it conflicted "greedy quantifier". I did not realize this was yet another point where POSIX regex and some others differ. – Joseph Stack Dec 02 '19 at 13:05

1 Answers1

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In the description of term matched, POSIX states that

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.

This statement clearly anwers your question. The string aab contains two substrings beginning at the same position matching the ERE a+(ab)?; these are aa and aab. The latter is the longest, thus it's matched.

oguz ismail
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