how do you specify non-capturing groups in sed?

Question

is it possible to specify non-capturing groups in sed?

if so, how?

Answer 1

Parentheses can be used for grouping alternatives. For example:

sed 's/a\(bc\|de\)f/X/'

says to replace "abcf" or "adef" with "X", but the parentheses also capture. There is not a facility in sed to do such grouping without also capturing. If you have a complex regex that does both alternative grouping and capturing, you will simply have to be careful in selecting the correct capture group in your replacement.

Perhaps you could say more about what it is you're trying to accomplish (what your need for non-capturing groups is) and why you want to avoid capture groups.

Edit:

There is a type of non-capturing brackets ((?:pattern)) that are part of Perl-Compatible Regular Expressions (PCRE). They are not supported in sed (but are when using grep -P).

Answer 2

I'll assume you are speaking of the backrefence syntax, which are parentheses ( ) not brackets [ ]

By default, sed will interpret ( ) literally and not attempt to make a backrefence from them. You will need to escape them to make them special as in \( \) It is only when you use the GNU sed -r option will the escaping be reversed. With sed -r, non escaped ( ) will produce backrefences and escaped \( \) will be treated as literal. Examples to follow:

POSIX sed

$ echo "foo(###)bar" | sed 's/foo(.*)bar/@@@@/'
@@@@

$ echo "foo(###)bar" | sed 's/foo(.*)bar/\1/'
sed: -e expression #1, char 16: invalid reference \1 on `s' command's RHS
-bash: echo: write error: Broken pipe

$ echo "foo(###)bar" | sed 's/foo\(.*\)bar/\1/'
(###)

GNU sed -r

$ echo "foo(###)bar" | sed -r 's/foo(.*)bar/@@@@/'
@@@@

$ echo "foo(###)bar" | sed -r 's/foo(.*)bar/\1/'
(###)

$ echo "foo(###)bar" | sed -r 's/foo\(.*\)bar/\1/'
sed: -e expression #1, char 18: invalid reference \1 on `s' command's RHS
-bash: echo: write error: Broken pipe

Update

From the comments:

Group-only, non-capturing parentheses ( ) so you can use something like intervals {n,m} without creating a backreference \1 don't exist. First, intervals are not apart of POSIX sed, you must use the GNU -r extension to enable them. As soon as you enable -r any grouping parentheses will also be capturing for backreference use. Examples:

$ echo "123.456.789" | sed -r 's/([0-9]{3}\.){2}/###/'
###789

$ echo "123.456.789" | sed -r 's/([0-9]{3}\.){2}/###\1/'
###456.789

Answer 3

As said, it is not possible to have non-capturing groups in sed. It could be obvious but non-capturing groups are not a necessity. One can just use the desired capturing ones and ignore the non-desired ones as if they were non-capturing. For reference, nested capturing groups are numbered by the position-order of "(".

E.g.,

echo "apple and bananas and monkeys" | sed -r "s/((apple|banana)s?)/\1x/g"

applex and bananasx and monkeys (note: "s" in bananas, first bigger group)

vs

echo "apple and bananas and monkeys" | sed -r "s/((apple|banana)s?)/\2x/g"

applex and bananax and monkeys (note: no "s" in bananas, second smaller group)