Timeline for sed command that replaces number and word by two

Current License: CC BY-SA 3.0

11 events

when toggle format	what		by	license	comment
Jan 22, 2018 at 3:44	vote	accept	Tinler
Jan 22, 2018 at 1:29	comment	added	cas		@Tinler the difference between the two forms is that a `&` in the replacement inserts the entire match, while `\1`, `\2`, etc insert only a matched capture group. e.g. with `s/the $capture group$/&/` vs `s/the $capture group$/\1/`. the `&` inserts "the capture group" into the replacement, `\1` inserts only "capture group". There can be multiple capture groups in a search pattern, and they are referred to in numerical order - `\1` is the first capture group, `\2` is the second, etc.
Jan 22, 2018 at 1:12	comment	added	George Udosen		This will also work: `sed -E 's/([[:digit:]])/\1\1/'`
Jan 22, 2018 at 1:02	comment	added	George Vasiliou		OK. in gnu sed using `\0` refers to the previously matched regex. In bsd sed this is done using `&`. So in your case since `\0` does not work, you can use `sed 's/[0-9]/&&/'`. Actuall using `&` will work even in gnu sed.
Jan 22, 2018 at 0:51	comment	added	Tinler		Yes. It becomes hew4r -> hew00r. I want hew44r. Also, why should I use that instead of "sed s/[0-9]/00/"? For this and the first example?
Jan 22, 2018 at 0:42	comment	added	Tinler		"sed s/[0-9]/\0\0/" does the exact same as "sed s/[0-9]/00/" or am I missing something?
Jan 22, 2018 at 0:25	answer	added	John Smith		timeline score: 2
Jan 22, 2018 at 0:20	history	edited	Tinler	CC BY-SA 3.0	deleted 1 character in body
Jan 22, 2018 at 0:20	comment	added	Tinler		My bad, I meant $sed s/[0-9]/00/ as you can see, it has to be zero. I'm looking for a way for the digit to be the one found in the [0-9].
Jan 22, 2018 at 0:13	comment	added	don_crissti		How about reading the manual... You'll learn how to reference the matched portion, regardless of what's in the LHS. ;)
Jan 22, 2018 at 0:08	history	asked	Tinler	CC BY-SA 3.0