What does \0 mean in sed?

Question

I have the following code:

echo "12. Chapter Name" | sed -n -E "s/([0-9]{2})\.[[:space:]].*/\1/p"

It prints 12 as expected, since \1 refers to the first capturing group.

However, if \0 is used instead of \1, the output is 12. Chapter Name, the entire input string is printed.

It seems that as long as the regex found a match, \0 prints the entire input string. Is this correct?

I'm running Debian 10.2.

I think you have `sed -n -E "s/([0-9]{2})\.[[:space:]].*/\1/p"` — Wiktor Stribiżew, May 31 '20 at 22:06
`\0` or (`$0` for the tools that use `$n` to identify the capturing groups) is the string that matches the entire `regex`. Not only in `sed`. — axiac, May 31 '20 at 22:09
@Wiktor Stribiżew Indeed I mistyped. Thanks for pointing that out. — Peppershaker, May 31 '20 at 22:42

Wiktor Stribiżew · Accepted Answer · 2020-05-31T22:36:58.867

The \0 is a placeholder for the whole match value.

Note that capturing group indices start with 1 and go up until 9 in POSIX regex patterns. The zeroth group is the whole match.

Also, instead of \0, it is more common - and probably more portable - to use & in sed.

echo "12. Chapter Name" | sed -n -E "s/([0-9]{2})\.[[:space:]].*/\0/p"

is equal to

echo "12. Chapter Name" | sed -n -E "s/([0-9]{2})\.[[:space:]].*/&/p"

1 Answers1