Appending 0 (zero) to a grep match reference

Question

tl;dr

I want to add a 0 at the end of the first backreference (\1) from a grep search. How can I differentiate this from the tenth backreference (\10).

---

Using grep in bbedit, I want to replace a part of a string with a part of that part of a string and a zero.

For example, find and replace this part of a string:

195.22.126.x

in this string:

195.22.126.x - - [13/Jul/2018:19:16:49 -0200] "GET /file.txt HTTP/1.1" 301 243

with part of that part of a string and a zero:

195.22.126.0

If I wanted to replace the "x" at the end of that part of a string with a "y", the grep code would look like this:

Find:
^(\d+\.\d+\.\d+\.)x

Replace:
\1y

But if I want to replace the "x" with a "0" (zero), the program tries to extract the tenth match, instead of appending a zero to the first match, because \10 looks like "tenth match" to it, not like "first match and a zero".

So how do I append a zero to a grep match?

---

Please note that I cannot simply replace "x" with "0", because there are other "x"s in the string. Therefore I have to match the correct part of the string.

Answer 1

EDIT Do this:

Find:
^(\d+\.\d+\.\d+\.)x

Replace:
\010

From the manual:

If more than two decimal digits follow the backslash, only the first two are considered part of the backreference. Thus, “\111” would be interpreted as the 11th backreference, followed by a literal “1”. You may use a leading zero; for example, if in your replacement pattern you want the first backreference followed by a literal “1”, you can use “\011”. (If you use “\11”, you will get the 11th backreference, even if it is empty.)

---

Well, if this is the actual string you want to work with, there’s a simple solution: just move the last dot out of the group.

Find:
^(\d+\.\d+\.\d+)\.x

Replace:
\1.0

If you want to replace something else where this workaround doesn’t work, another simple workaround would be like this:

First find:
^(\d+\.\d+\.\d+\.)x

First replace:
\1zzfoobarzz0

Second find:
zzfoobarzz

Second replace:
(empty)

Where zzfoobarzz is some such weird string that appears nowhere in the file. Ugly but workable.

Another idea would be to use a lookbehind assertion to match 195.22.126., leaving it out of the group. It would not be replaced, that way.