Trying to replace pattern _1_ with _01_ but its not working

Question

I am using Vim version 9.0.1280.

The search and replace command I am using is -

:%s/\(_\)\([1-9]\{1\}\)\(_\)/_0\1_/gc

When I run this vim offers to replace _1_ with _0\1_.

What am I doing wrong?

Answer 1

I would do:

:%s/\(_\)\([1-9]\{1\}\)\(_\)/_0\2_/gc

Or more compact:

:%s/_\zs\d\{1\}\ze_/0&/gc

I limit the part of the regular expression I want to replace with \zs and \ze and I can use the full pattern \0 replacement expression.

Or as @gildux mention:

:%s/\v_(\d{1})_/_0\1_/gc

Answer 2

When Vim says, replace with _0\1_ ..., it hasn't replaced the \1 you typed with the first group yet.

The first group in your expression is \(_\), so the resulting text after the substitution is _0__.

You want the matched number which is the second group. So to fix your original regexp, just use \2 instead of \1:

:%s/\(_\)\([1-9]\{1\}\)\(_\)/_0\2_/gc

However as gildux points out, there's no need to create groups for the underscores. You can just use a single group for the bit you actually want to re-use:

:%s/_\([1-9]\)_/_0\1_/gc
                          " The search pattern is:
    _                     " an underscore followed by:
     \(     \)            " group \1, containing
       [1-9]              " a non-zero digit
              _           " followed by another underscore

Note that I've also removed the \{1\} from your original query. [1-9] already matches one atom, so specifying you want to match one of them doesn't make much sense in this context! (Perhaps your example is slimmed down from your real use-case.)