Is it possible to back-reference the result of an outer :g command in a recursive call (or an inner :s call?)
For instance, let's say I want to search for something above the cursor, and modify it when it appears below the cursor:
:1,.g/Hello, \(\w\+\)!/.,$s/\1/\1-San
This should replace:
Hello, Luciano!
Goodbye, Luciano!
with
Hello, Luciano!
Goodbye, Luciano-San!
If you place the cursor on the empty line.
However, it complains about an invalid back-reference. This is in reference to the last \1 (note that , maybe surprisingly the first . The reason is likely that the \1 in the search pattern -but not the replacement - works!):s expression does not have any capture groups.
So my question is, whether there is a way to 'escape' the inner context of the command to execute in :g to be able to back-reference a match group from the outer g?
(I appreciate there are other ways to achieve the desired effect in the above contrived example - but my question relates specifically to escaping back references, not writing a better regex - the example here is just for illustration)
Back references can be used within a search pattern:
before: xoxoxo foobarfoo xoxoxo
/\(foo\)bar\1
after: xoxoxo foobarfoo xoxoxo
^^^^^^^^^
as well as in the replacement part of a substitution:
before: xoxoxo foobarfoo xoxoxo
:s/\(foo\)bar\1/baz\1baz
after: xoxoxo bazfoobaz xoxoxo
This is possible because the capture group and the back reference occupy the same "pattern space" in a single atomic "command".
When you do :g/<pattern>/<cmd>, <cmd> is a separate "command" that doesn't share :g's "pattern space" so it is impossible to reference a capture group from <pattern> in <cmd>.
What is possible, though, is to reuse <pattern> in <cmd>:
:g/\(foo\)bar\1/s//baz\1baz
but you are still not referencing a capture group from :g in :s even if it looks like it. You are making a new search with the same pattern and referencing the capture group in that new search.