Question

How to exactly repeat the n matched pattern in result string?

Example if I have the folowing text:

++ '[' -f /etc/bashrc ']'
++ . /etc/bashrc
+++ '[' '[\u@\h \W]\$ ' ']'
+++ '[' -z 'printf "\033]0;%s@%s:%s\007" "${USER}" "${HOSTNAME%%.*}" "${PWD/#$HOME/~}"' ']'
+++ shopt -s checkwinsize
+++ '[' '[\u@\h \W]\$ ' = '\s-\v\$ ' ']'
+++ shopt -q login_shell
+++ '[' 506 -gt 199 ']'
++++ id -gn

Now I want to substitute every ‘+’ for 3 spaces, but it can only happen at the begining of the pattern. I would use :<range>s/^<pattern> :%s/+/ /g, but if it there were a ‘+’ in the rest of the text I would simply mess it up.

The question:
How to match every + at begining and repeat the same count of found + in the result string?
expected:

^   ++$  -> ^         $
^   +++$ -> ^            $
^   +$   -> ^      $

Thanks

Answer 1

Try this:

:%s/^+*/\=repeat('   ',strlen(submatch(0)))/

submatch(0) contains all the matched + at the start of the line, strlen counts them. So for every plus sign at the start of the line three spaces are inserted using repeat.

For more information:

:help sub-replace-expression
:help repeat()
:help submatch()
:help strlen()