Regex to change the number of spaces in an indent level - regex

Question

Let's say you have some lines that look like this
1 int some_function() {
2 int x = 3; // Some silly comment
And so on. The indentation is done with spaces, and each indent is two spaces.
You want to change each indent to be three spaces. The simple regex
s/ {2}/ /g
Doesn't work for you, because that changes some non-indent spaces; in this case it changes the two spaces before // Some silly comment into three spaces, which is not desired. (This gets far worse if there are tables or comments aligned at the back end of the line.)
You can't simply use
/^( {2})+/
Because what would you replace it with? I don't know of an easy way to find out how many times a + was matched in a regex, so we have no idea how many altered indents to insert.
You could always go line-by-line and cut off the indents, measure them, build a new indent string, and tack it onto the line, but it would be oh so much simpler if there was a regex.
Is there a regular expression to replace indent levels as described above?

Answer 1

In some regex flavors, you can use a lookbehind:
s/(?<=^ *) / /g
In all other flavors, you can reverse the string, use a lookahead (which all flavors support) and reverse again:
s/ (?= *$)/ /g

Answer 2

Here's another one, instead utilizing \G which has NET, PCRE (C, PHP, R…), Java, Perl and Ruby support:
s/(^|\G) {2}/ /g
\G [...] can match at one of two positions:
✽ The beginning of the string,
✽ The position that immediately follows the end of the previous match.
Source: http://www.rexegg.com/regex-anchors.html#G
We utilize its ability to match at the position that immediately follows the end of the previous match, which in this case will be at the start of a line, followed by 2 whitespaces (OR a previous match following the aforementioned rule).
See example: https://regex101.com/r/qY6dS0/1

Answer 3

I needed to halve the amount of spaces on indentation. That is, if indentation was 4 spaces, I needed to change it to 2 spaces.
I couldn't come up with a regex. But, thankfully, someone else did:
//search for
^( +)\1
//replace with (or \1, in some programs, like geany)
$1
From source: "^( +)\1 means "any nonzero-length sequence of spaces at the start of the line, followed by the same sequence of spaces. The \1 in the pattern, and the $1 in the replacement, are both back-references to the initial sequence of spaces. Result: indentation halved."

Answer 4

You can try this:
^(\s{2})|((?<=\n(\s)+))(\s{2})
Breakdown:
^(\s{2}) = Searches for two spaces at the beginning of the line
((?<=\n(\s)+))(\s{2}) = Searches for two spaces
but only if a new line followed by any number of spaces is in front of it.
(This prevents two spaces within the line being replaced)
I'm not completely familiar with perl, but I would try this to see if it work:
s/^(\s{2})|((?<=\n(\s)+))(\s{2})/\s\s\s/g
As #Jan pointed out, there can be other non-space whitespace characters. If that is an issue, try this:
s/^( {2})|((?<=\n( )+))( {2})/ /g