# Regular expression counter
**URL:** https://boards.straightdope.com/t/regular-expression-counter/526897
**Category:** Factual Questions
**Created:** [January 29, 2010, 3:19pm UTC](https://boards.straightdope.com/t/regular-expression-counter/526897 "2010-01-29T15:19:00Z")
**Posts on this page:** 6
**Page:** 1
### Author:  [@Rune](https://boards.straightdope.com/u/Rune)
#### Post date: [January 29, 2010, 3:19pm UTC](https://boards.straightdope.com/t/regular-expression-counter/526897/1 "2010-01-29T15:19:00Z")
I want to replace the string “A B C D E F” with “A1 B2 C3 D4 E5” in a single regular expression. Is it possible to have a counter in regular expressions?
---
### Author:  [@friedo](https://boards.straightdope.com/u/friedo)
#### Post date: [January 29, 2010, 3:56pm UTC](https://boards.straightdope.com/t/regular-expression-counter/526897/2 "2010-01-29T15:56:40Z")
Depends what flavor of regex you’re using. In Perl you can say:
```auto
my $s = "A B C D E F";
my $n = 0;
$s =~ s/(\w)/$1 .++$n/eg;
```
…because Perl allows you to execute arbitrary code in the replacement string with the /e flag. But there isn’t a way to do it with “standard” regex primitives that I can think of.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [January 29, 2010, 4:01pm UTC](https://boards.straightdope.com/t/regular-expression-counter/526897/3 "2010-01-29T16:01:50Z")
You can’t do it with plain vanilla regular expressions, so you’ll need something like what **friedo** posted.
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### Author:  [@Rune](https://boards.straightdope.com/u/Rune)
#### Post date: [January 29, 2010, 6:40pm UTC](https://boards.straightdope.com/t/regular-expression-counter/526897/4 "2010-01-29T18:40:50Z")
Ok. Thanks. It’s Javascript. I tried:
var i = 0;
“A B C D E F”.replace(/ /g, ++i + " "));
- but i doesn’t get incremented before after all the replacements have been made. I now made a version:
var res = “”;
var tmp = “A B C D E F”.split(/ /g);
for (var i = 0; i \< tmp.length; i++) res += tmp\* + i + " ";
- not so nice.
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### Author:  [@croc\_hunter](https://boards.straightdope.com/u/croc_hunter)
#### Post date: [May 28, 2012, 9:54pm UTC](https://boards.straightdope.com/t/regular-expression-counter/526897/5 "2012-05-28T21:54:49Z")
Does anyone know why this reg exp in perl doenst work;
If there are two carriage returns in my string, this doens’t work.
$message =~ s/
/\/g;
But if i use just one
in the replace it does work…
$message =~ s/
/\/g;
any idea how i can serach for double
's in string.?
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### Author:  [@tellyworth](https://boards.straightdope.com/u/tellyworth)
#### Post date: [May 28, 2012, 10:23pm UTC](https://boards.straightdope.com/t/regular-expression-counter/526897/6 "2012-05-28T22:23:21Z")
Your question is unrelated to the first in the thread.
Nevertheless: your string probably has DOS-style CRLF line ends. Try
instead of
.