\s matches newline in regex?

Only1KW has asked for the wisdom of the Perl Monks concerning the following question:

I'm trying to write a regex replace to be run on a line containing a newline and I want the newline character to not be replaced. It seems like \s is matching the newline character, and I have no idea why or how to prevent that. I'm not using '$' or '.' in my regex, so I am not using a 'm' or 's' modifier. Any ideas?

Example script:

#!/usr/bin/perl

use strict;
use warnings;

my $line = "Hi\n";
print length($line) . "\n";
$line =~ s/Hi\s*//;
print length($line) . "\n";
[download]

Output I get:

3
0
[download]

I'm running v5.18.2 if it matters.

Comment on \s matches newline in regex? Select or Download Code

Replies are listed 'Best First'.
Re: \s matches newline in regex? ([^\S\n]) by tye (Sage) on Jul 22, 2015 at 16:56 UTC
`s/Hi[^\S\n]*//` [download] - tye	[reply] [d/l]
Re: \s matches newline in regex? by toolic (Bishop) on Jul 22, 2015 at 16:32 UTC
It seems like \s is matching the newline character, Yes. Tip #9 from the Basic debugging checklist: YAPE::Regex::Explain The regular expression: (?-imsx:Hi\s) matches as follows: NODE EXPLANATION ---------------------------------------------------------------------- (?-imsx: group, but do not capture (case-sensitive) (with ^ and $ matching normally) (with . not matching \n) (matching whitespace and # normally): ---------------------------------------------------------------------- Hi 'Hi' ---------------------------------------------------------------------- \s whitespace (\n, \r, \t, \f, and " ") (0 or more times (matching the most amount possible)) ---------------------------------------------------------------------- ) end of grouping ---------------------------------------------------------------------- [download] This works for your simple input: `$line =~ s/Hi[^\n]*//;` [download]	[reply] [d/l] [select]
Re^2: \s matches newline in regex? by Only1KW (Sexton) on Jul 22, 2015 at 16:49 UTC
So if \s matches more than a space, how do I match just a space? While I agree your suggestion works for my simple example, it won't work for my actual workload.	[reply]
Re^3: \s matches newline in regex? by Corion (Patriarch) on Jul 22, 2015 at 17:04 UTC
You can use `[ ]`, or `\x{20}` or `\U{0020}` I guess.	[reply] [d/l] [select]
Re^3: \s matches newline in regex? by GrandFather (Saint) on Jul 22, 2015 at 20:58 UTC
If you want just a space use just a space! If you want to match everything except some specific white space characters use a negated character class with \S and the other white space characters you want to keep: `s/Hi[^\S\n\r\f]+//g` for example. Premature optimization is the root of all job security	[reply] [d/l]
Re: \s matches newline in regex? by stevieb (Canon) on Jul 22, 2015 at 16:41 UTC
You can also use a positive lookahead in the regex to avoid replacing the newline char. `$line =~ s/Hi\s*(?=\n)//;` [download]	[reply] [d/l]
Re^2: \s matches newline in regex? by AnomalousMonk (Archbishop) on Jul 22, 2015 at 23:31 UTC
If there were repeated, contiguous newlines, that would only avoid replacing the final newline. `c:\@Work\Perl\monks>perl -wMstrict -le "my $line = qq{HeHi\n\n\nHo}; print qq{[$line]}; ;; $line =~ s/Hi\s*(?=\n)//; print qq{[$line]}; " [HeHi Ho] [He Ho]` [download] Give a man a fish: `<%-(-(-(-<`	[reply] [d/l] [select]
Re^3: \s matches newline in regex? by stevieb (Canon) on Jul 22, 2015 at 23:38 UTC
Yes, I'm aware of that, but it would have been prudent for me to have pointed that out. Thanks AnomalousMonk. -stevieb	[reply]
Re: \s matches newline in regex? by kroach (Pilgrim) on Jul 22, 2015 at 17:41 UTC
You can use non-greedy matching and $ to match whitespace up until a newline. `$line =~ s/Hi\s*?$//m;` [download]	[reply] [d/l]