Whitespace is contained in [^\w]. Maybe you want to use \S instead of [^\w]? Or, if you just want to remove all spaces, why don't you use s/\s//g?

Thanks for answer. This regexp does the thing: s/(\W)\s+/$1/g

That's probably not doing what you want, at least according the your original problem specification which called for removing spaces between, rather than after, characters. Consider:

use strict;
use warnings;

my $str = "The quick & the dead.";

$str =~ s/(\W)\s+/$1/g;
print $str;
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Prints:

The quick &the dead.
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True laziness is hard work

... why output is "&& &" but not "&&&"?

After the first match on  '& &' (which is replaced with '&&'), the 'match point' in the string is sitting just after the second '&' and just before the second space. The regex pattern needs at least three characters to match, and at that point there remain only two characters in the string.

In general, the regex of the OP will only match successive triplets of  '& &' leaving a single  \s between each substituted  '&&' pair because after each match the regex engine must scan forward to find, at the very least, another  \W\s\W triplet. (Heed Corion's caution that  \W includes all  \s characters!)

Like this?

$s = 'a b & &   *   * * * a * * * *& && & xy a';;

$s =~ s[(?<=\W)\s+(?=\W)][]g;;

 print $s;;
a b &&**** a ****&&&& xy a
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