The second regex doesn't work because you're not capturing anything to
$1 or
$2. To capture something to
$1, etc, use parens.
Further there's no need to escape the space in the replacement.
But, ignoring the issue of "non-English" (do you mean a different character set, like Greek or Japanese by any chance?), you're close. Try this:
#!/usr/bin/perl
use strict;
use warnings;
my $in = "foo bar blivitz done";
$in =~ s/\s+/\n/g; # replace spaces with \n
print "after first regex: $in \n";
# $in =~ s/[a-zA-Z]\n[a-zA-Z]/$1\ $2/g;
$in =~ s/([a-zA-Z])\n([a-zA-Z])/$1 $2/g; # remove newlines, restore s
+paces
print "after second regex: $in \n";
=head execution:
ww@GIG:~/pl_test$ perl 675941.pl
after first regex: foo
bar
blivitz
done
after second regex: foo bar blivitz done
ww@GIG:~/pl_test$
=cut
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