perl regex negative-lookbehind detect file lacking final linefeed - regex

Question

The following code uses tail to test whether the last line of a file fails to culminate in a newline (linefeed, LF).
> printf 'aaa\nbbb\n' | test -n "$(tail -c1)" && echo pathological last line
> printf 'aaa\nbbb' | test -n "$(tail -c1)" && echo pathological last line
pathological last line
>
One can test for the same condition by using perl, a positive lookbehind regex, and unless, as follows. This is based on the notion that, if a file ends with newline, the character immediately preceding end-of-file will be \n by definition.
(Recall that the -n0 flag causes perl to "slurp" the entire file as a single record. Thus, there is only one $, the end of the file.)
> printf 'aaa\nbbb\n' | perl -n0 -e 'print "pathological last line\n" unless m/(?<=\n)$/;'
> printf 'aaa\nbbb' | perl -n0 -e 'print "pathological last line\n" unless m/(?<=\n)$/;'
pathological last line
>
Is there a way to accomplish this using if rather than unless, and negative lookbehind? The following fails, in that the regex seems to always match:
> printf 'aaa\nbbb\n' | perl -n0 -e 'print "pathological last line\n" if m/(?<!\n)$/;'
pathological last line
> printf 'aaa\nbbb' | perl -n0 -e 'print "pathological last line\n" if m/(?<!\n)$/;'
pathological last line
>
Why does my regex always match, even when the end-of-file is preceded by newline? I am trying to test for an end-of-file that is not preceded by newline.

Answer 1

/(?<=\n)$/ is a weird and expensive way of doing /\n$/.
/\n$/ means /\n(?=\n?\z)/, so it's a weird and expensive way of doing /\n\z/.
A few approaches:
perl -n0777e'print "pathological last line\n" if !/\n\z/'
perl -n0777e'print "pathological last line\n" if /(?<!\n)\z/'
perl -n0777e'print "pathological last line\n" if substr($_, -1) ne "\n"'
perl -ne'$ll=$_; END { print "pathological last line\n" if $ll !~ /\n\z/ }'
The last solution avoids slurping the entire file.
Why does my regex always match, even when the end-of-file is preceded by newline?
Because you mistakenly think that $ only matches at the end of the string. Use \z for that.

Answer 2

Do you have a strong reason for using a regular expression for his job? Practicing regular expressions for example? If not, I think a simpler approach is to just use a while loop that tests for eof and remembers the latest character read. Something like this might do the job.
perl -le'while (!eof()) { $previous = getc(\*ARGV) }
if ($previous ne "\n") { print "pathological last line!" }'
PS: ikegami's comment about my solution being slow is well-taken. (Thanks for the helpful edit, too!) So I wondered if there's a way to read the file backwards. As it turns out, CPAN has a module for just that. After installing it, I came up with this:
perl -le 'use File::ReadBackwards;
my $bw = File::ReadBackwards->new(shift #ARGV);
print "pathological last line" if substr($bw->readline, -1) ne "\n"'
That should work efficiently, even very large files. And when I come back to read it a year later, I will more likely understand it than I would with the regular-expression approach.

Answer 3

The hidden context of my request was a perl script to "clean" a text file used in the TeX/LaTeX environment. This is why I wanted to slurp.
(I mistakenly thought that "laser focus" on a problem, recommended by stackoverflow, meant editing out the context.)
Thanks to the responses, here is an improved draft of the script:
#!/usr/bin/perl
use strict; use warnings; use 5.18.2;
# Loop slurp:
$/ = undef; # input record separator: entire file is a single record.
# a "trivial line" looks blank, consists exclusively of whitespace, but is not necessarily a pure newline=linefeed=LF.
while (<>) {
s/^\s*$/\n/mg; # convert any trivial line to a pure LF. Unlike \z, $ works with /m multiline.
s/[\n][\n]+/\n\n/g; # exactly 2 blank lines (newlines) separate paragraphs. Like cat -s
s/^[\n]+//; # first line is visible or "nontrivial."
s/[\n]+\z/\n/; # last line is visible or "nontrivial."
print STDOUT;
print "\n" unless m/\n\z/; # IF detect pathological last line, i.e., not ending in LF, THEN append LF.
}
And here is how it works, when named zz.pl. First a messy file, then how it looks after zz.pl gets through with it:
bash: printf ' \n \r \naaa\n \t \n \n \nbb\n\n\n\n \t'
aaa
bb
bash:
bash:
bash: printf ' \n \r \naaa\n \t \n \n \nbb\n\n\n\n \t' | zz.pl
aaa
bb
bash: