Why is '$_' the same as $ARGV in a Perl one-liner?

Question

I ran into this problem while trying to print single quotes in a Perl one-liner. I eventually figured out you have to escape them with \'\\\'\'. Here\'s some code t

Answer 1

I try to avoid quotes in one liners for just this reason. I use generalized quoting when I can:

% perl -ne 'chomp; print qq($_\n)'

Although I can avoid even that with the -l switch to get the newline for free:

% perl -nle 'chomp; print $_'

If I don't understand a one-liner, I use -MO=Deparse to see what Perl thinks it is. The first two are what you expect:

% perl -MO=Deparse -ne 'chomp; print "$_\n"' shortlist.txt
LINE: while (defined($_ = )) {
    chomp $_;
    print "$_\n";
}
-e syntax OK

% perl -MO=Deparse -ne 'chomp; print "$ARGV\n"' shortlist.txt
LINE: while (defined($_ = )) {
    chomp $_;
    print "$ARGV\n";
}
-e syntax OK

You see something funny in the one where you saw the problem. The variable has disappeared before perl ever saw it and there's a constant string in its place:

% perl -MO=Deparse -ne 'chomp; print "'$_'\n"' shortlist.txt

LINE: while (defined($_ = )) {
    chomp $_;
    print "shortlist.txt\n";
}
-e syntax OK

Your fix is curious too because Deparse puts the variable name in braces to separate it from the old package specifier ':

% perl -MO=Deparse -ne 'chomp; print "'\''$_'\''\n"' shortlist.txt
LINE: while (defined($_ = )) {
    chomp $_;
    print "'${_}'\n";
}
-e syntax OK