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

To your shell, 'chomp; print "'$_'\n"' results in a string that's the concatenation of

1. chomp; print " (the first sequence inside single quotes),
2. the value of its variable $_, and
3. \n" (the second sequence inside single quotes).

In bash, $_ "... expands to the last argument to the previous command, after expansion. ...". Since this happens to be shortlist.txt, the following is passed to perl:

chomp; print "shortlist.txt\n"

For example,

$ echo foo
foo

$ echo 'chomp; print "'$_'\n"'
chomp; print "foo\n"

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Note that the above mechanism shouldn't be used to pass values to a Perl one-liner. You shouldn't be generating Perl code from the shell. See How can I process options using Perl in -n or -p mode? for how to provide arguments to a one-liner.