Perl ARGV value in scalar context

Question

Given the following Perl script:

# USAGE: ./flurp -x -vf file1 file2 file3 file4

# e.
$a = shift;
$b = shift;
$c = shift;
@d = <>;

# ei.  value of $b = -vf
# eii. value of @d = content of file2, file3, file4

print "$b\n";
print "@d\n";
print "$ARGV\n";

This is the output:

-vf
{contents of file2, file3, file4}
file4

I am puzzled by the output of print "$ARGV\n". If I try to do print "$ARGV[-1]\n", an empty line is printed out to STDOUT. If I directly reference $ARG[2], I get an empty line as well.

Why is the script printing file4 when $ARGV is used?

As a counter-example, I tried print "$d\n", expecting to get the last line of file4. Instead of the expected output I got an empty line. How does $ARGV work?

I thought that `@LIST` and `$LIST` referred to the same variable since `$LIST[0]` is the first index of `@LIST`. Perl is a confusing beast. — Alan W, May 02 '17 at 23:18
`$ARGV[n]` is accessing `@ARGV`. `$ARGV` is a separate variable. — Matt Jacob, May 02 '17 at 23:19

martin clayton · Accepted Answer · 2017-05-02T22:11:21.737

4

In answer to your specific question: "How does $ARGV work?"

$ARGV Contains the name of the current file when reading from <> .

from Variables related to filehandles in the Perl docs.

Although at the point you print $ARGV you've finished reading from file4, the variable still holds the name of the file.

edited May 02 '17 at 22:11

answered May 02 '17 at 22:05

martin clayton

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Perl ARGV value in scalar context

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