ARGV how to use it?

So i am trying to read in file
readFile <GivenFile> modFile
looking for a regular file under the directories in the GivenFile and print them out is my over all goal.
basically I am looking for anything that looks like a directory in the given file and printing it out.
Since I am trying to do this in baby steps because I have never worked with perl.
in my file I have something like

!#/usr/bin/perl -w
#for starters I just want to see something I dont care what it is
while(< >){
print;
print "yass something kinda happen \n";
}

as of now you can see I dont have anything remotely checking to see if this is regular file or not.
Ive been reading all over the net and my book notes, feeling a little bit of heat here because ive been running into little hick ups here and there.
THanks for reading.

$ARGV is passed in from the command line. Not by stdin or a file.

Program:

perl -e 'print $ARGV[0], "\n"; '   test

test is a string on the command line.

Hi.

I usually look at perldoc perlvar for questions like this. In this case:

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

    @ARGV   The array @ARGV contains the command-line arguments intended for
            the script. $#ARGV is generally the number of arguments minus one,
            because $ARGV[0] is the first argument, not the program's command
            name itself. See "$0" for the command name.

    ARGV    The special filehandle that iterates over command-line filenames
            in @ARGV. Usually written as the null filehandle in the angle
            operator "<>". Note that currently "ARGV" only has its magical
            effect within the "<>" operator; elsewhere it is just a plain
            filehandle corresponding to the last file opened by "<>". In
            particular, passing "\*ARGV" as a parameter to a function that
            expects a filehandle may not cause your function to automatically
            read the contents of all the files in @ARGV.

Best wishes ... cheers, drl