shift @ARGV removes the first element from @ARGV and returns it. @ARGV contains all the arguments provided to your program.

I was asssuming you wanted to call your program like this:

my_program inputfile outputfile
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anyway,

my $whatever = <>;
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will read a single line from the file(s) as specified in @ARGV. in other words, $whatever will not be the filename but the next (or first) line of the file you passed as an argument. if you don't pass any arguments <> will read from STDIN, which means it will usually wait for you to enter a line. See perlop (quoted below:) Note that lines read usually end in a newline character ("\n") while filenames usually do not.

           while (<>) {
               ...                     # code for each line
           }

       is equivalent to the following Perl-like pseudo code:

           unshift(@ARGV, ’-’) unless @ARGV;
           while ($ARGV = shift) {
               open(ARGV, $ARGV);
               while (<ARGV>) {
                   ...         # code for each line
               }
           }
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$! will contain the last filesystem error, which will make it a lot easier to see what's going wrong if you pass the wrong filename to open(), or don't have permissions etc. See also perlvar.