New splicer function operates on external referent (although I'm not sure what this gets you: it's almost exactly like splice) (update: needs Perl 5.10+ for // operator):

c:\@Work\Perl\monks>perl -wMstrict -MData::Dump -le
"use 5.010;
 ;;
 sub my_splicer {
   my ($ar, $offset, $len, @list) = @_;
   $offset //= 0;
   $len    //= @$ar - $offset;
   return splice @$ar, $offset, $len, @list;
   }
 ;;
 use constant TEST => qw(one two three four five six seven eight);
 ;;
 my $ar;
 my @spl;
 my @args;
 ;;
 $ar = [ TEST ];
 @args = ();
 @spl = my_splicer($ar, @args);
 dd \@spl, $ar;
 ;;
 $ar = [ TEST ];
 @args = (2);
 @spl = my_splicer($ar, @args);
 dd \@spl, $ar;
 ;;
 $ar = [ TEST ];
 @args = (2, 3);
 @spl = my_splicer($ar, @args);
 dd \@spl, $ar;
 ;;
 $ar = [ TEST ];
 @args = (2, 3, qw(FOO BAR));
 @spl = my_splicer($ar, @args);
 dd \@spl, $ar;
"
(
  ["one", "two", "three", "four", "five", "six", "seven", "eight"],
  [],
)

(
  ["three", "four", "five", "six", "seven", "eight"],
  ["one", "two"],
)

(
  ["three", "four", "five"],
  ["one", "two", "six", "seven", "eight"],
)

(
  ["three", "four", "five"],
  ["one", "two", "FOO", "BAR", "six", "seven", "eight"],
)
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Update: Note this code will also work as expected with something like
   @args = ($ar, 2, 3, qw(FOO BAR));
   @spl = my_splicer(@args);

I haven't tested this, but I believe that in tybalt89's code (and in your OP code), because the splice is operating on a copy of the  $list array reference argument, so a fourth argument  LIST would be assigned to the lexical array  @copy and then almost immediately lost along with that array.

Do you mean that you want your function to operate on the external referent of the  $list array reference?