my @mac_addrs = ("0015FAA3F03A", "0015FAA3F03B", "0015FAA3F03C");

sub convert {
    return map { "AP" . join ".", unpack "(A4)*", lc $_ } @_;
}

print "$_\n" for convert(@mac_addrs);
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•  @answer = map { do_something_with("$_") } @array transforms one @array into another array @answer where each element of @answer is the result of the transform do_something_with(). Try this:

  $ perl -le '@a = (2, 3, 4, 5); @b = map { 2 * $_ } @a; print for @b'
  4
  6
  8
  10
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  Notice that @b has each of the elements of @a, doubled. The result of the block given to map can be an array, too, so:

  $ perl -le '@a = (2, 3, 4, 5); @b = map { $_ % 2 ? () : ($_, $_ / 2) }
  + @a; print for @b'
  2
  1
  4
  2
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  Notice that it printed the number and its half -- for the even numbers in @a.
• So, let's try the knowledge of map:

  $ perl -le '@mac_addrs = ("0015FAA3F03A", "0015FAA3F03B", "0015FAA3F03
  +C"); @a = map { lc $_ } @mac_addrs; print for @a'
  0015faa3f03a
  0015faa3f03b
  0015faa3f03c
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• Now, let's add unpack"(A4)*" to the equation (unpack any number of groups of four alphanumeric chars):

  $ perl -le '@mac_addrs = ("0015FAA3F03A", "0015FAA3F03B", "0015FAA3F03
  +C"); @a = map { unpack "(A4)*", lc $_ } @mac_addrs; print for @a'
  0015
  faa3
  f03a
  0015
  faa3
  f03b
  0015
  faa3
  f03c
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• Now, let's join the parts with ".":

  $ perl -le '@mac_addrs = ("0015FAA3F03A", "0015FAA3F03B", "0015FAA3F03
  +C"); @a = map { join ".", unpack "(A4)*", lc $_ } @mac_addrs; print f
  +or @a'
  0015.faa3.f03a
  0015.faa3.f03b
  0015.faa3.f03c
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• And then, add the "AP" in the beginning:

  @mac_addrs = ("0015FAA3F03A", "0015FAA3F03B", "0015FAA3F03C"); @a = ma
  +p { "AP" . join ".", unpack "(A4)*", lc $_ } @mac_addrs; print for @a
  +'
  AP0015.faa3.f03a
  AP0015.faa3.f03b
  AP0015.faa3.f03c
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• voilą!! you already have your answer. Now, package everything in a neat subroutine:

  $ perl -le '
  > use strict;
  > my @mac_addrs = ("0015FAA3F03A", "0015FAA3F03B", "0015FAA3F03C");
  > sub convert {
  >   map { "AP" . join ".", unpack "(A4)*", lc $_ } @_
  > }
  > print for convert @mac_addrs
  > '
  AP0015.faa3.f03a
  AP0015.faa3.f03b
  AP0015.faa3.f03c
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The first argument of map is code executed for every other argument of map.

print map uc($_),    qw( a b c d );  # Prints ABCD
print map { uc($_) } qw( a b c d );  # Same thing
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unpack '(A4)*', $val splits the value into as many 4 byte blocks as possible. I would have used "W" instead of "A", though. unpack '(W4)*', $val splits the value into as many 4 character blocks as possible.

@_ is not unamed —its name is @_— and nothing is being stored in it —its contents are being passed to map.

@_ holds the arguments of the current function (convert).

Could you point me at the documentation for the 'W' template char, cos I'm darned if I can find it?

---

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

"Too many [] have been sedated by an oppressive environment of political correctness and risk aversion."

I would have used "W" instead of "A"

The reason I used "A" is because it works in older Perls, too, like 5.8.8.  "W" is available in 5.10 only.

"AP" . join "." - I understand what this is doing, but I don't understand from any documentation what the first argument to map is, generally speaking (everything before the first comma)

For example

my @in  = 1 .. 5;
my @out = map $_ * 2, @in;    # doubles each element of @in
   @out = map $_ * 2, 1 .. 5; # same thing - any list will do
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