Converting pack to perl6

Question

I would like to convert the following from perl5 to perl6,

$salt = pack "C*", map {int rand 256} 1..16;

It create a string of 16 characters where each character has a randomly picked value from 0 to 255. Perl5 doesn't assign any semantics to those characters, so they could be bytes, Unicode Code Points, or something else.

I think I can get by with

$salt = (map {(^256).pick.chr},^16).join;

But I got stuck on using pack, here is my attempt,

use experimental :pack;

my $salt = pack("C*",(map {(^256).pick} , ^16)).decode('utf-8');

say $salt;
say $salt.WHAT;

and results can be either an error,

Malformed termination of UTF-8 string  
      in block <unit> at test.p6 line 3

or something like,

  j
  (Str)

My line of thought is that packing the integer List would return a Buf then decoding that should produce the required Str.

Update:

As suggested on comment and answer Buf is the correct object to use. Now to follow up on the pack part,

perl6 -e 'use experimental :pack; my $salt = pack("C*",(map {(^256).pick} , ^16));say $salt;say $salt.WHAT;'

Buf:0x<7D> (Buf)

that only packed one unit

On the other hand, using P5pack (suggested by Scimon) returns an error

perl6 -e 'use P5pack; my $salt = pack("C*",(map {(^256).pick} , ^16));say $salt;say $salt.WHAT;'
Cannot convert string to number: base-10 number must begin with valid digits or '.' in '⏏*' (indicated by ⏏)
  in sub one at /home/david/.rakudobrew/moar-master/install/share/perl6/site/sources/D326BD5B05A67DBE51C279B9B9D9B448C6CDC401 (P5pack) line 166
  in sub pack at /home/david/.rakudobrew/moar-master/install/share/perl6/site/sources/D326BD5B05A67DBE51C279B9B9D9B448C6CDC401 (P5pack) line 210
  in block <unit> at -e line 1

Update 2:

I didn't spot the difference.

perl6 -e 'say (map {(^256).pick}, ^16).WHAT;'
(Seq)
perl6 -e 'say Buf.new((^256).roll(16)).WHAT;'  
(Buf)

Now make them lists,

perl6 -e 'use experimental :pack; my $salt = pack("C*",(Buf.new((^256).roll(16)).list));say $salt;say $salt.WHAT;'
    Buf:0x<39>
    (Buf)

And

perl6 -e 'use P5pack; my $salt = pack("C*",Buf.new((^256).roll(16)).list);say $salt;say $salt.WHAT;'
    Cannot convert string to number: base-10 number must begin with valid digits or '.' in '⏏*' (indicated by ⏏)
      in sub one at /home/david/.rakudobrew/moar-master/install/share/perl6/site/sources/D326BD5B05A67DBE51C279B9B9D9B448C6CDC401 (P5pack) line 166
      in sub pack at /home/david/.rakudobrew/moar-master/install/share/perl6/site/sources/D326BD5B05A67DBE51C279B9B9D9B448C6CDC401 (P5pack) line 210
      in block <unit> at -e line 1

Reference:

Buffers and Binary IO

A first approach to pack/unpack in Perl 6

Thanks in advance for the help.

Answer 1

As ikegami says in a comment to your question, you really should use a Buf, which is basically a “string” of bytes.

my $salt = Buf.new((^256).roll(16));

You can write this to a file with something like:

spurt 'foo', $salt, :bin;

or encode it in base-64 with:

use MIME::Base64;
my $encoded = MIME::Base64.encode($salt);

But if you need this to be reasonably secure, have a look at Crypt::Random

use Crypt::Random;
my $salt = crypt_random_buf(16);

Answer 2

The easiest way is to probably to use https://modules.perl6.org/dist/P5pack:cpan:ELIZABETH which allows you to use Perl5 pack syntax.