From @cognominal

15​:48​:09] <cognominal> rakudo​: say eval(("a","b").Seq.perl ).WHAT
[15​:48​:13] <p6eval> rakudo 3370f0​: OUTPUT«Parcel()␤»
[15​:51​:13] <moritz_> cognominal​: it's a bug, but a small one. It's
rarely necessary to reproduce the exact list type, as they can all be
coerced very easily (and often automatically)
[15​:51​:26] <moritz_> rakudo​: my $x = (1, 2, 3).Seq; say $x.WHAT
[15​:51​:31] <p6eval> rakudo 3370f0​: OUTPUT«Seq()␤»
15​:53​:58] <cognominal> What is the composer for a Seq?
[15​:54​:08] <moritz_> @​list.Seq
[15​:54​:12] <cognominal> ...if any

Probably a ".Seq" should be appended to the current result returned
by for the Seq .perl method to coerce it to a Seq as it should be.

--
cognominal stef

From @cognominal

I note that eqv considers rightly (1,2).seq non equivalent to (1,2)

  $ perl6
  > (1,2).Seq eqv (1,2)
  0

On Tue, Jul 20, 2010 at 4​:03 PM, perl6 via RT
<perl6-bugs-followup@​perl.org> wrote​:

Greetings,

This message has been automatically generated in response to the
creation of a trouble ticket regarding​:
       "the .perl method for Seq returns a string that eval to a Parcel",
a summary of which appears below.

There is no need to reply to this message right now.  Your ticket has been
assigned an ID of [perl #​76596].

Please include the string​:

        [perl #​76596]

in the subject line of all future correspondence about this issue. To do so,
you may reply to this message.

                       Thank you,
                       perl6-bugs-followup@​perl.org

-------------------------------------------------------------------------
15​:48​:09]  <cognominal> rakudo​:  say eval(("a","b").Seq.perl ).WHAT
[15​:48​:13]  <p6eval> rakudo 3370f0​: OUTPUT«Parcel()␤»
[15​:51​:13]  <moritz_> cognominal​: it's a bug, but a small one. It's
rarely necessary to reproduce the exact list type, as they can all be
coerced very easily (and often automatically)
[15​:51​:26]  <moritz_> rakudo​: my $x = (1, 2, 3).Seq; say $x.WHAT
[15​:51​:31]  <p6eval> rakudo 3370f0​: OUTPUT«Seq()␤»
15​:53​:58]  <cognominal> What is the composer for a Seq?
[15​:54​:08]  <moritz_> @​list.Seq
[15​:54​:12]  <cognominal> ...if any

Probably a ".Seq"  should be appended to the current result returned
by for  the  Seq .perl method  to coerce it to a Seq as it should be.

--
cognominal stef

--
cognominal stef

@coke - Status changed from 'new' to 'open'

From @coke

10​:43 < pmichaud> I call obsolete. The meaning of Seq and .perl have
changed
  since that ticket was filed.

Rejecting ticket; If you can come up with a similar case against modern-
day specs, please open a new ticket.

Thanks!

--
Will "Coke" Coleda

@coke - Status changed from 'open' to 'rejected'