在 Sep 24, 2006 12:21 AM 時，Audrey Tang 寫到：
在 Sep 23, 2006 8:36 PM 時，Markus Laire 寫到：
On 9/23/06, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
@args = [EMAIL PROTECTED],1,2,3;
- push [,] @args; # same as push @foo,1,2,3
+ push [,] @args; # same as push(@foo: 1,2,3)
I don't quite understand this. Shouldn't C<[,] @args> be
equivalent to
C<[EMAIL PROTECTED],1,2,3> just as C<[+] 0,1,2,3> is equivalent to C<0+1+2+3>?
So why is there C<:> instead of C<,> after C<@foo>?
Does this have something to do with the fact that C<@args> is
C<[EMAIL PROTECTED],1,2,3> and not C<@foo,1,2,3>?
Exactly. Per this interpretation, [EMAIL PROTECTED] is shorthand for \(@foo :),
and
[,] would first flatten the contents of @arg, and then process each
one;
if an element is Capture, it is joined into the current arglist; if
it's not,
then it's made a simple positional.
I wasn't sure about this treatment, so I checked on #perl6 with Larry;
an alternative is to treat the elements of @foo always as positional
arguments, but that will make the two [,] calls below nonequivalent:
my @args = [EMAIL PROTECTED], 1, 2, 3;
[,] @args;
[,] [EMAIL PROTECTED], 1, 2, 3;
I'd prefer to make them equivalent, on the principle that all listops
conceptually flatten+concat their arguments first, and then process
each element regardless of its origin.
On the other hand, it can also be argued that
[,] TERM, TERM, TERM;
is shorthand for
|(TERM), |(TERM), |(TERM)
and the treatment will depend solely on TERM's scalar-context type
without considering list flattening. Per that interpretation, indeed
these
two need not be equivalent:
my @args = [EMAIL PROTECTED], 1, 2, 3;
[,] @args; # four positionals, same as |@args
[,] [EMAIL PROTECTED], 1, 2, 3; # one invocant, three positionals, same as |
([EMAIL PROTECTED]), |(1), |(2), |(3)
however, having a listop that does not do flatten+concat doesn't feel
quite right, so I'd still like the flatten-concat-explode model as
currently specced.
Thanks,
Audrey
