Deborah Ariel Pickett:
# > > ..., and someone pointed out that it had a problem
# > > with code like "{ some_function_returning_a_hash()
# }". Should it give a
# > > closure? Or a hash ref? ...
# > Oh, well now that it's stated this way... (something went
# wrong in my
# > brain when I read the actual message) It returns a closure
# :(. A4 says
# > that as a term, { } define a closure, unless it contains a pair
# > constructor at the top level. But, thanks to Perl 6's
# smartness, that would
# > be excessive syntax anyway:
# > $hashref = some_function_returning_a_hash()
# > would do what you wanted.
#
# Would it always? What if I had two functions (or more), all returning
# part of the hash I want to package up? Can I do:
# $hashref = some_function_returning_a_hash(),
# some_other_function_returning_more_of_the_hash();
# and get the result of both functions into the anonymous hash?
Not directly--just use the hash {} constructor.
# Besides, does
# $hashref = some_function_returning_a_hash()
# make $hashref simply refer to the result of the function, or does it
# make $hashref refer to a hash containing a *copy* of the result of the
# function? If Perl6 is going to do fancy things with
# functions returning
# lvalues, which looks like the case, those two things aren't
# necessarily
# the same thing.
#
# Or, saying the same thing another way, does this:
# $href = %hash;
# which I presume will be legal Perl6, mean the same as this Perl5:
# $href = \%hash; #A
# or this Perl5:
# $href = { %hash }; #B
# and how would I say each of A and B in Perl6 unambiguously?
A. And unambiguously:
$href = \%hash; #A
$href = hash { %hash }; #B
# Automatic referencing and dereferencing is all well and good, and it
# appears that it's here to stay in Perl6 (it's been in most
# Apocalypses),
# but I don't think anyone's actually sat down yet to thrash out exactly
# under what circumstances it happens.
Autodereferencing happens whenever we have a scalar but we need an array
or hash; autoreferencing happens whenever we have an array or hash but
need a scalar (usually because of scalar assignment, but not
necessarily).
--Brent Dax <[EMAIL PROTECTED]>
@roles=map {"Parrot $_"} qw(embedding regexen Configure)
He who fights and runs away wasted valuable running time with the
fighting.
