Back to this again . .
..., 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
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
On Monday 15 July 2002 11:22 pm, Deborah Ariel Pickett wrote:
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
Brent Dax wrote:
$href = hash { %hash }; #B
Why the curlies? if Chash is a function (ctor), then surely these should
be parentheses. In this context, parentheses are optional, so this could be
written
$href = hash %hash;
Dave.
David Whipp:
# Brent Dax wrote:
# $href = hash { %hash }; #B
#
# Why the curlies? if Chash is a function (ctor), then surely
# these should be parentheses. In this context, parentheses are
# optional, so this could be written
#
#$href = hash %hash;
Chash is not a function. It's a
On Tue, 16 Jul 2002, Deborah Ariel Pickett wrote:
I still have my vote on %() as a hash constructor in addition to {}. :)
The problem I see with that is that % as a prefix implies a
*dereferencing*, though years of Perl5 conditioning like this:
%{ $mumble } = return_a_hash();
Using %(...) to create a hashref, as { ... } does in Perl5, would go
against all that, because the purpose of making a hashref is to
*reference* something. Now a unary % operator/sigil/prefix might mean
referencing, or it might mean dereferencing, depending on whether the
symbols
On Tuesday 16 July 2002 01:01 am, Deborah Ariel Pickett wrote:
If %(...) makes a shallow copy of its innards, as Perl5's { ... } does,
then how do you impose hash context onto something without doing the
copy?
%{} forces hash context. What else could it do?
%{ foo() } calls foo() in hash
Sorry, I was being too terse in my original message, I guess some of the
meaning got lost.
When I said:
If %(...) makes a shallow copy of its innards, as Perl5's { ... } does,
then how do you impose hash context onto something without doing the
copy?
What I meant to say was:
Speaking
