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:
> > Spea
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 c
> > 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
> > symb
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();
> p
> 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();
print_hash( %{ $mumble } );
(Yes, the braces are optional; I'm
David Whipp:
# Brent Dax wrote:
# > $href = hash { %hash }; #B
#
# Why the curlies? if C is a function (ctor), then surely
# these should be parentheses. In this context, parentheses are
# optional, so this could be written
#
#$href = hash %hash;
C is not a function. It's a keyword
Brent Dax wrote:
> $href = hash { %hash }; #B
Why the curlies? if C 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.
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
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 wh
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
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