Ashley Winters:
# > You've got a point. There's an easy way to say "I want a sub":
# >
# > my $sub = -> { ... }
# >
# > But I can't think of a similarly punctuation-intensive way
# to say "I
# > want a hash." (someone please step in and correct me).
#
# I nominate:
#
# $() == scalar()
# %()
On Monday 15 July 2002 07:52 am, Brent Dax wrote:
> Ashley Winters:
> # > You've got a point. There's an easy way to say "I want a sub":
> # >
> # > my $sub = -> { ... }
> # >
> # > But I can't think of a similarly punctuation-intensive way
> # to say "I
> # > want a hash." (someone please step
> ..., 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 clos
In Apocalypse 2 Larry Wall wrote:
> RFC 082: Arrays: Apply operators element-wise in a list context
>
> APL, here we come... :-)
>
> This is by far the most difficult of these RFCs to decide, so I'm going
> to be doing a lot of thinking out loud here. This is research--or at
> least, a search.
On Mon, 15 Jul 2002, Karl Glazebrook wrote:
> In Apocalypse 2 Larry Wall wrote:
>
> > RFC 082: Arrays: Apply operators element-wise in a list context
> >
> > APL, here we come... :-)
> >
> > This is by far the most difficult of these RFCs to decide, so I'm going
> > to be doing a lot of thinkin
>Karl Glazebrook <[EMAIL PROTECTED]> disgusted:
>
> @solution = (^-@b + sqrt(@b^**2 ^+ 4^*@a^*@c) ) ^/ (2^*@a);
>[Stuff]
>If I was forced to write vector code like this I *WILL* give up on perl,
>and resort to Numerical
>Python or IDL instead.
>
You can always use the map and foreach lik
On Fri, 2002-07-12 at 13:22, Thomas A. Boyer wrote:
> Aaron Sherman wrote:
> > An example:
> >
> > $pid = fork() // -1;
> > if $pid < 0 {
> > # error ...
> > } else unless $pid {
> > # Parent
> > } else if $pid > 0 {
> > # Child
> > } else {
> >
On Mon, 15 Jul 2002, Luke Palmer wrote:
> On Mon, 15 Jul 2002, Karl Glazebrook wrote:
>
> > @solution = (^-@b + sqrt(@b^**2 ^+ 4^*@a^*@c) ) ^/ (2^*@a);
>
> That would not be very pretty, indeed. It would also not be very
> efficient. (BTW, its b**2 - 4ac, not + :)A more efficient, pretty,
On Mon, 2002-07-15 at 11:29, Karl Glazebrook wrote:
> complex formulae. Imagine:
>
> @solution = (^-@b + sqrt(@b^**2 ^+ 4^*@a^*@c) ) ^/ (2^*@a);
>
> (or would it be ^sqrt() ?) - This looks like sendmail :-)
I would imagine that non-binary operators would simply have a hyper-form
(which could
[several replies flattened into one]
On Monday, July 15, 2002, at 01:45 PM, Sean O'Rourke wrote:
> On Mon, 15 Jul 2002, Luke Palmer wrote:
>
>> On Mon, 15 Jul 2002, Karl Glazebrook wrote:
>>
>>> @solution = (^-@b + sqrt(@b^**2 ^+ 4^*@a^*@c) ) ^/ (2^*@a);
>>
>> That would not be very pretty, ind
At 2:09 PM -0400 7/15/02, Karl Glazebrook wrote:
>
>>On Monday, July 15, 2002, at 01:52 PM, Aaron Sherman wrote:
>>Sure, that's always an option. I think Perl has a lot going for it other
>>than the way vectorization happens, and with the ability to define your
>>own array behavior, you can prett
Karl Glazebrook:
# On Monday, July 15, 2002, at 01:45 PM, Sean O'Rourke wrote:
#
# > On Mon, 15 Jul 2002, Luke Palmer wrote:
# >
# >> On Mon, 15 Jul 2002, Karl Glazebrook wrote:
# >>
# >>> @solution = (^-@b + sqrt(@b^**2 ^+ 4^*@a^*@c) ) ^/ (2^*@a);
# >>
# >> That would not be very pretty, indeed
In a message dated Mon, 15 Jul 2002, Brent Dax writes:
> With explicit, you just get the result of Inf ** 2 (which presumably is
> still Inf) in $bar. Perhaps neither is what you want, but at least it
> doesn't take forever to run.
Yes. This is my fear of hyperoperation being the default for no
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
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
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 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.
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
> 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
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
Trey Harris wrote:
> Yes. This is my fear of hyperoperation being the default for normal
> aggregates. Loops--and large, multiply-nested, potentially-infinite
> ones--can spring out of code that doesn't look loopy at all. Erm... you
> know what I mean. :-)
>
> Karl, do you have any objection
> > 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 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
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
24 matches
Mail list logo
