> > @fib[1, 5, 10..15, 20]
> (1 8 (89 144 233 377 610 987) 10946)
> ^
>
> Why???
As Joe noted in his last email:
> remember, back in perl-land the default behavior is to flatten,
> in raku ... by default [raku is] oriented toward building up
> complex structures like
I didn't understood
> my @fib = 1,1, * + * … *;
[...]
> @fib[1]
1
> @fib[5]
8
> @fib[1..5]
(1 2 3 5 8)
*> @fib[1, 5, 10..15, 20](1 8 (89 144 233 377 610 987) 10946)*
* ^*
Why???
On Mon, Dec 7, 2020 at 2:27 AM William Michels via perl6-users <
On Sun, Nov 29, 2020 at 6:38 PM Ralph Mellor
wrote:
> > Zen slicing as a possible way of 'de-containerizing' :
> > https://docs.raku.org/language/subscripts#index-entry-Zen_slices
>
> A zen-slice only affects the single reference it's applied to.
>
> And it is a no op when applied to anything
ToddAndMargo via perl6-users wrote:
> I am a little late to this conversation, but `,=`
> looks a lot like `push` to me.
Yes that was my first impression, if you read ahead a bit in the
discussion you'll see it explained.
In summary: the = shortcuts all work in a precisely parallel way, so
@r
Hi All,
I am a little late to this conversation, but `,=`
looks a lot like `push` to me. Am I missing
something?
-T
William Michels wrote:
> Joe, what would you expect the code below to produce?
> %h<> ,= c => 3;
> @a[] ,= 'd';
Well *I* expect it to error out, but that's my p5 brain talking.
The Raku approach is if you ask for nothing it gives you
everything, so an empty index like that essentially
> Ralph Mellor wrote:
> >> > @r = @r , 'd';
> >>
> >> There isn't anything very useful in this behavior though, is there?
>
> Just to be clear, I wasn't saying I didn't think circular references
> should be forbidden, I just specifically meant that you weren't likely
> to want the ",=" operator
> Zen slicing as a possible way of 'de-containerizing' :
> https://docs.raku.org/language/subscripts#index-entry-Zen_slices
A zen-slice only affects the single reference it's applied to.
And it is a no op when applied to anything other than a `Scalar`.
So it'll have no effect when applied
On Sun, Nov 29, 2020 at 9:16 AM Joseph Brenner wrote:
>
> William Michels wrote:
> >> > "Perhaps more importantly, what improvement do you propose?"
> >
> > Apologies for top-posting, but what immediately comes to my mind upon
> > encountering the creation of a self-referential
William Michels wrote:
>> > "Perhaps more importantly, what improvement do you propose?"
>
> Apologies for top-posting, but what immediately comes to my mind upon
> encountering the creation of a self-referential (circular/infinite)
> object is proverbially 'going-down-a-level' and trying again.
Joseph Brenner wrote:
> Just to be clear, I wasn't saying I didn't think circular references
should be forbidden,
Sorry about the double-negative. It could use another "not" to triple it.
Ralph Mellor wrote:
>> > @r = @r , 'd';
>>
>> Okay, that makes sense. So the circular reference I thought I
>> was seeing is really there, and it's working as designed.
>>
>> There isn't anything very useful in this behavior though, is there?
>
> Yes.
>
> Here are some relevant results from a
P.S. My apologies for top-posting in the quoted text, and my apologies
to William for the duplication.
On 11/29/20, Parrot Raiser <1parr...@gmail.com> wrote:
> Having a consistent ("regular", in the linguistic sense), structure
> for something like the op= form is obviously very desirable. It's
> > "Perhaps more importantly, what improvement do you propose?"
Apologies for top-posting, but what immediately comes to my mind upon
encountering the creation of a self-referential (circular/infinite)
object is proverbially 'going-down-a-level' and trying again. So I
tried 1. 'decontainerizing'
> > @r = @r , 'd';
>
> Okay, that makes sense. So the circular reference I thought I
> was seeing is really there, and it's working as designed.
>
> There isn't anything very useful in this behavior though, is there?
Yes.
Here are some relevant results from a search for "self referential" in
About the documentation in general...
> > that particular pair-input syntax is my least favorite.
> > Flipping around the order of key and value when the value is a numeric...?
> >
> > And it isn't needed to demo the operator, any pair input syntax works.
> > I might argue that examples should
First off, much thanks to Ralph Mellor for his detailed explanations.
Ralph Mellor wrote:
> @r ,= 'd';
>
> The above expands to:
>
> @r = @r , 'd';
Okay, that makes sense. So the circular reference I thought I
was seeing is really there, and it's working as designed.
There isn't anything
@r ,= 'd';
The above expands to:
@r = @r , 'd';
That in turn passes a list of two values to the LHS receiver.
That receiver is `@r`, an array, and what arrays do with `=`
is to empty themselves and then assign the list of elements
on the RHS of the `=` into corresponding `Scalar`s stored in
Hi Joe,
I can reproduce your results on Rakudo_2020.10, but I'm afraid I don't
have much more to say about the ",=" operator since I'm unfamiliar
with it.
Do the "docs" page(s) make more sense changing the phrase
"class-dependent" behavior to "hash-dependent" behavior?
Best, Bill.
On Thu, Nov
On 10/12/19 3:08 AM, William Michels via perl6-users wrote:
Inline:
On Fri, Oct 11, 2019 at 8:33 PM ToddAndMargo via perl6-users
wrote:
On 10/11/19 8:09 PM, William Michels via perl6-users wrote:
Hi Todd, Per the REPL, $x looks to be a List:
mbook:~ homedir$ perl6
To exit type 'exit' or
Inline:
On Fri, Oct 11, 2019 at 8:33 PM ToddAndMargo via perl6-users
wrote:
>
> On 10/11/19 8:09 PM, William Michels via perl6-users wrote:
> > Hi Todd, Per the REPL, $x looks to be a List:
> >
> > mbook:~ homedir$ perl6
> > To exit type 'exit' or '^D'
> >>
> >> my $x = (44,66)
> > (44 66)
> >>
On 10/11/19 8:09 PM, William Michels via perl6-users wrote:
Hi Todd, Per the REPL, $x looks to be a List:
mbook:~ homedir$ perl6
To exit type 'exit' or '^D'
my $x = (44,66)
(44 66)
say $x.WHAT
(List)
say $x.^name
List
my $y = < 55 77 >
(55 77)
say $y.WHAT
(List)
say $y.^name
List
Hi Todd, Per the REPL, $x looks to be a List:
mbook:~ homedir$ perl6
To exit type 'exit' or '^D'
>
> my $x = (44,66)
(44 66)
> say $x.WHAT
(List)
> say $x.^name
List
>
> my $y = < 55 77 >
(55 77)
> say $y.WHAT
(List)
> say $y.^name
List
>
> say $*VM
moar (2019.07.1)
HTH, Bill.
On Fri, Oct 11,
On 10/11/19 2:46 AM, William Michels via perl6-users wrote:
Below works:
mbook:~ homedir$ perl6 -e 'my $x = (44, 66); say $x; say $x.any < 43'
(44 66)
any(False, False)
#
mbook:~ homedir$ perl6 -e 'my $x = (44, 66); say $x; say $x.any < 50'
(44 66)
any(True, False)
#
mbook:~ homedir$ perl6 -e
Below works:
mbook:~ homedir$ perl6 -e 'my $x = (44, 66); say $x; say $x.any < 43'
(44 66)
any(False, False)
#
mbook:~ homedir$ perl6 -e 'my $x = (44, 66); say $x; say $x.any < 50'
(44 66)
any(True, False)
#
mbook:~ homedir$ perl6 -e 'my $x=0; my $any=2|4|8; $x==$any ?? put "x
exists, value= $x"
On 10/8/19 10:53 AM, Brad Gilbert wrote:
Most operations with Junctions produce Junctions.
> 1 + any(2, 3)
any(3, 4)
$ p6 'say 4 + any(44,66);'
any(48, 70)
Sweet! But what would you ever use it for?
Would this be the intended use: add a number to all
values in an array?
$ p6
Thank you very much Brad!!
Let's redirect all conversation on this particular issue here:
https://github.com/rakudo/rakudo/issues/3221
Best Regards, Bill.
On Tue, Oct 8, 2019 at 10:53 AM Brad Gilbert wrote:
>
> Most operations with Junctions produce Junctions.
>
> > 1 + any(2, 3)
>
Most operations with Junctions produce Junctions.
> 1 + any(2, 3)
any(3, 4)
> any(1, 2) + 3
any(4, 5)
> any() ~~ /./
any(｢a｣, ｢c｣)
In the case of the following line, `$/` gets assigned a junction of the
results.
> if any(@genus) ~~ m/Hama/ { put "Matches at least
Am 03.04.2012 17:10, schrieb Daniel Carrera:
(1..10).WHAT # = Range()
@foo = 1..10;
@foo.WHAT # = Array()
When you assign a range to @foo, the result is an array. Something
similar happens, for example, if you assign a scalar to @foo... The
context of the assignment causes Perl 6 to
On 3 April 2012 17:24, Moritz Lenz mor...@faui2k3.org wrote:
You can, very nearly. You just need to write
my @vec is Vector;
because you really want to change the type of the container, not just of the
contents (my Vector @vec would be an array containing Vector objects).
Another option
On 04/03/2012 08:24 PM, Daniel Carrera wrote:
On 3 April 2012 17:24, Moritz Lenz mor...@faui2k3.org wrote:
You can, very nearly. You just need to write
my @vec is Vector;
because you really want to change the type of the container, not just of the
contents (my Vector @vec would be an array
On 3 April 2012 20:38, Moritz Lenz mor...@faui2k3.org wrote:
which version of Rakudo are you using? (I've tried on the last
development version from git)
Rakudo Star 2012.02
% perl6 --version
This is perl6 version 2012.02 built on parrot 4.1.0 revision 0
Hmm... So you'd have to mess with
On 2010-07-23, at 4:25 am, Moritz Lenz wrote:
I'm still not convinced. [that there should be a special index variable]
Yes, it would be convient, but I've yet to see a non-contrived example where
it's actually necessary, and which can't be implemented trivially with other
Perl 6 tools.
I
If we expected Perl6 to be able to recognize these series and continue
them based on nothing but the first few elements, that would be a
dwimmy OEIS. (And an OEIS module that did that by consulting the real
OEIS would be cool, outside of core.) But that's not what this is
about. This is just
Am 23.07.2010 00:29, schrieb Damian Conway:
However, those *are* clunky and nigh unreadable, so I certainly wouldn't
object to having the index of the next generated element readily
available as an explicit variable in a series' generator block.
That would make all manner of evil both easier
Hi,
Am 22.07.2010 17:18, schrieb Jon Lang:
When I last reviewed the writeup for the series operators, I noticed two issues:
First, why is the RHS argument a list? You only ever use the first
element of it; so why don't you just reference a single value?
The idea is that you can continue
On Thu, Jul 22, 2010 at 11:41 AM, Moritz Lenz mor...@faui2k3.org wrote:
The difficulty you're running into is that you're trying to use the wrong
tool for the job. Just don't use the series operator when it's not easy to
use. Perl 6 has other mechanism too, which are better suited for these
On Thu, Jul 22, 2010 at 9:25 AM, Aaron Sherman a...@ajs.com wrote:
On Thu, Jul 22, 2010 at 11:41 AM, Moritz Lenz mor...@faui2k3.org wrote:
The difficulty you're running into is that you're trying to use the wrong
tool for the job. Just don't use the series operator when it's not easy to
use.
On Thu, Jul 22, 2010 at 1:13 PM, Jon Lang datawea...@gmail.com wrote:
I also think it's doable without a special tool:
0, { state $i = 1; $^a + $i++ } ... *
Kludgey; but possibly doable.
Well, it's kind of what state is there for.
But what I'd really like to see would be for the
On Thu, Jul 22, 2010 at 11:35 AM, Aaron Sherman a...@ajs.com wrote:
On Thu, Jul 22, 2010 at 1:13 PM, Jon Lang datawea...@gmail.com wrote:
Yes, it would be a
special tool; but it would be much more in keeping with the keep
simple things easy philosophy that Perl 6 tends to promote:
0, {
On Thu, Jul 22, 2010 at 4:52 PM, Jon Lang datawea...@gmail.com wrote:
I do have to admit that that's awfully clean-looking, but the
implementation
would force a closure in a series to behave differently from a closure
anywhere else.
How so?
Unlike some of you, I haven't managed to
On 23 July 2010 01:41, Moritz Lenz mor...@faui2k3.org wrote:
Use the right tool for the right job:
square numbers: 0, 1, 4, 9, 16, 25, 36, etc.
(1..10).map(* ** 2)
Or even just:
(1..10) »**» 2
Note that you can also get most of the effects you want by using
@_ in the series'
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 7/22/10 11:18 , Jon Lang wrote:
Second, I'm trying to think of a simple and intuitive way to write up
a series expression for:
triangle numbers: 0, 1, 3, 6, 10, 15, 21, etc.
square numbers: 0, 1, 4, 9, 16, 25, 36, etc.
factorials:
On Wed, Mar 25, 2009 at 12:25:59AM +0100, Moritz Lenz wrote:
: Ruud H.G. van Tol wrote:
: I wonder if there is place for a decision operator.
:
: This looks like a perfect place for a (possibly user defined) operator
: with an attribute, something like
:
: 1 ~~ 0 :dop(7)
To me it looks more
Ruud H.G. van Tol wrote:
I wonder if there is place for a decision operator.
(search terms: FPGA, evolvable hardware, Xilinx, PLB, LUT, NESW)
dop:| 0 1 2 3 4 5 6 7 8 9 A B C D E F
+
0 0 | 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 1 | 0 0 0 0 1 1 1 1 0 0 0 0 1
Moritz Lenz wrote:
Ruud H.G. van Tol wrote:
I wonder if there is place for a decision operator.
(search terms: FPGA, evolvable hardware, Xilinx, PLB, LUT, NESW)
dop:| 0 1 2 3 4 5 6 7 8 9 A B C D E F
+
0 0 | 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 1 | 0 0 0
HaloO,
Larry Wall wrote:
Then [X]() also is ()? How about (0,1) X ([]) === (0,1)?
No, that's (0,[]), (1,[1]). [] *doesn't* flatten in list context.
I guess you meant (0,[]), (1,[]). And you didn't answer what
[X]() returns. Following your arguments this would be undef.
I am genuinely
-Original Message-
From: Mark A. Biggar [mailto:[EMAIL PROTECTED]
Sent: Sunday, April 13, 2008 11:22 PM
To: Miller, Hugh
Cc: Moritz Lenz; p6l
Subject: Re: cross operator and empty list
Miller, Hugh wrote:
From: Moritz Lenz [mailto:[EMAIL PROTECTED]
[EMAIL PROTECTED] wrote
HaloO,
Xavier Noria wrote:
{0, 1} X {{}} = {(0, {}), (1, {})}
which, you see, is different from {0, 1}. They have different elements.
The fact that there's a clear mapping that sort of identifies them has
nothing to do with set equality.
But X is cooperating with , in Perl 6:
(0,1) X
On Mon, Apr 14, 2008 at 12:05:15PM +0200, TSa wrote:
But X is cooperating with , in Perl 6:
(0,1) X (()) === ((0,()),(1,())) === (0,1)
That is, X strips the outer list and comma concatenates the
inner empty list away.
No, the inner () is also in list context, and () in list context
always
HaloO,
Larry Wall wrote:
No, the inner () is also in list context, and () in list context
always just disappears.
And 0,1 X () is going to be (). Perl 6's infix:X is defined over
lists, not sets. If you want to overload X for set types, you may.
Then [X]() also is ()? How about (0,1) X
HaloO,
I wrote:
Then [X]() also is ()? How about (0,1) X ([]) === (0,1)?
The original question was sort of about how to write a list
that has .elems == 1 but no content.
Other ideas are: [[]] and @@() with the latter not very likely
because it implies any multidimensional array somehow having
On Mon, Apr 14, 2008 at 06:28:06PM +0200, TSa wrote:
HaloO,
Larry Wall wrote:
No, the inner () is also in list context, and () in list context
always just disappears.
And 0,1 X () is going to be (). Perl 6's infix:X is defined over
lists, not sets. If you want to overload X for set
At 09:58 -0700 4/14/08, Larry Wall wrote:
By the way, you don't need to put parens around the arguments to X. It takes a
list on either side. We made it tall so that it would stand out visually
anyway:
$a,$b,$c X $x,$y,$z
How long before some engineer or 3D graphic artist gets really
On Mon, Apr 14, 2008 at 11:47:04AM -0600, Doug McNutt wrote:
: At 09:58 -0700 4/14/08, Larry Wall wrote:
: By the way, you don't need to put parens around the arguments to X. It takes
a list on either side. We made it tall so that it would stand out visually
anyway:
:
: $a,$b,$c X $x,$y,$z
Doug McNutt douglist-at-macnauchtan.com |Perl 6| wrote:
At 09:58 -0700 4/14/08, Larry Wall wrote:
By the way, you don't need to put parens around the arguments to X. It takes a
list on either side. We made it tall so that it would stand out visually
anyway:
$a,$b,$c X $x,$y,$z
-Original Message-
From: Moritz Lenz [mailto:[EMAIL PROTECTED]
Sent: Saturday, April 12, 2008 10:37 AM
To: [EMAIL PROTECTED]
Cc: p6l
Subject: Re: cross operator and empty list
[EMAIL PROTECTED] wrote:
Technically the Cartesian cross operator doesn't have an
identity value.
It has
Miller, Hugh wrote:
From: Moritz Lenz [mailto:[EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Technically the Cartesian cross operator doesn't have an
identity value.
It has.
The set which contains only the emty set, or in perl terms ([]);
Or am I missing something?
Should be a (any) 1 point set
[EMAIL PROTECTED] wrote:
Technically the Cartesian cross operator doesn't have an identity value.
It has.
The set which contains only the emty set, or in perl terms ([]);
Or am I missing something?
Cheers,
Moritz
--
Moritz Lenz
http://moritz.faui2k3.org/ | http://perl-6.de/
signature.asc
HaloO,
Larry Wall wrote:
(@a X @b X @c).elems == @a.elems * @b.elems * @c.elems
Sorry, I was aiming at defining a neutral element of the X
operator. In cartesian products of sets this is achieved
by having a set that contains as sole member the empty tuple.
So how would that be written?
On Mon, Apr 7, 2008 at 4:50 AM, TSa [EMAIL PROTECTED] wrote:
HaloO,
Larry Wall wrote:
(@a X @b X @c).elems == @a.elems * @b.elems * @c.elems
Sorry, I was aiming at defining a neutral element of the X
operator.
A neutral element for the cross operator seems weird if that is to be
Adriano, I think perhaps what Tsa is trying to get at is the identity value
for the X operator, and I believe I know what it is.
In the relational model of data, both the version of the model where tuples
have unordered named attributes/elements (which I prefer), and the version
where tuples
Technically the Cartesian cross operator doesn't have an identity value. There
is no set X such that
A x X = A. Now any singleton set gives a result that is naturally isomorphic
to the original set, I.e, there is a obvious bijection between the two sets,
but they are not equal sets.
--
Mark
[EMAIL PROTECTED] wrote:
Technically the Cartesian cross operator doesn't have an identity value. There is no set X such that
A x X = A. Now any singleton set gives a result that is naturally isomorphic to the original set, I.e, there is a obvious bijection between the two sets, but they are
Cartesain product with the empty set is empty. A x B is the set of all pairs
(a,b) where a is in A and b is in B. If either is empty then there are no such
pairs and the result is also empty.
--
Mark Biggar
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
-- Original message
On Fri, Apr 04, 2008 at 06:51:20PM +0200, TSa wrote:
HaloO,
why is (1,2,3) X () defined to be the empty list
and not (1,2,3) as is the case with the cartesian
product of sets which X basically is with preserved
order.
(@a X @b X @c).elems == @a.elems * @b.elems * @c.elems
Larry
Cartesian product of anything with the empty set is empty. Which is
why SQL has outer joins.
On 4/4/08, TSa [EMAIL PROTECTED] wrote:
HaloO,
why is (1,2,3) X () defined to be the empty list
and not (1,2,3) as is the case with the cartesian
product of sets which X basically is with preserved
Fagyal Csongor skribis 2006-09-28 15:11 (+0200):
say $string xx $repeat;
List context.
my $add = $string xx $repeat;
Item context, for a list repetition operator. Doesn't really make sense,
and I think a warning or error message would be more appropriate.
I
A. Pagaltzis wrote:
I have the following code:
class MyStupidString {
method repeatit(Str $string, Int $repeat) {
say $string xx $repeat;
my $add = $string xx $repeat;
say $add;
}
};
my $obj = MyStupidString.new;
On Sat, 07 May 2005 01:47:08 -0400, Matt Creenan [EMAIL PROTECTED] wrote:
So here's some random ideas that probably make no sense ($ can be
optional.. don't know)
*snip*
That brings me to another idea. Is $_ as an array used? @_?
This relates back to the discussion on topics. Could be use @_
Mark A. Biggar skribis 2005-05-06 22:12 (-0700):
Actually if we define |...| at all, I'd prefer it mean abs(), its usual
mathmatical meaning.
No. We can't just use circumfix |...| with arbitrary expressions in it,
because | is taken as an infix operator. It has to be quoteish (like
(this is
Matt Creenan skribis 2005-05-07 1:47 (-0400):
I thought about $blockname = { ... }, but = is obviously taken, as is ==
$blockname =: for 1..5 {
$blockname := for 1..5 {
} $blockname;
} =: $blockname;
} $blockname;
$blockname for 1..5 {
$blockname
Matt Creenan skribis 2005-05-07 4:14 (-0400):
That brings me to another idea. Is $_ as an array used? @_?
The default signature of subs is ([EMAIL PROTECTED]).
Juerd
--
http://convolution.nl/maak_juerd_blij.html
http://convolution.nl/make_juerd_happy.html
On 5/6/05, Larry Wall [EMAIL PROTECTED] wrote:
The question is whether to treat the left arg the same way we treat
attribute defaults, with one free closure call. We could say that
{ rand 10 } x 100
{ rand 10 } xx 100
should just automatically call the closure on the left
On Sat, May 07, 2005 at 02:23:15PM +0200, Juerd wrote:
: Matt Creenan skribis 2005-05-07 1:47 (-0400):
: I thought about $blockname = { ... }, but = is obviously taken, as is ==
: $blockname =: for 1..5 {
: $blockname := for 1..5 {
: } $blockname;
: } =: $blockname;
: }
Luke Palmer skribis 2005-05-06 10:43 (-0600):
Why the %!@ would you ignore that!? :-)
I hate my brain. Now I wonder if Bool.does(Hash). Does it? :)
Juerd
--
http://convolution.nl/maak_juerd_blij.html
http://convolution.nl/make_juerd_happy.html
http://convolution.nl/gajigu_juerd_n.html
Luke Palmer skribis 2005-05-06 10:43 (-0600):
Thanks! Here's an annotated bit for each ?.
Only the triple-questionmarks were meant as questions. I should have
picked a better meta-operator for AVAILABLE?. But apparently, even
though I didn't mean to ask so many questions, there still are
On 5/6/05, Juerd [EMAIL PROTECTED] wrote:
Luke Palmer skribis 2005-05-06 10:43 (-0600):
!not none() ???
Nope. In order to create those, you just need to say none(). There
is no operator form.
Do we have postfix ! for factorials, or is it available?
No, it's
Luke Palmer skribis 2005-05-06 11:04 (-0600):
Because we're marking all of our singular nouns with $, and you have
to admit, the $ sigil in perl code is much more common than @ and %.
What good is a noun marker if you mark some of your verbs with it too?
But verbing doesn't weird language at
On Fri, May 06, 2005 at 06:24:00PM +0200, Juerd wrote:
To try and make it easier to pick (ASCII) operators, a simple table of
what's given away and what's available. Please let me know if there are
any mistakes.
If anyone knows how to fill in the ??? parts, be my guest!
[...]
\w+
Patrick R. Michaud skribis 2005-05-06 12:20 (-0500):
Ummm, what about Cnot and Ctrue ?
I'm sticking to non-words here, as I mentally parse not and true as
single-arg subs, single-arg subs as unary operators, etcetera. I can't
help it, but I have absolutely no idea how to determine the
I'm sticking to non-words here, as I mentally parse not and true as
single-arg subs, single-arg subs as unary operators, etcetera. I can't
help it, but I have absolutely no idea how to determine the difference.
Is it prefix:not or just not? I have no idea. I do know that it's
infix:x, not x.
On Fri, May 06, 2005 at 01:31:43PM -0400, Rob Kinyon wrote:
: I'm sticking to non-words here, as I mentally parse not and true as
: single-arg subs, single-arg subs as unary operators, etcetera. I can't
: help it, but I have absolutely no idea how to determine the difference.
: Is it
On Fri, May 06, 2005 at 06:49:44PM +0200, Juerd wrote:
: Luke Palmer skribis 2005-05-06 10:43 (-0600):
: Why the %!@ would you ignore that!? :-)
:
: I hate my brain. Now I wonder if Bool.does(Hash). Does it? :)
Any Object does Hash, and treats any argumentless method as a potential
hash key.
On Fri, May 06, 2005 at 11:25:31AM -0700, Larry Wall wrote:
: Any Object does Hash, and treats any argumentless method as a potential
: hash key.
I should also point out that the main reason for this is to allow
easier translation of Perl 5 idioms to Perl 6 without having to guess
whether $foo
Juerd skribis 2005-05-06 18:24 (+0200):
|AVAILABLE any()
We can use this for labels:
|foo| for ... {
while ... {
...;
next foo if ...;
}
}
It'll confuse the heck out of Ruby coders, but I do like this syntax. It
makes labels
On Fri, May 06, 2005 at 10:43:07AM -0600, Luke Palmer wrote:
: :: namespace ternary
:
: That's class sigil in term position. Separating namespaces never
: have preceding whitespace, so they're always part of some larger term.
Really more like a package sigil, which can be
On Fri, May 06, 2005 at 06:24:00PM +0200, Juerd wrote:
: {} href|closure hash (deref+)subscript (no ws)
: {}? (clash) AVAILABLE (ws)
s/AVAILABLE/statement block/
Actually, I'd try to find a way to combine all the paired ws-dependent
entries onto the same
Larry Wall skribis 2005-05-06 18:22 (-0700):
(But then you need to put postfix first in the heading.)
The heading uses junctions, and junctions are unordered ;)
Juerd
--
http://convolution.nl/maak_juerd_blij.html
http://convolution.nl/make_juerd_happy.html
Juerd wrote:
Juerd skribis 2005-05-06 18:24 (+0200):
|AVAILABLE any()
We can use this for labels:
|foo| for ... {
while ... {
...;
next foo if ...;
}
}
It'll confuse the heck out of Ruby coders, but I do like this syntax. It
makes
On Sat, 07 May 2005 01:12:02 -0400, Mark A. Biggar [EMAIL PROTECTED] wrote:
Actually if we define |...| at all, I'd prefer it mean abs(), its usual
mathmatical meaning.
I agree. I think || is just confusing.
I thought about $blockname = { ... }, but = is obviously taken, as is ==
So here's
Roger Hale wrote:
One set of cases that doesn't seem to have come up in discussion:
(1, 3, 2) - (83, 84, 81, 80, 85)
Should this give
(-82, -81, -79, -80, -85)
From an arithmetic point of view it should be exactly that. The
implementation might need to morph the code though, see below.
as
Thomas Sandla wrote:
John Williams wrote:
Good point. Another one is: how does the meta_operator determine the
identity value for user-defined operators?
Does it have to? The definition of the identity value---BTW, I like
the term neutral value better because identity also is a relation
between
John Williams wrote:
Good point. Another one is: how does the meta_operator determine the
identity value for user-defined operators?
Does it have to? The definition of the identity value---BTW, I like
the term neutral value better because identity also is a relation
between two values---is that
I found where Damain explains the rule as basically replicate dimensions,
extend lengths, using an identity value when extending the length.
http://www.mail-archive.com/perl6-language@perl.org/msg08304.html
On Wed, 13 Apr 2005, Thomas Sandlaß wrote:
Brent 'Dax' Royal-Gordon wrote:
IIRC, it's
On Thu, Apr 14, 2005 at 11:08:21AM -0600, John Williams wrote:
: Good point. Another one is: how does the meta_operator determine the
: identity value for user-defined operators?
:
: (1,2,3,4,5) my_infix_op (3,2,4)
:
: Maybe we should say that the excess length is simply copied unchanged.
:
Aldo Calpini skribis 2005-03-09 12:12 (+0100):
my @a = 1,2,3;
my $a = 1,2,3;
These are
(my @a = 1), 2, 3;
(my $a = 1), 2, 3;
if I understand precedence correctly. (S03)
my $a = [EMAIL PROTECTED];
my $a = *(1,2,3); # or is this a syntax error?
my $a = *(list 1,2,3);
my
Juerd wrote:
my @a = 1,2,3;
my $a = 1,2,3;
These are
(my @a = 1), 2, 3;
(my $a = 1), 2, 3;
if I understand precedence correctly. (S03)
right, sure. I vaguely remember something about comma instead of parens
being the list constructor, but maybe it was just in my fantasy.
and thanks for
Aldo Calpini writes:
my @a = [1,2,3]; # or does it make @a[0] = (1,2,3)?
Yes, @a[0] = [1,2,3];
and I have absolutely no clue about the following:
my *$a = @a;
my *$a = [EMAIL PROTECTED];
my *$a = (1,2,3);
my *$a = [1,2,3];
Those are all illegal. You need to use binding for
On Wed, Mar 09, 2005 at 12:22:37PM +0100, Juerd wrote:
:my $a = [EMAIL PROTECTED];
:my $a = *(1,2,3); # or is this a syntax error?
:my $a = *(list 1,2,3);
:my $a = *[1,2,3];
:
: I hope this will emit some kind of too-many-arguments warning in
: addition to assigning 1 to $a.
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