On Tue, Apr 1, 2008 at 5:39 AM, Mark J. Reed <[EMAIL PROTECTED]> wrote:
> Perl5, all the builtin "functions" are really defiend as operators,
"defined", even. (However fiendishly.)
Anyway, "function" vs "operator" is mostly a difference in terminology
that makes no difference in practice, but I
HaloO,
Mark J. Reed wrote:
Anyway, "function" vs "operator" is mostly a difference in terminology
I'm not sure what the defined difference between function and operator
is in Perl 6 but I make a very clear distinction. An operator is acting
an *one* type, that is &op:(::T,T-->T) while a functi
I've read Synopsis and I wondered why to treat max and min as
operator. IMHO, view them as list functions is more reasonable. Like
below:
@test.max
is clearer than
@test[0] max @test[1] or [max] @test.
Any reply would be really appreciated and will much help me learn
perl6. Thanks in advance!
On Tue, Apr 1, 2008 at 1:44 AM, Xiao Yafeng <[EMAIL PROTECTED]> wrote:
> I've read Synopsis and I wondered why to treat max and min as
> operator. IMHO, view them as list functions is more reasonable. Like
> below:
>
> @test.max
Which is how you would probably call it in Perl6. Or else
max(@t
#[ are there people paying attention to these issues on other mailing lists? ]
= on Compact structs
revision 1, initial posting
What functions serialize/deserialize to the C view?
If these are to be member functions, they would be applicable only if the
struct is compact, and erroneous to call
On Tue, Apr 01, 2008 at 05:39:36AM -0400, Mark J. Reed wrote:
> On Tue, Apr 1, 2008 at 1:44 AM, Xiao Yafeng <[EMAIL PROTECTED]> wrote:
> > I've read Synopsis and I wondered why to treat max and min as
> > operator. IMHO, view them as list functions is more reasonable. Like
> > below:
> >
> > @te
I sit corrected. Guess that's one of the places pugs is out of date.
On 4/1/08, Patrick R. Michaud <[EMAIL PROTECTED]> wrote:
> On Tue, Apr 01, 2008 at 05:39:36AM -0400, Mark J. Reed wrote:
> > On Tue, Apr 1, 2008 at 1:44 AM, Xiao Yafeng <[EMAIL PROTECTED]> wrote:
> > > I've read Synopsis and I
Damian Conway (Wednesday, March 26, 2008 3:39 AM):
> I have already contacted Uri and expressed my dismay at his entirely
> inappropriate interjection of an advertisement for our Perl College
> event into
> this discussion about funding for critical Perl projects and personnel.
There are at least
On Apr 1, 2008, at 6:13 , John M. Dlugosz wrote:
Meanwhile, how do I use it?
my Buf $temp = $record;
$stream.print ($temp);
$stream.print (Buf $record);
> $stream.print($record.pack) # I would think?
--
brandon s. allbery [solaris,freebsd,perl,pugs,haskell
HaloO,
John M. Dlugosz wrote:
So a function can only supply values for one dimension? Given
@array[foo;bar]
each function is called in list context and returns a list for one of the
dimensions.
But what if you wanted a function to replace the literal subscript in:
@array[1,2;1;3]
@a
Brandon S. Allbery KF8NH allbery-at-ece.cmu.edu |Perl 6| wrote:
$stream.print (Buf $record);
> $stream.print($record.pack) # I would think?
Thank you so much for the reply. I was beginning to wonder if Perl 6
interest is dead, or if there is another secret lair for current
enthusiasts t
On Tue, Apr 01, 2008 at 12:25:50PM -0500, John M. Dlugosz wrote:
> Thank you so much for the reply. I was beginning to wonder if Perl 6
> interest is dead, or if there is another secret lair for current
> enthusiasts that I'm unaware of.
Sorry, many of us have to live under budgetary constraint
On Apr 1, 2008, at 13:25 , John M. Dlugosz wrote:
Brandon S. Allbery KF8NH allbery-at-ece.cmu.edu |Perl 6| wrote:
$stream.print (Buf $record);
> $stream.print($record.pack) # I would think?
A .pack member function on a Compact struct is indeed my first gut
feeling, but at the end of the
I take that as encouragement, and confirmation that this is the right
place to post (and archive) my musings. HOPEFULLY I can get questions
answered too.
I credit the success of C++, in part, to Stroustrup's personal training
of the first gurus. He would explain things that made me see the
TSa wrote:
Mark J. Reed wrote:
Anyway, "function" vs "operator" is mostly a difference in terminology
I'm not sure what the defined difference between function and operator
is in Perl 6 but I make a very clear distinction. An operator is acting
an *one* type, that is &op:(::T,T-->T) while a fu
In S29, there are definitions like
our Capture method shape (@array: ) is export
But in S12 there is no mention as to what an "our" method is. It states that
"my" is used to make private methods, and "^" to make class methods.
I think this is a doc relic and should be fixed globally in that
TSa Thomas.Sandlass-at-barco.com |Perl 6| wrote:
Now my question: could slice context be a runtime feature that acts
before the dispatch to &postcircumfix:<[ ]> by retrieving the shape
of the @array and handing it over to &foo as context, capture the
shape of the slice returned and hand over the
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