On Tue, 8 Feb 2005, Matt Fowles wrote:
pipe dreams
Juerd wondered if he could mix = and == in a sane way. The answer
appears to be no. Once you bring in == you should stick with it.
Huh?!? It doesn't seem to me that the answer is 'no'. In fact C ==
is supposed to be yet another operator,
Michele Dondi wrote:
On Tue, 8 Feb 2005, Matt Fowles wrote:
pipe dreams
Juerd wondered if he could mix = and == in a sane way. The answer
appears to be no. Once you bring in == you should stick with it.
Huh?!? It doesn't seem to me that the answer is 'no'. In fact C ==
is supposed to be
On Wed, Feb 09, 2005 at 10:04:48AM +0100, Michele Dondi wrote:
: On Tue, 8 Feb 2005, Matt Fowles wrote:
:
: pipe dreams
:Juerd wondered if he could mix = and == in a sane way. The answer
:appears to be no. Once you bring in == you should stick with it.
:
: Huh?!? It doesn't seem to me
On Wed, 9 Feb 2005, Larry Wall wrote:
Yes, you can certainly intermix them as long as you keep your
precedence straight with parentheses. Though I suppose we could go
as far as to say that = is only scalar assignment, and you have to
use == or == for list assignment. That would
Larry Wall wrote:
On Wed, Feb 09, 2005 at 10:04:48AM +0100, Michele Dondi wrote:
: On Tue, 8 Feb 2005, Matt Fowles wrote:
:
: pipe dreams
:Juerd wondered if he could mix = and == in a sane way. The answer
:appears to be no. Once you bring in == you should stick with it.
:
: Huh?!? It
Does
($k, $v) == pop %hash;
or
($k, $v) == %hash.pop;
make sense to anyone except me?
Since we now have an explicit concept of pairs, one could consider a
hash to be nothing but an unordered (but well indexed) list of pairs.
So, C pop %hash would be a lot like C each , except, of course,
that
Rod Adams wrote:
Does
($k, $v) == pop %hash;
or
($k, $v) == %hash.pop;
make sense to anyone except me?
Makes sense to me. Although I would be more inclined to think of pop as
returning a pair - but does a pair in list context turn into a list of
key, value? If so then the above makes lots of
Matthew Walton [EMAIL PROTECTED] writes:
Rod Adams wrote:
Does
($k, $v) == pop %hash;
or
($k, $v) == %hash.pop;
make sense to anyone except me?
Makes sense to me. Although I would be more inclined to think of pop
as returning a pair - but does a pair in list context turn into a list
of
On Wed, 2005-02-09 at 06:04, Rod Adams wrote:
Larry Wall wrote:
Yes, you can certainly intermix them as long as you keep your
precedence straight with parentheses. Though I suppose we could go
as far as to say that = is only scalar assignment, and you have to
use == or == for list
--- Matt Fowles [EMAIL PROTECTED] wrote:
Logic Programming in Perl 6
Ovid asked what logic programming in perl 6 would look like. No
answer
yet, but I suppose I can pick the low hanging fruit: as a
limiting case
you could always back out the entire perl 6 grammar and insert
On Wed, 2005-02-09 at 14:57, Ovid wrote:
--- Matt Fowles [EMAIL PROTECTED] wrote:
Logic Programming in Perl 6
Ovid asked what logic programming in perl 6 would look like. No
answer yet, but I suppose I can pick the low hanging fruit: as a
limiting case
you could always back
On Wed, Feb 09, 2005 at 11:57:17AM -0800, Ovid wrote:
: --- Matt Fowles [EMAIL PROTECTED] wrote:
:
: Logic Programming in Perl 6
: Ovid asked what logic programming in perl 6 would look like. No
: answer
: yet, but I suppose I can pick the low hanging fruit: as a
: limiting case
:
On 2005.02.08.19.07, Matt Fowles wrote:
| Brock~
|
|
| On Tue, 8 Feb 2005 12:08:45 -0700, Brock [EMAIL PROTECTED] wrote:
|
| Hm. I take that back... it was a silly comment to make and not very
| mathematically sound. Sorry.
|
| --Brock
|
| - Forwarded message from Brock [EMAIL
On Wed, Feb 09, 2005 at 11:57:17AM -0800, Ovid wrote:
--- Matt Fowles [EMAIL PROTECTED] wrote:
Logic Programming in Perl 6
Ovid asked what logic programming in perl 6 would look like. No
answer
yet, but I suppose I can pick the low hanging fruit: as a
limiting case
Matt Fowles wrote:
All~
On Tue, 08 Feb 2005 17:51:24 +0100, Miroslav Silovic [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] wrote:
Well, we see the same kind of thing with standard interval arithmetic:
(-1, 1) * (-1, 1) = (-1, 1)
(-1, 1) ** 2 = [0, 1)
The reason that junctions behave this way is
All~
On Wed, 09 Feb 2005 22:48:00 +, Matthew Walton
[EMAIL PROTECTED] wrote:
Matt Fowles wrote:
All~
On Tue, 08 Feb 2005 17:51:24 +0100, Miroslav Silovic [EMAIL PROTECTED]
wrote:
[EMAIL PROTECTED] wrote:
Well, we see the same kind of thing with standard interval arithmetic:
In article [EMAIL PROTECTED], [EMAIL PROTECTED] (Luke
Palmer) wrote:
Well, we see the same kind of thing with standard interval arithmetic:
[...]
It didn't bother me that junctions weren't ordered transitively.
(Ordering had better work transitively for ordinary numbers, but
junctions aren't
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