Hello all,
I have been using Perl for years. It is the language of choice for most
of my needs. This is my first time posting. I have read Apocalypse 3
and many of your responses.
Issues:
1) Binary _or string concatenation
2) RFC 082: Arrays: Apply operators element-wise in a
Erik Lechak wrote:
2) RFC 082: Arrays: Apply operators element-wise in a list context
(hyper operators)
and
3) Special variable representing index of array in foreach structure
$# maybe (not in apocolypse3, I think)
Hi Eric. Thanks for your comments. Unfortunately it's a little early for
Thanks for the info,
Bryan,
Thank you for the info. I am more engineer than computer scientist, so
please excuse the ignorance behind these questions.
Except that the operator truly is simply an underscore. But it's also a
valid identifier character, so where it may be confused
On Saturday 06 October 2001 03:47 pm, Erik Lechak wrote:
Thank you for the info. I am more engineer than computer scientist,
so please excuse the ignorance behind these questions.
No problem.
Except that the operator truly is simply an underscore. But it's also a
valid identifier
Bryan,
I guess my biggest complaint was that underscore would be the only single
character operator that would be in danger of being lumped into the variable
name. I forgot about x . I use it all the time. If x has that constraint
why not the underscore. I worked at a place where
David M. Lloyd wrote:
On Thu, 4 Oct 2001, Michael G Schwern wrote:
Backtracking is at the heart of Logic Programming (or Declarative
Programming, if you like). This is one of the 3 main programming paradigms
(along with procedural and functional). The most popular Declarative
Hello!
Is this going to concat $a,$b and $c?
$foo = _($a,$b,$c);
(One way to save underlines and spaces.)
Or would that be:
$foo = _@($a,$b,$c);
BTW: what will these do?
$a _=_ ($b,$c);
$a ^_= ($b,$c); # (better with hypo-operator?, see below)
(WIM in
Edwin Steiner:
# Is this going to concat $a,$b and $c?
#
# $foo = _($a,$b,$c);
#
# (One way to save underlines and spaces.)
# Or would that be:
#
# $foo = _@($a,$b,$c);
That would be C$foo=join('', $a, $b, $c), just like in Perl 5.
# BTW: what will these do?
#
# $a _=_
Are these the same thing?
print _@{$data.{costs}};
print _ $data{costs};
-John
On Fri, 28 Sep 2001 21:27:48 +0200, Johan Vromans wrote:
Michael G Schwern [EMAIL PROTECTED] writes:
so if $key does not exist you'll get 'some default' instead of undef.
Except that a more common case is
my $foo = $hash{foo} || 'some default';
my $bar = $hash{bar} || 'some other
Okay, so this:
100 -s $filepath = 1e6
really means this:
100 -s $filepath-s $filepath = 1e6
which means that this:
$a == $b != NaN
really means this:
$a == $b $b != NaN
But $a == $b != NaN is supposed to [solve] the problem of numerical
comparisons between
From EX3:
A subroutine's adverbs are specified as part of its normal parameter list, but
separated from its regular parameters by a colon:
my sub operator:
is prec(\operator:+($)) ( *@list : $filter //= undef)
{ ...
This specifies that operator:
can take a single scalar adverb, which is
On 10/6/01 10:27 PM, Damian Conway wrote:
Doesn't that mean:
hello == 0 0 != NaN
will evaluate to true?
No. The step you're missing is that the non-numeric string hello,
when evaluated in a numeric context, produces NaN. So:
hello == 0 0 != NaN
is:
Nan ==
Damian == Damian Conway [EMAIL PROTECTED] writes:
Too much typing:
Damian module PAIR;
Damian method car { return .key }
Damian method cdr { return .value }
Damian method AUTOVIVIFY (default, $name) {
Damian if ($name =~ m/^c([ad])([ad]*)r$/) {
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