On Fri, 4 Feb 2005, Juerd wrote:
Does this make sense?
my @words = gather {
for =(open '/usr/share/dict/words' err die) {
.=chomp;
next if /-[a-z]/;
/$re/ and take { word = $_, score = %scores{ .letters }.sum };
}
} == sort { .score } is
Will the relative precedence of grouping versus anchors for beginning and
end of line remain the same in Perl6 rules?
The error of writing
/^(?:free|net|open)bsd|bsdos|interix$/
when you mean
/^(?:(?:free|net|open)bsd|bsdos|interix)$/
is rather too easy to make. This is not the first time
Michele Dondi skribis 2005-02-07 11:45 (+0100):
With some effort I managed to understand _which_ sense it should make up
to this:
} == sort { .score } is descending, { .word.length }, { .word };
I mean: everything of what is gather()ed gets 'piped' into sort() which
sorts according to C
Nicholas Clark skribis 2005-02-07 12:10 (+):
Will the relative precedence of grouping versus anchors for beginning and
end of line remain the same in Perl6 rules?
There currently is no such thing as precedence in regexes. Changing this
would make understanding regexes a lot harder, I think.
On Mon, Feb 07, 2005 at 01:21:00PM +0100, Juerd wrote:
Nicholas Clark skribis 2005-02-07 12:10 (+):
Will the relative precedence of grouping versus anchors for beginning and
end of line remain the same in Perl6 rules?
There currently is no such thing as precedence in regexes. Changing
[EMAIL PROTECTED] wrote:
pugs ? 4 (0 | 6) 2
(#t|#f)
Here's my take on it.
Compare
my $a = (0 | 6);
say 4 $a and $a 2;
vs
say 4 (0 | 6) and (0 | 6) 2;
The difference is that in the first case the junction refers to the same
object, and the result should probably be expanded only
On Fri, Feb 04, 2005 at 10:39:30PM +0100, Juerd wrote:
: Does this make sense?
:
: my @words = gather {
: for =(open '/usr/share/dict/words' err die) {
: .=chomp;
: next if /-[a-z]/;
: /$re/ and take { word = $_, score = %scores{ .letters }.sum };
:
On Mon, Feb 07, 2005 at 05:33:06PM +0100, Miroslav Silovic wrote:
my $a = (0 | 6);
say 4 $a and $a 2;
Yup. My mathematic intuition cannot suffer that:
4 X 2
to be true in any circumstances -- as it violates associativity.
If one wants to violate associativity, one should presumably
On Tue, 8 Feb 2005 11:12:40 +0800, Autrijus Tang [EMAIL PROTECTED] wrote:
On Mon, Feb 07, 2005 at 05:33:06PM +0100, Miroslav Silovic wrote:
my $a = (0 | 6);
say 4 $a and $a 2;
Yup. My mathematic intuition cannot suffer that:
4 X 2
to be true in any circumstances -- as it
On Tue, Feb 08, 2005 at 11:12:40AM +0800, Autrijus Tang wrote:
: This way, both associativity and junctive dimensionality holds, so
: I think it's the way to go. Please correct me if you see serious
: flaws with this approach.
Feels right to me.
Larry
10 matches
Mail list logo
