John Williams wrote:
Back in October I suggested that $a ^+= b would act like reduce,
but in discussion
it was decided that it would act like length
I now pose the question: Is ^+= a hyper assignment operator or an
assignment hyper operator?
with a scalar involved
the method and
Apologies for trying to resuscitate this old horse, but a new idea
occurred to me.
Back in October I suggested that $a ^+= b would act like reduce,
but in discussion
it was decided that it would act like length, by the interpretation:
$a ^+= b
$a = $a ^+ b
$a = ($a, $a, $a, ...)
Damian Conway [EMAIL PROTECTED] writes:
Colin exemplifies:
$a = 1;
@a = (1);
@b = (1, 2, 3);
@c = (4, 5, 6);
$a = $a ^+ @b;
@a = @a ^+ @b;
print $a; # 7
No. It will (probably) print: 4. Because:
$a = $a ^+ @b;
becomes:
Given:
$a = 1;
@b = (1, 2, 3);
Damian suggested that:
$a = $a ^+ @b
becomes:
$a = ($a, $a, $a) ^+ (1, 2, 3)
$a = (1, 1, 1) ^+ (1, 2, 3)
$a = (2, 3, 4)
$a = 4;
Whereas Piers thought that:
$a = $a ^+ @b
becomes:
$a = [$a, $a,
[EMAIL PROTECTED] writes:
: Given:
:
: $a = 1;
: @b = (1, 2, 3);
:
: Damian suggested that:
:
: $a = $a ^+ @b
:
: becomes:
:
: $a = ($a, $a, $a) ^+ (1, 2, 3)
: $a = (1, 1, 1) ^+ (1, 2, 3)
: $a = (2, 3, 4)
: $a = 4;
:
: Whereas Piers thought that:
:
I just read Apocalypse and Exegesis 3, and something stuck out at me
because of its omission, namely using hyper operators for reduction.
$a ^+= @list; # should sum the elements of @list
Larry says @a ^+ 1 will replicate the scalar value for all a's, and Damian
talks about doing summation with
John observed:
I just read Apocalypse and Exegesis 3, and something stuck out at me
because of its omission, namely using hyper operators for reduction.
$a ^+= @list; # should sum the elements of @list
Larry says @a ^+ 1 will replicate the scalar value for all a's,
On Wed, Oct 10, 2001 at 09:42:58PM +1000, Damian Conway wrote:
John observed:
I just read Apocalypse and Exegesis 3, and something stuck out at me
because of its omission, namely using hyper operators for reduction.
$a ^+= @list; # should sum the elements of @list
On Wed, Oct 10, 2001 at 01:27:35PM -0700, Colin Meyer wrote:
On Wed, Oct 10, 2001 at 09:42:58PM +1000, Damian Conway wrote:
John observed:
I just read Apocalypse and Exegesis 3, and something stuck out at me
because of its omission, namely using hyper operators for
On Wed, Oct 10, 2001 at 01:27:35PM -0700, Colin Meyer wrote:
On Wed, Oct 10, 2001 at 09:42:58PM +1000, Damian Conway wrote:
John observed:
I just read Apocalypse and Exegesis 3, and something stuck out at me
because of its omission, namely using hyper operators for
On Wed, Oct 10, 2001 at 02:23:33PM -0700, Colin Meyer wrote:
[ @a ^+= @b ]
What I'd expect is more like:
foreach my $elem (@a) {
$elem ^+= @b;
}
Hrm. Why would you expect that when you'd have written as you just
did? Would you also expect
@a = @b ^+ @c;
to mean
On Wed, 10 Oct 2001 14:23:33 -0700, Colin Meyer wrote:
Does this mean that
@a ^+= @b;
will add every value of @b to every value of @a?
What I'd expect is more like:
foreach my $elem (@a) {
$elem ^+= @b;
}
If you want that effect, apply scalar on the RHS.
--
On Wed, Oct 10, 2001 at 04:34:14PM -0500, Jonathan Scott Duff wrote:
On Wed, Oct 10, 2001 at 02:23:33PM -0700, Colin Meyer wrote:
[ @a ^+= @b ]
What I'd expect is more like:
foreach my $elem (@a) {
$elem ^+= @b;
}
Hrm. Why would you expect that when you'd have written as
On Wed, Oct 10, 2001 at 03:41:18PM -0700, Colin Meyer wrote:
Maybe this illustrates my confusion:
$a = 1;
@a = (1);
@b = (1, 2, 3);
@c = (4, 5, 6);
$a = $a ^+ @b;
@a = @a ^+ @b;
print $a; # 7
print @a; # 7 or 2?
Or, after re-reading the apocolypse again, it seems:
print @a;
Aaron asked:
I'm wondering about some other operators Will there be a ^?? operator?
I believe that there will be a hyperoperator for *every* operator.
I might, for example, say:
@a = @b ^?? 'yea' :: 'nay';
Which I would expect to be the same as:
Colin exemplifies:
$a = 1;
@a = (1);
@b = (1, 2, 3);
@c = (4, 5, 6);
$a = $a ^+ @b;
@a = @a ^+ @b;
print $a; # 7
No. It will (probably) print: 4. Because:
$a = $a ^+ @b;
becomes:
$a = ($a,$a,$a) ^+ @b;
which is:
$a =
$a ^+= @list; # should sum the elements of @list
Does this mean that
@a ^+= @b;
will add every value of @b to every value of @a?
No. The rule is that a hyperoperator replicates its lower-dimensional
operand up to the same number of dimensions as its
According to Damian:
Colin exemplifies:
$a = 1;
@a = (1);
@b = (1, 2, 3);
@c = (4, 5, 6);
$a = $a ^+ @b;
@a = @a ^+ @b;
print $a; # 7
No. It will (probably) print: 4. Because:
print @a; # 7 or 2?
Prints: 2 2 3. Because:
So, does that mean:
$a = ($a) ^+ @b;
print $a; # prints: 3
# $a = ($a,undef,undef) ^+ @b ...
Yes.
My new confusion has to do with why does the hyperoperator expand $a to
($a,$a,$a), but (1) to (1,undef,undef)? Oh, it's because hyper-ops
expand an arg if it is
Damian Conway wrote:
@a ^+= reduce {$^a+$^b} @b;
What's this? Are positional args to HOFs now alphabetic rather than numeric?
cf. http://dev.perl.org/rfc/23.html#Positional_placeholders
@a ^+= reduce {$^a+$^b} @b;
What's this? Are positional args to HOFs now alphabetic rather than
numeric?
No. I just chose to used named placeholders, since they're more easily
followed in this example.
cf. http://dev.perl.org/rfc/23.html#Positional_placeholders
cf.
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