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, ...) ^+ b
$a = ($a+$b[0], $a+$b[1], ...)
$a = (a list in scalar context )
$a = length(b)
I now pose the question: Is ^+= a hyper assignment operator or an
assignment hyper operator?
An assignment hyper operator would be interpreted as above, or with
two arrays as:
a ^+= b
a = a ^+ b
a = ($a[0] + $b[0], $a[1] + $b[1], ... )
A hyper assignment operator would be interpreted like this (doing the
hyper part first):
a ^+= b
$a[0] += $b[0], $a[1] += $b[1], ...
$a[0] = $a[0] + $b[0], $a[1] = $a[1] + $b[1], ...
With two arrays the method is different, but the end result is the
same. (One might be more
efficient than the other, but I won't speculate on that here.) But with
a scalar involved
the method and the result is different. $a = length(b) is the
assignment hyper operator
interpretation. The hyper assignment operator interpretation looks
like this:
$a ^+= b
($a, $a, $a, ...) ^+= b
$a += $b[0], $a += $b[1], $a += $b[2], ...
which is DWIM: $a = reduce(b), assuming, of course that $a is
initially zero.
For a = $b, the end result also appears to be the same.
assignment hyper operator
a ^+= $b
a = a ^+ $b
a = a ^+ ($b, $b, $b, ...)
a = ($a[0]+$b, $a[1]+$b, $a[2]+$b, ...)
hyper assignment operator
a ^+= $b
a ^+= ($b, $b, $b, ...)
$a[0] += $b, $a[1] += $b, $a[2] += $b, ...
~ John Williams
