> > > $a ^+= @list; # should sum the elements of @list
>
> Does this mean that
>
> @a ^+= @b;
>
> will add every value of @b to every value of @a?
No. The rule is that a hyperoperator replicates its lower-dimensional
operand up to the same number of dimensions as its higher-dimensional
operand, then applies the hyped operator element-by-element across the
two operands.
Since @a and @b are (presumably) of the same dimensionality:
@a ^+= @b;
means:
@a[0] += @b[0]
@a[1] += @b[2]
@a[2] += @b[2]
# etc.
To add every value of @b to every value of @a, you'll need either:
@a ^+= reduce {$^a+$^b} @b;
or:
@a = @(any(@a) + any(@y));
(depending on your definition of "every").
Damian
