> > right, and what does this all mean? I have yet to see a good meaning
> > for
> > @array ^[+]= @array2 ...
>
> I think it's this:
>
> @a [+=] @b -> @a[x] += @b[x]
>
> @a [+]= @b -> @temp = @a [+] @b; a = @temp;
>
Ok, so the '=' isn't being explicitly vectorized. So -
@a ^[+]= @b;
@a = (1,2,3,4); @b = (5,6,7,8);
@temp = @a ^+ @b; = (1+5,2+6,3+7,4+8) = (6,8,10,12);
@temp = (6,8,10,12);
@a = (6,8,10,12);
wheras:
@a ^+= @b;
$a[0] += $b[0];
$a[1] += $b[1];
$a[2] += $b[2];
$a[3] += $b[3];
$a[0] = 1 + 5;
$a[1] = 2 + 6;
$a[2] = 3 + 7;
$a[3] = 4 + 8;
@a = (6,8,10,12);
ie: they are exactly the same. I'd say that '=' has *implicit* vectorization
here in the array case. In the scalar case:
@a ^[+]= 1;
@temp = @a ^+ 1;
it is exactly the same.
@a ^+= 1;
$a[0] +=1;
....
$a[3] +=;
So again, I don't see the difference between the two. ^[+]= and ^+= are synonyms
as far as I can see, and hence no need for the first form.
Ed
