On Fri, 1 Nov 2002, Ed Peschko wrote: > @a ^[+]= @b;
compared to > @a ^+= @b; > > ie: they are exactly the same. You are right, you get the same answer whether you do the hyper or the assignment first, except in the "scalar ^op= list", in which case doing the assignment last gets you the length of the list or some other useless result. > I'd say that '=' has *implicit* vectorization > here in the array case. In the scalar case: Right. ^= is rather pointless, because = already understands list context. OTOH, you can get some different effects out of ^= by virtue of the "dimensionally replicate, quantitatively undef-extend" rule for vectoring operators. @a ^= @b # @a.length == max( @a.length, @b.length ) @a ^= $b # all currently existing elements of @a are set to $b $b ^= @a # Yuck! $b = last element of @a. ~ John Williams
