actually , ones we decide that ^ *is necessary for vectorization , we
can allow other brackets , optional brackets ( where unambiguous ) ,
and spaces inside the brackets :
a ^+= b
a ^[+]= b
a ^(+)= b
a ^( + )= b
a ^{ + }= b
a ^{+}= b
a ^[ + ]= b
right, and what does this
On Sat, Nov 02, 2002 at 02:18:44AM +0200, [EMAIL PROTECTED] wrote:
snip ...
in that case the vectorization is *compleatly* orthogonal to the
details of op and we even can have something like
@a ^[{ $^a $^b ?? 1 :: ($^a,$^b) := ($^b,$^a) }] @b
I agree with all that you said
I'm probably opening up a whole new can of worms here, but if we said
that the following were both vector operators:
^ == intersection operator
v == union operator
then these could have potentially useful meanings on their *own* as set
operators, as well as modifying other
