Juerd skribis 2005-05-14 17:23 (+0200):
Markus Laire skribis 2005-05-14 18:07 (+0300):
[>>+^=<<] (@a, @b, @c)
These arrays flatten first (otherwise [+] @foo could never calculate the sum of the elements), so imagine that you have
$foo, $bar, $baz, $quux, $xyzzy
to let >>+^=<< operate on.
Is this then ok?
[>>+^=<<] (@a ; @b ; @c)
[>>+^=<<] ([EMAIL PROTECTED], [EMAIL PROTECTED], [EMAIL PROTECTED])
As S09 says that:
At the statement level, a semicolon terminates the current expression. Within any kind of bracketing construct, semicolon notionally produces a list of lists, the interpretation of which depends on the context. Such a semicolon list always provides list context to each of its sublists. The following two constructs are structurally indistinguishable:
(0..10; 1,2,4; 3) ([0..10], [1,2,3,4], )
If not, how then would I use hyper-reduction ops like [>>+^=<<] with several arrays?
i.e. How do I write
@a >>+^=<< @b >>+^=<< @c
using the [>>+^=<<] op?
-- Markus Laire <Jam. 1:5-6>