Juerd wrote:
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)



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], [3])

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>

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