On Sun, Apr 24, 2005 at 03:02:16PM +1000, Brad Bowman wrote:
: Hi,
:
: I'm trying to understand the following section in S03:
:
: S03/"Junctive operators"
:
: Junctions are specifically unordered. So if you say
: for all(@foo) {...}
: it indicates to the compiler that there is no coupling between loop
: iterations and they can be run in any order or even in parallel.
:
: Is this a "for" on a one element list, which happens to
: be a junction, or does the all() flatten?
No, S03 is probably just wrong there. Junctions are scalar values, and
don't flatten in list context. Maybe we need something like:
for =all(@foo) {...}
to iterate the junction.
: Is the whole block run once with 1,2 and 3, or does the
: junction go into the block and autothread each operation?
I expect =all(@foo) would do the former, while all(@foo) would do
the latter, in which case you might as well have used "given" instead.
: for all(1,2,3) {
: next if $_ < 2; # testing 1 or all(1,2,3) ?
: %got{$_} = 1;
: }
: say %got.perl; # "(('2', 1), ('3', 1))" or "()" ?
Well, { 2 => 1, 3 => 1 } is the more likely notation.
: The "no coupling" in s03 suggests to me that the right
: answer is "(('2', 1), ('3', 1))", but I'm just guessing.
I think =all(@foo) should do what you expect there. Without the =
it should return { 1 => 1, 2 => 1, 3 => 1 } since there's only one
loop iteration, and it is *not* true that all(1,2,3) < 2. If you'd
said
for any(1,2,3) {...}
then it would have done the "next", because 1 < 2.
I should say that I don't see that =all() is different from =any().
They each just produce a list in "random" order. Though I suppose,
if we say that =one(1,2,3) should randomly pick one value, then
=any(1,2,3) should pick anywhere from 1 to 3 values. And, of course,
=none(1,2,3) should return a list of all the things that aren't 1, 2,
or 3 in random order. Maybe a lazy implementation will be beneficial
at that point. :-)
Larry
