On 8/16/06, David Green wrote:
$a=[1, 2, [EMAIL PROTECTED];
$c=[1, 2, [EMAIL PROTECTED];
$d=[1, 2, [EMAIL PROTECTED];
$a =:= $c; #false, different variables
$a === $c; #true, same elements make up $a and $c
$a eqv $c; #true, same elements therefore same values
$a === $d; #false, [EMAIL PROTECTED] and [EMAIL PROTECTED] are
different refs
So $a, $c, and $d may all have the same *value*
(or "snapshot", when evaluated all the way down
through nesting and references), i.e. they might
be eqv, but only $a and $c are === because they
have the same contents [unevaluated contents]
and $d doesn't.
(Actually $a===$c above should be false.) Given
that === answers the question "are these things
the same object", what's the solution for my
original motivation of comparing two items for
their unevaluated contents? "Eqv" evaluates
everything (both references and nested
containers) down to immutable values; I want to
follow nested structures all the way down, but
not evaluate/deref any variable references.
For the one-dimensional $a and $c given above, I could do something like:
?all(@[EMAIL PROTECTED], grep {$^a === $^c} zip(@$a; @$c))
For multidimensional/nested arrays, I could check
that they're the same size with $a.shape eqv
$b.shape, but I believe grep (or map, etc.) work
only one-dimensionally. I don't think using
hyperoperators would work either, because $a
»===« $c would deref the contained @x before
applying ===, right?
Plus hyperops return a nested structure, and I'm
looking for a single bool -- I think hyperising
"all" around the whole result would work....
Hyperops also upgrade dimensions that don't match
between the RHS and LHS, which is not always what
you want.
So perhaps what I'm looking for is more syntactic
sugar for easily traversing nested data
structures in different ways.
-David
