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Re: === and array-refs

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From:
David Green
Date:
August 17, 2006 11:19
Subject:
Re: === and array-refs
Message ID:
a06230904c10a4df732bb@[172.27.1.7]
On 8/16/06, David Green wrote:
>    $a=[1, 2, \@x];
>    $c=[1, 2, \@x];
>    $d=[1, 2, \@y];
>
>  	$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, \@x and \@y 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 $c given above, I could do something like:
	?all(@$a==@$c, grep {$^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
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