On Fri, Jul 01, 2005 at 07:11:26AM +0000, Smylers wrote: > > The question you have to ask yourself is why should a reference be > > treated different from any other value? It is a VALUE. > > Except it isn't. Or at least, not all the time: it depends how you wish > to look at it. If you just consider a reference to be a value > (effectively a pointer, a memory address) then you aren't examining a > data structure _deeply_; you're just doing a _shallow_ comparision of it > as a reference.
I think this hits the nail on the head. References are treated differently because we don't look at the reference we look at the data that reference points to. is_deeply() does this fairly consistently ignoring blessing, ties and overloading. Under the proposed logic, this test fails: my $a = []; my $b = []; my $c = []; my $x = [$a, $a]; my $y = [$b, $c]; is_deeply($x, $y); because $x and $y can be altered in exactly the same way yet come out differently. $x->[0][0] = 1; # [[1],[1]] $y->[0][0] = 1; # [[1],[]] There's plenty of other cases where this can happen. If $x is tied. If $x is a string overloaded object. If $x is an object. These can lead to situations where the same operation is applied to both structures yet results in different structures. But we ignore this because is_deeply() has to draw a line between internal and external information somewhere. Its drawn at the reference value. is_deeply() is not about exact equivalence. Its about making a best fit function for the most common uses. I think most people expect [$a, $a] and [$b,$c] to come out equal. Test::Deep is for tweaked deep comparisons. -- Michael G Schwern [EMAIL PROTECTED] http://www.pobox.com/~schwern Reality is that which, when you stop believing in it, doesn't go away. -- Phillip K. Dick
