On 8/16/06, Darren Duncan <[EMAIL PROTECTED]> wrote:
I'll try saying what I meant differently here:
The difference between === and eqv is that, if you have 2 symbols, $a
and $b, and $a === $b returns true, then that result is guaranteed to
be eternal if you don't assign to either symbol [or other symbols
aliased to either] afterwards.
The idea is that, the degree to which === examines 2 variables to
consider them equal or not is only so far as they are immutable. So
if you say "$foo = $bar", and then "$baz === $foo" returns true, then
a subsequent assignment to or type-allowed mutation of $bar won't
invalidate that $baz === $foo, but an assignment to $foo would.
IMHO the text "a subsequent assignment to or" is useless here because
I don't think any subsequent assignment to $bar could ever affect
$foo, even if they were mutable types:
$bar = [1,2];
$foo = $bar;
...
$bar = 123; # This doesn't affect $foo
Of course, type-allowed mutation of $bar will affect $foo if $bar is
mutable type.
Still, thanks for clarification - I misunderstood what you meant with
"someone else holding another symbol".
--
Markus Laire
