On Wed, 2005-04-06 at 14:37, Larry Wall wrote:
> On Wed, Apr 06, 2005 at 08:24:23PM +0200, Juerd wrote:
> : Larry Wall skribis 2005-04-06 11:10 (-0700):
> : >     $$ref follow the ref list to the actual object.
> : 
> :     my $foo;
> :     my $bar = \$foo;
> :     my $quux = \$bar;
> :     my $xyzzy = \$quux;
> : 
> : How then, with only $xyzzy, do you get $bar? $$xyzzy would follow until
> : $foo. I don't like this at all.
> You can't get at $bar anyway.  You can only get at its thingy.  Otherwise
> you're talking symbolic refs.

I'm not sure I understand what "symbolic" means in this context. Can
someone explain? I'm specifically confused because I thought that:

                Given a parameter, create a new value of type Ref, make
                it refer to the operand and then return the newly
                created object.

This, it turns out is my delusion because neither S03 nor A03 define
such an operator (though it's used in examples a few times in both, and
it is given a precedence).

So, what specifically does unary:\ do?

If it does indeed create a new object (value if you will), then why is
it that we would never want to modify that object?

What if I want to add a mixin, for example:

        $r = \$a but Ref::Weak;

Now, someone gets a ref to $r, and wants to call a method defined in
Ref::Weak. How should they do that?

Aaron Sherman <[EMAIL PROTECTED]>
Senior Systems Engineer and Toolsmith
"It's the sound of a satellite saying, 'get me down!'" -Shriekback

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