On Fri, May 19, 2006 at 12:53:29PM -0700, Larry Wall wrote:
> and, in fact, the Foo.^bar syntax is just short for Foo.meta.bar.

So, you anticipated my half-question.

> The type of metaobject Foo.meta might be called "Class" if that's what the
> metaobject protocol decides it should be, but Perl the Language doesn't
> care.  If so, then Foo.meta.isa(Class) would be true.  But Foo.isa(Class)
> is still false.

OK, in my previous message, you should apparently read "metaobject" for
"type object".  But I think the questions still apply, as does the proposal
that all _metaobjects_ that currently are correlated with packages should
instead just _do_ Package.
Chip Salzenberg <[EMAIL PROTECTED]>

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