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]>
