Re: Another Matlab test

[Ben Abbott]

On May 18, 2013, at 9:44 AM, Michael Goffioul wrote:

> On Fri, May 17, 2013 at 9:37 PM, Ben Abbott <address@hidden> wrote:
> On May 18, 2013, at 9:32 AM, Michael Goffioul wrote:
> 
> > On Fri, May 17, 2013 at 9:26 PM, Ben Abbott <address@hidden> wrote:
> > On May 18, 2013, at 7:49 AM, Michael Goffioul wrote:
> >
> > > What's the result of the following:
> > >
> > > meta.class.fromName('meta.class').Name
> > >
> > > Thanks,
> > > Michael.
> >
> > If this isn't what you're looking for, just give me some more detail.
> >
> > Hmmm, I really only wanted the result of:
> >
> > meta.class.fromName('meta.class').Name
> >
> > Michael.
> 
> Oh! ... you were being literal!
> 
> Yes :).
>  
> >> meta.class.fromName('meta.class').Name
> 
> ans =
> 
> meta.class
> 
> 
> Thanks. I suppose the same will happen with subpackages, the "Name" property 
> will contain the fully-qualified name. That is, if you have a package 
> structure as +pack/+subpack/myfun.m, then
> 
>     meta.package.fromName('pack.subpack').Name
> 
> returns
> 
>     pack.subpack
> 
> Would you mind confirming this?
> 
> Michael.

You are correct!

matlab> meta.package.fromName('pack.subpack').Name

ans =

pack.subpack

Ben