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Re: Another Matlab test
From: |
Ben Abbott |
Subject: |
Re: Another Matlab test |
Date: |
Sat, 18 May 2013 09:49:09 +0800 |
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