Re: Another Matlab test

[Michael Goffioul]

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.