Re: whitespace and structure access operator

[John W. Eaton]

On 11/25/2015 11:23 AM, Rik wrote:
> On 11/25/2015 04:43 AM, John W. Eaton wrote:
> > On 11/25/2015 03:59 AM, Sebastian Schöps wrote:
> > > Rik-4 wrote
> > > > Can someone test whether the Matlab interpreter accepts the following
> > > > code?
> > > >
> > > > b .x = 1
> > > >
> > > > The key is the space between the structure access operator "." and the
> > > > initially undefined variable b.  Without the space Octave parses this just
> > > > fine.
> > > >
> > > > --Rik
> > >
> > >
> > > Yes it does (R2015b, Mac):
> > >
> > > > > b .x = 1
> > >
> > > b =
> > >
> > >       x: 1
> >
> > Thanks.
> >
> > Rik, do you want me to take a look at fixing this, or do you want to do it?
>
> jwe,
>
> I'm sure this is in the parser which means you are likely to find the
> solution faster than me.  I hope it is as simple as having some sort of
> rule to eat whitespace in the lexer.  The original bug report that brought
> this to my attention was https://savannah.gnu.org/bugs/?42559.

This problem is due to the lexer recognizing things like

  b .x

as commands instead of structure references, so the syntax is valid and 
it is being treated as a function call equivalent to

  b ('.x')

That's incorrect, but I don't see an immediate fix.

What does Matlab do for things like

  b . x = 1

Does the behavior differ if there is a function called 'b' in the path 
(I hope not, but I have to ask)?

jwe