Re: Arguments out?

[Vic Norton]

Yeah, I thought about that later, Thorsten. Thanks for pointing it out.

So when you write something like

   function   [U, S, V] = svd(A)

and A comes in as an m x n, you might do the computations to get

   U - orthogonal m x m
   S - diagonal m x n
   V - orthogonal n x n

satisfying

   A = U * S * V'

Then you look at nargout.

You will only accept nargout = 0, 1, or 3. If nargout = 0 or 1, you 
want to return diag(S). If nargout = 3, you want to return the triple 
[U, S, V]. Apparently your final lines should read something like this

   if (nargout < 2)
      U = diag(S);
   elseif (nargout != 3)
      error("you don't know what you are doing, dummy!");
   endif

This is what bothers me. U and diag(S) are completely different kinds 
of objects. Writing U = diag(S) simply goes against my grain.


Regards,

Vic


At 6:10 PM +0100 12/11/03, Thorsten Meyer wrote:
> Hi Vic,
>
> try it like this:
>
> function [a,b,c] = tete()
>  if nargout == 0
>    a = "zero";
>  elseif nargout == 1
>    a = "one";
>  elseif nargout == 2
>    a = "two";
>    b = "one";
>  elseif nargout == 3
>    a = "three";
>    b = "two";
>    c = "one";
>  end
> endfunction
>
> you'll get:
> octave:9> tete
> ans = null
> octave:10> tete
> ans = zero
> octave:11> a=tete
> a = one
> octave:12> [a,b]=tete
> a = two
> b = one
> octave:13> [d,e,f]=tete
> d = three
> e = two
> f = one
>
> Vic Norton wrote:
>
> > I'm afraid I don't understand how output arguments work in octave. 
> > I am writing a routine, simp, to solve the linear programming 
> > problem
> >
> >    minimize
> >                  c * x
> >    subject to
> >                A * x = b
> >           prod( x(I) >= 0 ) == 1
> >
> > The function should look like this
> >
> >     [ x, pi, fail ] = simp( A, b, c, I )    #  pi * b == c * x
> >
> > with nargout = 3. But, for nargout = 0, 1, 2, I would like it to return
> >
> >     simp( A, b, c, I )                  #  nargout == 0, return ans = x
> >     x = simp( A, b, c, I )              #  nargout == 1
> >     [ x, fail ] = simp( A, b, c, I )    #  nargout == 2
> >
> > Is this possible? Or do I have to use
> >
> >     [ x, fail, pi ] = simp( A, b, c, I )
> >
> > in order that the nargout = 0, 1, 2 returns be what I want?
> >
> >
> > Regards,
> >
> > Vic




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