Re: Arguments out?

[Thorsten Meyer]

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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