From help-request at octave dot org Tue Mar 28 22:11:47 2006
Subject: Re: Forcing Variable Outputs
From: Bill Denney <denney at seas dot upenn dot edu>
To: Geordie McBain <geordie dot mcbain at aeromech dot usyd dot edu dot au>
cc: help at octave dot org
Date: Tue, 28 Mar 2006 23:11:31 -0500 (EST)

Not exactly.  As I tried that, it just gave the output as the same as the 
input.  The problem is that I want this to work correctly (without having 
to do a lot of if blocks).

function [varargout] = fxn1(varargin)

   nargin
   nargout

   varargout{:} = fxn2(varargin{:});

endfunction

function [varargout] = fxn2(varargin)

   nargin
   nargout

   varargout = {varargin};

endfunction

As it is, if I call

[a b c] = fxn1(1, 2, 3)

I get that nargin for fxn1 is 3, and nargin for fxn2 is 3, but nargout for 
fxn1 is 3 and nargout for fxn2 is 1.

Bill

On Wed, 29 Mar 2006, Geordie McBain wrote:

> Does this do what you want?
>
>  function varargout = fxn1 (varargin)
>    varargout = fxn2 (varargin{:});
>  endfunction
>
>  function varargout = fxn2 (varargin)
>    varargout = {varargin};
>  endfunction
>
> I see:
>
> octave2.9:1> a = fxn1 (1)
> a = 1
> octave2.9:2> [a, b] = fxn1 (1, 2)
> a = 1
> b = 2
>
>
> On Tue, 2006-03-28 at 20:53 -0500, Bill Denney wrote:
>> I have a function that I want to force to have a variable number of output
>> arguements based on its calling function's number of output arguements.
>> In other words, I want to do something like
>>
>> function [varargout] = fxn1(varargin)
>>
>>    varargout{:} = fxn2(varargin{:});
>>
>> endfunction
>>
>> and I want fxn2 to see the same number of output arguements that fxn1
>> sees.
>>
>> I thought of initializing varargout like
>>
>> varargout = cell(1, nargout);
>>
>> but that didn't work.  Is there a way to do this?
>>
>> Bill
>>
>

-- 
"I love America more than any other country in this world, and, exactly
for this reason, I insist on the right to criticize her perpetually."
   -- James Baldwin



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