Help with Varargin

I have a function called MySolve and i am looking to adapt it to use Varargin.

As far as i am aware Varargin allows defaults for parameters to be set. but can be over-ridden if an input is given. I am having trouble understanding the syntax for using Varargin, here is the function MySolve.m that i have written:

function [x,converged]=MySolve(f,x0,tol,maxit)

%Set Converged intial value to 0 (false)
%Set initial value of x as x0
converged=0;
x=x0;

%run a loop from 1 to maxit
for k=0:maxit

      x1=x;
      r=f(x);
      J=MyJacobian(f,x,1e-6);
      x=x-(J\r);
      if(max(abs(x-x1)))<tol && max(abs(f(x)))<tol
          converged=1;
      end
  end

I am looking to have defaults for tol and maxit as 1e-10 and 100 respectively.

any help is much appreciated.

Answer 1

[Sarah Wait Zaranek]

I would suggest the following. This way, the input values are reflected. Hardwire the inputs you definitely want, and use varargin to hold the optional inputs.

function [x,converged]=MySolve(f,x0,varargin)

optargin = size(varargin,2);

if optargin == 0
    tol = 1e-10;
    maxit = 100;
elseif optargin == 1
    tol = varargin{1};
    maxit = 100;
else
    tol = varargin{1};
    maxit = varargin{2};
end

[Joseph Hughes]

works perfectly much appreciated :)

[Sarah Wait Zaranek]

No problem. Once you get the hang of it, varargin is sooo useful.

Answer 2

[Jarrod Rivituso]

Actually you could use nargin to do this. Essentially, nargin tells you the number of the arguments, and you can use it add some code at the beginning of your function to compensate for missing inputs.

An example:

function twoInputs(x,y)

if nargin == 0
    x = 1;
    y = 2;
end

if nargin == 1
    y = 2;
end

disp(x)
disp(y)