Set default value if no function input given

I want my function to use default values if no input variables are passed. I have done this in the past using the 'nargin' function. I prefer this approach to using the 'isempty' function. For some reason my code now returns an error when I only input 't' because the varargin{} arrays are technically empty. This error makes sense, but I don't understand how to get around it as I thought the if/else statements would take care of that. How can I debug this?

Error: Index exceeds the number of array elements (0).

function [V, varargout] = HH(t, varargin)

if nargin < 1
    V0 = -60;
else 
    V0 = varargin{1};
end

if nargin < 2
    I = 0;
else 
    I = varargin{2};
end

Answer 1

[Stephen23]

You did not count the input t when using nargin. You can include the number of inputs before varargin in the logical comparisons:

if nargin < 2 % not 1 !!!!
    V0 = -60;
else 
    V0 = varargin{1};
end
if nargin < 3 % not 2 !!!
    I = 0;
else 
    I = varargin{2};
end

Or change the operator to <= (assuming exactly one input before varargin):

if nargin <= 1 % not <
    V0 = -60;
else 
    V0 = varargin{1};
end
if nargin <= 2 % not <
    I = 0;
else 
    I = varargin{2};
end

Another approach is to simply count the number of elements in varargin, which means that you can write code that is robust against future changes too, because you can change how many inputs there are before varargin.

if numel(varargin) < 1 % not NARGIN
    V0 = -60;
else 
    V0 = varargin{1};
end
if numel(varargin) < 2 % not NARGIN
    I = 0;
else 
    I = varargin{2};
end

[Anna Jacobsen]

The second approach worked! Thank you!

[Stephen23]

@Anna Jacobsen: please accept my answer if it helped you!