MATLAB: Varargin

varargin

what is varargin used for ?and how to use it? and can I change it to an ordinary input variable?

Best Answer

http://www.mathworks.com/help/techdoc/ref/varargin.html

There are situations in which it is not possible to use any fixed number of arguments, so conversion depends upon what your code does.

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The dot between adaptfilt and lms is actually to depict that lms is a method that belongs to the adaptfilt class (the dot as in other languages is used as a separator between package, class and method names). For example, you can run:

>> which adaptfilt.lms

To see where the definition is located.

So you can directly use the adaptfilt.lms as described here.

MATLAB: Possible bug in Parse function input

This does seem to be a limitation in the implementation of inputParser. Following code show a case where the parser will fail. In this code, the second optional input is a char array.

function a = findArea(width, varargin)
   defaultHeight = 1;
   defaultUnits = 'inches';
   defaultShape = 'rectangle';
   expectedShapes = {'square','rectangle','parallelogram'};
   p = inputParser;
   validScalarPosNum = @(x) isnumeric(x) && isscalar(x) && (x > 0);
   validScalarPosNum2 = @(x) ischar(x); % <---- the optional parameter is char array
   addRequired(p,'width',validScalarPosNum);
   addOptional(p,'height',defaultHeight,validScalarPosNum2);
   addParameter(p,'units',defaultUnits,@isstring);
   addParameter(p,'shape',defaultShape,...
                 @(x) any(validatestring(x,expectedShapes)));
   parse(p,width,varargin{:});
   
   a = p.Results; 
end

now call it like this

>> findArea(7, 'shape', 'square')
Error using findArea (line 14)
No value was given for 'square'. Name-value pair arguments require a name
followed by a value.

I skipped the optional parameter height, but the parser failed to note that.

I guess this is done intentionally by MATLAB to avoid some kind of ambiguity in parsing the inputs. For example, in this case, it is easy to tell since we are using a single optional input. But consider a function that takes several optional inputs and parameters. In that case, it will become difficult for the parser to guess which values are optional inputs and which are parameters.

Note that the parser will only fail if you are taking character array as inputs for optional parameters. I guess the MATLAB expects that the character arrays should be input using parameters.

Best Answer

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