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Possible bug in Parse function input

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I am using findArea as an example in documentation to parse and validate required,optional and pair-value arguments of a function. I have copied findArea function here to demonstrate the possible bug in matlab parse function.
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);
addRequired(p,'width',validScalarPosNum);
addOptional(p,'height',defaultHeight,validScalarPosNum);
addParameter(p,'units',defaultUnits,@isstring);
addParameter(p,'shape',defaultShape,...
@(x) any(validatestring(x,expectedShapes)));
parse(p,width,varargin{:});
a = p.Results.width*p.Results.height;
end
Now I am calling the findArea function as given in documentation:
1) a = findArea(7,3,'shape','square');
As expected Validation of 7 (first input) and 3 (second input) is done with validScalarPosNum. For Name-Value pair, validation of value of "shape" is done with validatestring. This is fine.
2) a = findArea(7,'shape','square');
Now the problem is here. Validation of 7 (first input) is done with validScalarPosNum. However, the validation of second input "shape" is also done with validScalarPosNum. This is possibly bug. Then as in 1) Validation of value of "shape" is done with validatestring. The problem arises when the optional input is defined in the function (in this example "height"), but not passed in the function arguments.
Does anybody has an idea what is going on here?
Thank you very much.

Accepted Answer

Ameer Hamza
Ameer Hamza on 2 Apr 2020
Edited: Ameer Hamza on 3 Apr 2020
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.
  3 Comments
Ameer Hamza
Ameer Hamza on 3 Apr 2020
FWIW, It is possible to get arbitrary length char arrays as optional input. The only limitation now is that it should not be the same as any other parameter.
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,p) ischar(x) && ~ismember(x, p.Parameters); % <---- the optional parameter is char array
addRequired(p,'width',validScalarPosNum);
addOptional(p,'height',defaultHeight,@(x) validScalarPosNum2(x,p));
addParameter(p,'units',defaultUnits,@isstring);
addParameter(p,'shape',defaultShape,...
@(x) any(validatestring(x,expectedShapes)));
parse(p,width,varargin{:});
a = p.Results;
end

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