inputParser with only a value, not a name-value pair

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Gregory Dachner
Gregory Dachner el 13 de Jul. de 2018
Comentada: Fangjun Jiang el 18 de Jul. de 2018
Is it possible to use inputParser with only a value, not a name-value pair. So that when you send a function a specific argument, the argument is automatically set to a variable, without a variable name needed to be given in the function call?
Instead of
a = findArea(13,'shape','square');
You could just send findArea 'square', like:
a = findArea(13,'square');
And within the function, it would recognize that square can only be a shape and set shape's value to 'square'.

Respuestas (1)

Fangjun Jiang
Fangjun Jiang el 13 de Jul. de 2018
Yes. You would make that a required argument, not an optional argument.
function a = findArea(width,Shape,varargin)
....
addRequired(p,'shape',defaultShape,... @(x) any(validatestring(x,expectedShapes)));
  2 comentarios
Gregory Dachner
Gregory Dachner el 17 de Jul. de 2018
I'm not sure what you expect this to do, but it breaks the function, causes all inputs to return the error:
Error using inputParser/addRequired Too many input arguments.
Fangjun Jiang
Fangjun Jiang el 18 de Jul. de 2018
should be
addRequired(p,'shape', @(x) any(validatestring(x,expectedShapes)));

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