Misunderstanding InputParser Validation Behavior

Question

H.A. Camp 2018-5-25

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评论： H.A. Camp 2018-5-26

I seem to be confused as to the proper implementation of the inputParser when it comes to input validation. Consider the following function:

 function out = testInputParser(varargin)
     ip = inputParser;
     ip.addParameter('x', [], @double);
     ip.parse(varargin{:});
     out = ip.Results.x;
 end

These cases work as expected:

 % Should pass:
 >> testInputParser('x', 123) 
 ans =
   123
 % Should fail:
 >> testInputParser('x', '123')
 Error using testInputParser (line 5)
 The value of 'x' is invalid. It must satisfy the function:
 double.

...But then why do these cases NOT work?

 % Expected to pass, but fails
 testInputParser('x', -123)
 Error using testInputParser (line 5)
 The value of 'x' is invalid. It must satisfy the function:
 double.
 % Expected to fail, but passes
 >> testInputParser('x', 'a')
 ans =
 a

I can avoid some of this heartburn by using @isnumeric for my input validator, but I also want to ensure I've got doubles and not, say, int16 values coming in.

So, this seems like strange behavior to me. What am I missing?

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Answer 1

Walter Roberson 2018-5-25

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@double is not an input validation function: it just takes the input and converts to double datatype. Whatever test is being done internally appears to accept positive results from the validation function.

You should be using a validation function such as @(x) isa(x, 'double') or @isfloat if single precision is also acceptable.

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H.A. Camp 2018-5-26

Ah, of course! Thanks, Walter, for your patience with such a simple problem. :) Somehow, I fooled myself into thinking that the act of type conversion would inherently perform the check when that's not really what it's intended for.

Much appreciated!

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