How do I get my code to recognise an empty function input?

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Jessica Parry
Jessica Parry am 7 Apr. 2013
Beantwortet: Garret Mans am 23 Mai 2019
function P = poly_input(y)
while
if isempty(y) == 1
y = input('Invalid, no value entered. Enter a value: ', 's');
end
end
end
Basically, i'd like for a re-entry of number message to appear if no value is entered for y!
Please and thanks!
  4 Kommentare
Jessica Parry
Jessica Parry am 7 Apr. 2013
Bearbeitet: Jessica Parry am 7 Apr. 2013
sadly this still isnt working how i hoped it would when typing in the command window the function name and a blank matlab still produces an error;
>> poly_input()
Error using Untitled (line 2)
Not enough input arguments.
Image Analyst
Image Analyst am 7 Apr. 2013
Bearbeitet: Image Analyst am 7 Apr. 2013
See my comment about how to handle nargin (number of input arguments) being zero.

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Antworten (3)

Wayne King
Wayne King am 7 Apr. 2013
isempty() should work
y = input('What is your favorite color? ','s');
if isempty(y)
y = input('Invalid, no value entered. Enter a value: ', 's');
end

Image Analyst
Image Analyst am 7 Apr. 2013
Try this:
while true
% Ask user for a number.
defaultValue = 45;
titleBar = 'Enter a value';
userPrompt = 'Enter the number ';
caUserInput = inputdlg(userPrompt, titleBar, 1, {num2str(defaultValue)});
if isempty(caUserInput) % Doesn't work. Not empty if user input is null.
return;
end; % Bail out if they clicked Cancel.
numericalValue = str2double(cell2mat(caUserInput));
% Check for a valid integer.
if isnan(numericalValue)
% They didn't enter a number.
% They clicked Cancel, didn't enter anything
% or entered a character, symbols, or something else not allowed.
message = sprintf('I said it had to be a number.\nTry again.');
uiwait(warndlg(message));
else
break;
end
end
  3 Kommentare
Matt Butts
Matt Butts am 19 Mai 2016
If I'm not mistaken, this will also require a modification to the function definition:
function P = poly_input(varargin)
if nargin < 1
y = input('Invalid, no value entered. Enter a value: ', 's');
else
y = varargin{1};
end
Steven Lord
Steven Lord am 19 Mai 2016
Nope. This works just fine. [Whether it's preferred or not is a different question. Personally I'd probably have poly_input use narginchk to throw an error if you didn't call it with 1 input.]
function P = poly_input(y)
if nargin < 1
y = input('Enter a value for y: ');
end
P = [y 0];
Call it like this and you will be prompted to enter a value.
P = poly_input()

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Garret Mans
Garret Mans am 23 Mai 2019
I know I'm only a few years late to this thread, but I'm confused why this question got so overcomplicated.
You just need the input and an initial while loop to catch the empty input. Maybe theres something I'm missing but as far as the question asked, this solves the problem just fine:
function gimmiStuff(y)
while isempty(y)
y = input('You gave me empty stuff. Gimmi stuff: ');
end
% Continue with function
.
.
.
end

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