nargin and the if statement
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Hi Im trying to understand how to use nargin but am having an issue. Hopefully someone can shed some light on. I copied this function from my text book
function z =  sqrtfun(x,y)
if (nargin==1)
   z = sqrt(x);
elseif (nargin==2)
   z = sqrt(x+y);
end
Then in another script I have put together this code to call the function
clc;
clear;
close all;
x = input('x');
y = input('y');
z = sqrtfun(x,y);
fprintf('Answer z ::: %4.2f\n', z);
The issue i'm having is that if i leave y blank no value is displayed for z. If i enter a value for x and y i get an output value for z. I don't know why this happens??
0 件のコメント
採用された回答
  Walter Roberson
      
      
 2016 年 9 月 22 日
        When you leave y blank in response to an input() prompt, what you get back is an empty array, and you provide that empty array as an argument. nargin tests the number of arguments passed, not what value they are, so it knows you are passing two arguments. This leads to the calculation
sqrt(x+y)
where x is not empty but y is empty. The sum of a non-empty array and an empty array is the empty array, so the result is empty.
You should modify your code to
function z =  sqrtfun(x,y)
if (nargin==1) || isempty(y)
   z = sqrt(x);
else
   z = sqrt(x+y);
end
2 件のコメント
その他の回答 (1 件)
  KSSV
      
      
 2016 年 9 月 22 日
        You change the function as follows:
function z = sqrtfun(x,varargin)
if (nargin==1) 
    z = sqrt(x); 
elseif (nargin==2) 
    y = varargin{1} ;
    z = sqrt(x+y); 
end
When you enter one value i.e. x nargin = 1 then it goes to if, if you enter two values i.e x,y nargin will be 2. y will be stored in varargin and we are calling it in else statement.
2 件のコメント
  Sultan Al-Hammadi
 2018 年 11 月 27 日
				what does nargin do?
and what is the functionality of "varargin{1}"?
  Steven Lord
    
      
 2018 年 11 月 27 日
				See the description and examples on the nargin function documentation page for more information about what it does and how to use it.
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