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??
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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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