Variable Number of Input Arguments

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My code:
function [too_young] = under_age(age,limit)
if age<21
too_young = true;
else
too_young = false;
end
if age<limit
too_young = true;
else
too_young = false;
end
Getting error when executing
too_young = under_age(20):
Not enough input arguments.
Error in under_age (line 7)
if age<limit
as.png

Accepted Answer

Ioannis Andreou
Ioannis Andreou on 1 Feb 2020
Edited: Ioannis Andreou on 1 Feb 2020
If you want variable number of inputs use nargin here
function too_young = under_age(age, limit)
if nargin < 2
limit = 21
end
...

More Answers (4)

Bhaskar R
Bhaskar R on 1 Feb 2020
Edited: Bhaskar R on 1 Feb 2020
You need provide two inputs but you have provided only one input 20. Give inputs functions as age and limit
[too_young] = under_age(15,21);
%Where 15 is age, 21 is limit
  1 Comment
somnath paul
somnath paul on 15 Aug 2020
In the instruction, it is clearly given that if limit is not given as a input argument then place the default value as 21.
If Code to call my function is
too_young = under_age(20)
Then the function should take the default value now the problem is how to place the default value for matlab.
function too_young = under_age(age,limit)
if nargin<2 && limit == 21
too_young = true
else
too_young = false
end
my answer in above but still they show me errors

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somnath paul
somnath paul on 15 Aug 2020
function too_young = under_age(age,limit)
if nargin == 1
% If number input argument is one then we will consider nargin as 1, that's why nargin == 1.
limit = 21;
if limit > age
too_young = true;
else
too_young = false;
end
end
if nargin == 2
% If number input argument is two then we will consider nargin as 2, that's why nargin == 2.
if limit > age
too_young = true;
else
too_young = false;
end
end
  1 Comment
Rik
Rik on 15 Aug 2020
As you expressed in your comment, the reasoning is to set a value for the limit if it isn't provided. Therefore it doesn't make sense to duplicate the rest of the code as well.

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PaaKwesi Anderson
PaaKwesi Anderson on 9 Oct 2020
Edited: Walter Roberson on 29 Mar 2021
function too_young = under_age(age, limit)
if nargin<2
limit=21;
end
if age<limit
too_young = true;
else
too_young = false;
  3 Comments
Rik
Rik on 9 Oct 2020
If you don't have a good reason to keep this solution, yes, I think you should delete it.
For how to fix the formatting: have a read here.

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Ganesh Vanave
Ganesh Vanave on 1 Jan 2021
function too_young = under_age(age,limit)
if nargin == 1
limit = 21;
if limit > age
too_young = true;
else
too_young = false;
end
else nargin == 2
if limit > age
too_young = true;
else
too_young = false;
end
end
  1 Comment
Rik
Rik on 2 Jan 2021
This is a suboptimal setup. You are repeating code, which means any change in algorithm will require you to remember to change two places.
Also, why did you post this answer? What does it teach?

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