Variable Number of Input Arguments

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

    
   
    

   

    
    
    
    

Answer 1

[Ioannis Andreou]

If you want variable number of inputs use nargin here

function too_young = under_age(age, limit)
    if nargin < 2
        limit = 21
    end
    ...

Answer 2

[Bhaskar R]

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

For more information on MATALAB functions

[somnath paul]

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

Answer 3

[somnath paul]

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

[Rik]

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.

Answer 4

[PaaKwesi Anderson]

function too_young = under_age(age, limit)
if nargin<2
    limit=21;
end
if age<limit
    too_young = true;
else
    too_young = false;

[Rik]

Why are you posting a complete solution to a homework question? That only encourages cheating.

You should also fix the formatting.

[PaaKwesi Anderson]

Sorry Sir. Please should I delete it?

Fix the formatting? I don't really get you Sir

[Rik]

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.

Answer 5

[Ganesh Vanave]

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

[Rik]

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?