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](https://www.mathworks.com/matlabcentral/answers/uploaded_files/269389/as.png)
0 个评论
采纳的回答
Ioannis Andreou
2020-2-1
编辑:Ioannis Andreou
2020-2-1
If you want variable number of inputs use nargin here
function too_young = under_age(age, limit)
if nargin < 2
limit = 21
end
...
0 个评论
更多回答(4 个)
Bhaskar R
2020-2-1
编辑:Bhaskar R
2020-2-1
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 个评论
somnath paul
2020-8-15
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
somnath paul
2020-8-15
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 个评论
Rik
2020-8-15
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.
PaaKwesi Anderson
2020-10-9
编辑:Walter Roberson
2021-3-29
function too_young = under_age(age, limit)
if nargin<2
limit=21;
end
if age<limit
too_young = true;
else
too_young = false;
3 个评论
PaaKwesi Anderson
2020-10-9
Sorry Sir. Please should I delete it?
Fix the formatting? I don't really get you Sir
Ganesh Vanave
2021-1-1
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 个评论
Rik
2021-1-2
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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