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Variable Number of Input Arguments

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Sai Hitesh Gorantla
编辑： Walter Roberson 2021-3-29
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

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采纳的回答

Ioannis Andreou
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
...

更多回答（4 个）

Bhaskar R
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
For more information on MATALAB functions
  1 个评论
somnath paul
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

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somnath paul
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
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.

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PaaKwesi Anderson
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
显示 1更早的评论
PaaKwesi Anderson
PaaKwesi Anderson 2020-10-9
Sorry Sir. Please should I delete it?
Fix the formatting? I don't really get you Sir
Rik
Rik 2020-10-9
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 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
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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