Variable Number of Input Arguments
조회 수: 8 (최근 30일)
이전 댓글 표시
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
댓글 수: 0
채택된 답변
Ioannis Andreou
2020년 2월 1일
편집: Ioannis Andreou
2020년 2월 1일
If you want variable number of inputs use nargin here
function too_young = under_age(age, limit)
if nargin < 2
limit = 21
end
...
댓글 수: 0
추가 답변 (4개)
Bhaskar R
2020년 2월 1일
편집: Bhaskar R
2020년 2월 1일
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
댓글 수: 1
somnath paul
2020년 8월 15일
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
2020년 8월 15일
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
댓글 수: 1
Rik
2020년 8월 15일
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
2020년 10월 9일
편집: Walter Roberson
2021년 3월 29일
function too_young = under_age(age, limit)
if nargin<2
limit=21;
end
if age<limit
too_young = true;
else
too_young = false;
댓글 수: 3
PaaKwesi Anderson
2020년 10월 9일
Sorry Sir. Please should I delete it?
Fix the formatting? I don't really get you Sir
Ganesh Vanave
2021년 1월 1일
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
댓글 수: 1
Rik
2021년 1월 2일
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?
참고 항목
카테고리
Help Center 및 File Exchange에서 Characters and Strings에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!