# Write a function called under_age that takes two positive integer scalar arguments: age that represents someone's age, and limit that represents an age limit. The function returns true if the person is younger than the age limit. If the second arg

151 views (last 30 days)
Abhishek singh on 22 Mar 2019
Commented: DGM on 4 Dec 2023
function too_young = under_age(age,limit)
limit = 21
if age <= limit
too_young = true;
elseif age >= limit
too_young = false;
else
fprintf('invalid\n')
end
Jan on 4 Apr 2022
This has become a strange thread. We find a pile of suboptimal implementations.
We need 6 lines of code (including the trailing "end"). John has condensed the IF block into 1 line, but without doubt his implementation is clean and exhaustive.
So what is the reason to post a lot of other less elegant and not working versions?
Many beginners suggest:
if limit > age
too_young = true;
else
too_young = false;
end
instead of the compact and efficient:
too_young = limit > age;
DGM on 21 Feb 2023
Edited: DGM on 4 Dec 2023
Sometimes when clearing out tangled dry underbrush, one is overcome with a profound appreciation for the simple elegance of fire. I think I'd better take a break.
If future me (or anyone else) wants to pick up where I left off, here are my notes on a subset of the answers which very closely follow a common form:
378292 the first of the type, has extra return statement, inconsistent indentation
380788 has extraneous ()
427771 randomly indented
428601 indented, but otherwise identical to 427771
423954 all variable names are replaced with letters
442503 has extra isempty() test, logic is flipped but correct
446375 identical to 428601 excepting whitespace (includes console dump)
477187 identical to 428601 excepting whitespace
477322 trying too hard to say "it's totally different"
478066 identical to 428601 excepting whitespace and unused code as comments
1074038 identical to 428601 excepting whitespace and a comment that says "the shortest way"
EDIT: these also constitute a set of trivial variations of a common answer:
Find the ones that look the most like copypasta, see if user posted other answers, find more threads full of samey copypasta, fall down rabbit hole.

John D'Errico on 23 Mar 2019
Edited: John D'Errico on 8 Nov 2020
This is not an error. It is a failure of your code to work as it is supposed to work by your goals.
What did I say? What has EVERYONE said so far? You cannot set a default as you did.
Even though you pass in the number 18 as the limit, the first thing you do inside is reset that value to 21.
Instead, you need to think about how to test to see if limit was provided at all.
Edit: (18 months later)
Now that there are dozens of answers, all of which are lengthy, I'll show how I would write it.
function tooyoung = under_age(age,limit)
% under_age: returns true if the person is under the age limit
% usage: tooyoung = under_age(age,limit)
%
% Arguments: (input)
% age - numeric. As written, age can be scalar, or any size array
%
% limit - optional argument, if not provided, the default is 21
%
% arguments: (output)
% tooyoung - a boolean variable, indicating if age was under the limit.
%
% NO tests are done to verify that age or limit, if provided are valid
% ages itself, or even if they is numeric at all. Better code would
% do these things.
% test for limit having been provided. if not, then the default is 21
if nargin < 2, limit = 21; end
% There is no need to use an if statement. The test itself is the desired result.
tooyoung = age < limit;
end
See that most of my code was actually comments. Help is hugely important, since it allows you to quickly use the code next year when you forget how it was written, or when you forget what the arguments mean.
Think of internal comments as reminders to yourself, for a year from now when you look at the code and need to change the code for some reason, or god forbid, you need to debug it. My recommended target is one line of comment per line of code, or at worst, one line of code per significant thing done in the code.
Comments are free! Good code should look positively green due to all of the comments. My solution code was lengthy only because of all of the comments.
What else?
Remember that white space is hugely important. It makes your code easy to read.
Use intelligent, meaningful variable names. Good mnemonic variable names help to make your code self-documenting, and again, easy to read. When you are scanning through the code, you don't want to continuously go back and be forced to remember what does the variable wxxyy do?
As I said, better code would have included tests to verify that both age and limit, if provided, were actually numeric variables. The best code is friendly code. When it fails, you want the code to fial in a friendly way, telling the person what was seen to be wrong. What you don't want to happen is the code does something screwy and unexpected, or returns some random difficult to understand error message. The best code makes it easy to use that code.
The final test in this code is a vectorized test, in that if you called the code like this:
tooyoung = under_age([12 5 29 75],21)
tooyoung =
1 1 0 0
it will work and return a vector of results. Vectorized code is generally good code, since it allows the user to not be forced to use a loop when they want to use the code many times in a row.
Finally, could I have written code that would have required only one line of code? Thus testing to see if limit was provided, and returning a comparison to age in just one line? Probably, but that would have been unnecessarily complicated, difficult to read and debug. There would have been no gain in the quality of the code or how fast it runs. Good code is simple, easy to read, easy to use, easy to debug.
Amal P on 5 Aug 2020
Why nargain < 2? Why not some other integer?
Rik on 5 Aug 2020
Have you read the documentation for the nargin function to see what it does? It returns the n[umber of] arg[uments you provide as] in[put]. So if it is smaller than 2, that means the limit was not provided as an input argument.

Saurabh Kumar on 28 Mar 2019
Write a function called under_age that takes two positive integer scalar arguments:
1. age that represents someone's age, and
2. limit that represents an age limit.
The function returns true if the person is younger than the age limit. If the second argument, limit, is not provided, it defaults to 21. You do not need to check that the inputs are positive integer scalars.
function x = under_age(age,limit)
if nargin < 2
limit = 21;
if age < limit
x = true;
else
x = false;
end
end
if nargin == 2
if age < limit
x = true;
else
x = false;
end
end
Quyen Le on 20 Jun 2021
Ofc I tried your answer and it worked. but i would want to understand clearly
Walter Roberson on 20 Jun 2021
MATLAB (and nearly all programming languages... but not all) are Procedural Languages, in which the order of statements is important.
Suppose you were climbing a ladder. For the most part, you take one step upward at a time. Now suppose you program it that way,
"Move foot 50 cm higher and put weight on it"
But what about when you get to the top?
"Move foot 50 cm higher and put weight on it"
"If there was no higher rung, don't take that step"
Too late. You already put your weight in mid-air before checking whether there was something to put your weight onto.
You instead need a check like
"If there is a higher rung, move foot 50 cm higher and put weight on it"
Check first, before relying on it being there.
The way you coded
if age < limit
if nargin < 2
limit = 21
end
end
is similar to not having checked for a higher rung existing before putting your weight where it would be.
===
It is better programming practice to check for exceptions first, check to see whether a calculation is going to be valid before doing the calculation.
However, there are some situations in which it is valid to "patch up" if you encounter an exception. For example, better programming practice for "1/x if x is not 0, 0 if x is 0" would be
y = zeros(size(x));
This code never does any division by 0.
But if you are sure you are using IEEE 754, you can count on the fact that IEEE 754 defines that 1/x has some result for 1/0 (rather than crashing the program), and you can instead do a patch-up-later version:
y = 1./x;
y(x == 0) = 0;
The patch-up-later version of the question would be like,
function too_young = under_age(age, limit)
too_young = age < 21;
if nargin > 1
too_young = age < limit;
end
end
Notice that the variable limit is not used until after it has been checked to be sure that it is present.

Astr on 8 Sep 2019
function too_young = under_age(age, limit)
if nargin == 1
limit = 21;
end
too_young = gt(limit, age);

mayank ghugretkar on 7 Jun 2019
this can work too...
A bit compact approach
function too_young = under_age(age, limit)
if nargin < 2
limit = 21;
end
if age < limit
too_young=true;
else
too_young=false;
return
end
end
Karina Medina Barzola on 3 Jun 2021
nargin is a variable? if i change it to an another variable, it sent me an error. what nargin is. Thanks
Rik on 3 Jun 2021
help nargin
NARGIN Number of function input arguments. Inside the body of a user-defined function, NARGIN returns the number of input arguments that were used to call the function. If the function uses an arguments validation block, then only positional arguments provided by the function call are included in this number. Optional arguments not provided by the caller are not included. Name-value arguments are never included, whether provided or not. NARGIN(FUN) returns the number of declared inputs for the function FUN. The number of arguments is negative if the function has a variable number of input arguments. If the function uses an arguments validation block, NARGIN returns the number of declared positional arguments on the function line as a nonnegative value. FUN can be a function handle that maps to a specific function, or a character vector or string scalar that contains the name of that function. See also NARGOUT, VARARGIN, NARGINCHK, INPUTNAME, MFILENAME. Documentation for nargin doc nargin Other functions named nargin fittype/nargin inline/nargin

function too_young= under_age(age, limit)
if (nargin<2)
limit=21;
end
if (age<limit)
too_young=true
else
too_young=false
end
end
Kilaru Venkata Krishna on 2 May 2020
function too_young = under_age(age,limit);
if (nargin<2)
limit=21;
end
if age < limit
too_young=true;
else age >= limit
too_young=false;
end
Kilaru Venkata Krishna on 2 May 2020
its taking age >= limit as older
.......and ......age<limit as young

Siddharth Joshi on 23 Apr 2020
function too_young = under_age(age,limit)
if nargin<2
limit=21;
end
if age<limit
too_young=true;
else
too_young=false;
end
end
too_young = under_age(18,18)
too_young = under_age(20)
too_young =
logical
0
too_young =
logical
1

sudeep lamichhane on 27 Apr 2020
function too_young = under_age(age, limit)
if nargin<2
limit=21;
end
if age < limit
too_young= true;
else
too_young= false;
end
Krashank Kulshrestha on 14 May 2020
correct but how u do it

Sai Hitesh Gorantla on 1 Feb 2020
Edited: Rik on 17 Jun 2020
function [too_young] = under_age(age,limit)
if nargin == 2
if age<limit
too_young = true;
else
too_young = false;
end
elseif nargin<2
if age<=21
too_young = true;
else
too_young = false;
end
end

mohammad elyoussef on 4 Apr 2020
function c = under_age(a,b)
if nargin < 2
b = 21;
end
if b > a
c = true;
else
c = false;
end

maha khan on 9 Apr 2020
function [too_young]= under_age(age,limit)
if (nargin < 2) || isempty(limit)
limit = 21;
end
if age>21
too_young=false;
elseif age < limit
too_young=true;
elseif age==age
too_young=false;
elseif age<=21
too_young=true;
elseif age < age
too_young=false;
elseif age<=21
too_young=true;
else
too_young=true;
end
Walter Roberson on 9 Apr 2020
Suppose the user passes in a limit of 35, such as testing for eligibility to be President of the United States. And suppose the age passed in is 29. Then if age>21 would be if 29>21 and that would be true, so you would declare too_young=false but clearly the answer should be true: 29 < the specified limit.

Mir Mahim on 7 May 2020
function a = under_age(age,limit)
if nargin < 2
limit = 21;
end
a = age < limit;
end

Aasma Shaikh on 26 May 2020
function too_young= under_age (age, limit)
if nargin<2
limit=21;
if (age<limit)
too_young = true;
else
too_young = false;
end
elseif ((nargin==2) && (age<limit))
too_young = true;
else
too_young = false;
end
end
% Copy, paste and Run
DGM on 21 Feb 2023
Returns a false result regardless of the age if three arguments are passed.

jaya shankar veeramalla on 29 May 2020
function too_young = under_age(age,limit)
if (nargin < 2) || isempty(limit)
limit = 21;
end
if age < limit
too_young = true;
else age >= limit
too_young = false;
end

AYUSH MISHRA on 4 Jun 2020
function too_young =under_age(age,limit)
if nargin <2
limit=21;
end
if age<limit
too_young=true;
else
too_young=false;
end
SOLUTION ;
under_age(18,18)
ans =
logical
0
under_age(20)
ans =
logical
1

Keshav Patel on 8 Jun 2020
function too_young =under_age(age,limit)
if nargin<2
limit=21;
if age<limit
too_young=true;
else
too_young=false;
end
end
if nargin ==2
if age<limit
too_young=true;
else
too_young=false;
end
end
DGM on 21 Feb 2023
This is essentially identical to @Saurabh Kumar's answer, except one of the variable names has been changed.

saurav Tiwari on 17 Jun 2020
Edited: Rik on 17 Jun 2020
function [a]=under_age(n,m)
a=n<m;
if a==1;
fprintf('true')
end
if nargin<2;
m=21;
end
end
saurav Tiwari on 17 Jun 2020
awesome
Rik on 17 Jun 2020
Why did you decide to put the part with nargin at the end?

AKASH SHELKE on 9 Aug 2020
Edited: AKASH SHELKE on 9 Aug 2020
function too_young = under_age(age,limit)
if nargin<2
limit = 21;
end
if age < limit
too_young = true;
else
too_young = false;
end

function too_young = under_age(age,limit)
if nargin < 2
limit = 21;
end
if age < limit
too_young = 1 == 1;
else
too_young = 1 ==2;
end
end
##### 3 CommentsShow 1 older commentHide 1 older comment
Rik on 10 Aug 2020
That's a creative way to write true and false.
But why did you decide to post it? It doesn't seem to show a new strategy to solve this problem. Note that Answers is not a homework submission service.
Walter Roberson on 10 Aug 2020
too_young = 1 == 1;
You should
doc true
doc false

Omkar Gharat on 11 Aug 2020
function [too_young] = under_age(age,limit);
% age = input('Enter age of applicant : ')
% limit = input('Enter age limit of applicant : ')
if nargin < 2
limit = 21;
end
if age < limit
too_young = true;
else
too_young = false;
end
end
This is my code .And it is 100% correct
Madan Kc on 2 Oct 2020
no its not correct for under_age(20)

Khom Raj Thapa Magar on 6 Sep 2020
Make sure your indentation is correct while coding
function too_young = under_age(age,limit)
if nargin<2
limit = 21;
if age < limit
too_young = true;
else
too_young = false;
end
end
if nargin == 2
if age < limit
too_young = true;
else
too_young = false;
end
end
Calling functions
>> too_young = under_age(18,18)
>> too_young = under_age(20)
Output:
too_young =
logical
0
too_young =
logical
1

Jessica Trehan on 21 Sep 2020
function too_young=under_age(age,limit)
if nargin<2
limit=21;
if age<limit
too_young=true;
else
too_young=false;
end
end
if nargin==2
if age<limit
too_young=true;
else
too_young=false;
end
end
%THE PERFECT CODE
Rik on 11 Jan 2021
The variable name is 'too_young'. Your code is claiming that 40 is too young if the age limit is 21. Either the variable name is misleading, or your comparator is the wrong way around.
Olha Skurikhina on 11 Jan 2021
Thank you. It is very stupid. I already solved harder problems along the course but this one couldn`t tackle. That was my problem and plus if the age= limit, it is still should return false so '<=' is incorrect.

amjad ali on 6 Sep 2021
function too_young=under_age(age,limit);
switch nargin
case 1
limit=21;
end
if (limit>age);
too_young=true;
else
too_young=false;
end
switch nargin
case 2
end
if limit>age ;
too_young=true;
else
too_young=false;
end
Walter Roberson on 6 Sep 2021
This code fails if nargin is 0 .
This code tests the age against the limit twice, and has a useless second swtich statement that does nothing.
amjad ali on 6 Sep 2021
i have no idea before that the second switch statement is useless,

VIGNESH B S on 13 Oct 2021
function [too_young] = under_age(age,limit)
if nargin == 2
too_young = compare(age,limit);
elseif nargin == 1
too_young = compare(age,21);
end
end
if x<y
else
end
end
##### 3 CommentsShow 1 older commentHide 1 older comment
VIGNESH B S on 14 Oct 2021
answer = 1>0 is same as declaring true .... the question too asks us about returning either true or false base on some condition... also i wanted to have functions in it to look a bit elegant.. so i just declared other function and did calculations!! thanks all no more than that ..
Image Analyst on 14 Oct 2021
But it doesn't look elegant. An elegant program would do
if x<y
else
end

Alexandar on 29 Jun 2022
This worked well for me. Very short but I sort of don't understand why the "nargout" part worked.
function too_young = under_age(age, limit)
if nargin < 2
limit = 21
end
too_young = age<limit;
if nargout == 2
too_young = true(1);
end
Rik on 29 Jun 2022
Why did you add those last lines? Your output is already a logical. Also, false(0) will create an empty array.
And the double = is a comparison. For the if branch that means no code is executed (and the ; is superfluous), and for the else branch that means the too_young variable is compared to an empty array. Since those aren't equal, the result of that is false. That result is not assigned to a variable, so the ; prevents you from seeing it is stored in ans.
Are you using Matlab itself to write this code? There should be several mlint warnings.
Alexandar on 29 Jun 2022
Edited: Alexandar on 29 Jun 2022
Yes, this was done for a matlab coursea assignment that said the code worked perfectly well which confused me. I started coding three weeks ago.

%Help required
Error showing=Output argument "x" (and possibly others) not assigned a value in the execution with "under_age" function.
function x = under_age(age,limit)
if nargin < 2
limit=21;
if age < limit
x=true;
end
end
if nargin==2
if age<limit
x=true;
end
end
##### 3 CommentsShow 1 older commentHide 1 older comment
In the statement above it's not asked for when x fails.That's why I never made statement when x fails. It simply shows nothing on command prompt when x fails.
Help me out and thanks for your response!
function too_young = under_age(age,limit)
if nargin < 2
limit=21;
if age < limit
too_young=true;
else
too_young=false;
end
end
if nargin==2
if age<limit
too_young=true;
else
too_young=false;
end
end
Thanks a lot Mr.Walter Roberson

MURSHED SK on 13 Oct 2022
Edited: MURSHED SK on 13 Oct 2022
This might help:
function too_young = under_age(age, limit)
if nargin <2
limit =21;
end
if age<limit
too_young = true;
else
too_young = false;
end
% the shortest way
DGM on 13 Oct 2022
Again, how is this different from the many existing answers?
Also, this isn't the shortest way. John's answer at the top of the thread is shorter, and is thoroughly explained.

JASON on 4 Dec 2023
function too_young=under_age(age,limit)
if nargin<2
limit = 21 ;
if age<limit
too_young=true;
else
too_young=false;
end
end
if nargin>1
if age < limit
too_young=true;
else
too_young=false;
end
end
Walter Roberson on 4 Dec 2023
DGM on 4 Dec 2023
This is the sixth duplicate of 367872. Trivial changes to variable names, spacing, or contextually-equivalent logical tests don't constitute new information.
Added to the notes I posted above. If I don't prune this sooner, it'll get pruned whenever I decide to clean up the thread -- whenever that might be.

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