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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
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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
16 commentaires
Jan
le 4 Avr 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
le 21 Fév 2023
Modifié(e) : DGM
le 4 Déc 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.
Réponse acceptée
John D'Errico
le 23 Mar 2019
Modifié(e) : John D'Errico
le 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.
5 commentaires
Rik
le 5 Août 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.
Plus de réponses (26)
Saurabh Kumar
le 28 Mar 2019
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 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
7 commentaires
Quyen Le
le 20 Juin 2021
Ofc I tried your answer and it worked. but i would want to understand clearly
Walter Roberson
le 20 Juin 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));
mask = x ~= 0;
y(mask) = 1./x(mask);
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
le 8 Sep 2019
function too_young = under_age(age, limit)
if nargin == 1
limit = 21;
end
too_young = gt(limit, age);
0 commentaires
mayank ghugretkar
le 7 Juin 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
5 commentaires
Karina Medina Barzola
le 3 Juin 2021
nargin is a variable? if i change it to an another variable, it sent me an error. what nargin is. Thanks
Nijita Kesavan Namboothiri
le 25 Juin 2019
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
6 commentaires
Kilaru Venkata Krishna
le 2 Mai 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
le 2 Mai 2020
its taking age >= limit as older
.......and ......age<limit as young
Siddharth Joshi
le 23 Avr 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
0 commentaires
sudeep lamichhane
le 27 Avr 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
1 commentaire
Sai Hitesh Gorantla
le 1 Fév 2020
Modifié(e) : Rik
le 17 Juin 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
0 commentaires
mohammad elyoussef
le 4 Avr 2020
function c = under_age(a,b)
if nargin < 2
b = 21;
end
if b > a
c = true;
else
c = false;
end
0 commentaires
maha khan
le 9 Avr 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
1 commentaire
Walter Roberson
le 9 Avr 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
le 7 Mai 2020
function a = under_age(age,limit)
if nargin < 2
limit = 21;
end
a = age < limit;
end
0 commentaires
Aasma Shaikh
le 26 Mai 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
1 commentaire
jaya shankar veeramalla
le 29 Mai 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
0 commentaires
AYUSH MISHRA
le 4 Juin 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
0 commentaires
Keshav Patel
le 8 Juin 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
1 commentaire
DGM
le 21 Fév 2023
This is essentially identical to @Saurabh Kumar's answer, except one of the variable names has been changed.
saurav Tiwari
le 17 Juin 2020
Modifié(e) : Rik
le 17 Juin 2020
function [a]=under_age(n,m)
a=n<m;
if a==1;
fprintf('true')
end
if nargin<2;
m=21;
end
end
2 commentaires
AKASH SHELKE
le 9 Août 2020
Modifié(e) : AKASH SHELKE
le 9 Août 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
0 commentaires
Muhammad Akmal Afibuddin Putra
le 10 Août 2020
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 commentaires
Rik
le 10 Août 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
le 10 Août 2020
too_young = 1 == 1;
You should
doc true
doc false
Omkar Gharat
le 11 Août 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
1 commentaire
Khom Raj Thapa Magar
le 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
0 commentaires
Jessica Trehan
le 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
7 commentaires
Rik
le 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
le 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
le 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
2 commentaires
Walter Roberson
le 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
le 6 Sep 2021
thankx for your comment
i have no idea before that the second switch statement is useless,
VIGNESH B S
le 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
function [answer] = compare(x,y)
if x<y
answer = (1>0);
else
answer = (0>1);
end
end
3 commentaires
VIGNESH B S
le 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
le 14 Oct 2021
But it doesn't look elegant. An elegant program would do
answer = x < y;
instead of the clunky
if x<y
answer = (1>0);
else
answer = (0>1);
end
Alexandar
le 29 Juin 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
4 commentaires
Rik
le 29 Juin 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.
Muhammad
le 23 Juil 2022
%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 commentaires
Muhammad
le 23 Juil 2022
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!
Muhammad
le 23 Juil 2022
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
le 13 Oct 2022
Modifié(e) : MURSHED SK
le 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
1 commentaire
DGM
le 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
le 4 Déc 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
% how about this ?
2 commentaires
Walter Roberson
le 4 Déc 2023
How does this differ from https://www.mathworks.com/matlabcentral/answers/451726-write-a-function-called-under_age-that-takes-two-positive-integer-scalar-arguments-age-that-repres#comment_2280620 other than the fact you tested nargin>1 instead of nargin==2 ?
DGM
le 4 Déc 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.
This question is locked.
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