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

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
    

[Abhishek singh]

getting an error

[Matt J]

Not me.

[Adam]

If you are getting an error then post the error message, otherwise how are you expecting people to help?

Also show how you are calling the function since that is rather fundamental to what the result of it will/should be.

In future please put your question in the body of the question. The title is supposed to be a summary, not a truncated version of a pasted homework question!

[Geoff Hayes]

Abhishek - why is limit being set to 21? It is also the second input parameter so whatever you are passing in (if you are doing so) will be overwritten by 21.

[Abhishek singh]

limit has default value of 21

[John D'Errico]

Because you are setting the variable limit to 21 inside the function, it will not matter what value you pass in for limit. limit will ALWAYS be 21. That is not a default that can be overridden as you want to do.

Think about it. Suppose you pass in the number 35 for limit. What will happen?

The very first thing is limit will be replaced by the number 21.

Anyway, if you think you are getting an error, always paste in the COMPLETE ERROR message, everything in red. Show us the complete error text.

[Abhishek singh]

[Narendran S]

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
    

[Walter Roberson]

No, this always uses 21 as the limit no matter what the user passes in.

[Rik]

Comment posted as answer by Natarajan M.A:

i have a question,i am a biggner and new to matlab.why do we put( nargin < 2) and (nargout == 2) in all of these programs.

[Rik]

Have you read the documentation for nargin? One of the requirements of this program is to work for either 1 or 2 inputs, so you need a way to distinguish those.

[Shagun Agarwal]

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 && age<limit
too_young=true;
else
too_young=false;
end

The above code is giving a wrong output at age=20, can someone please help me out in finding the mistake.

[Rik]

Did you step through the code line by line? Then you will notice that that it is possible for the second to last line to run when it shouldn't. It also helps to use the smart indent to see the nesting:

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 && age<limit
    too_young=true;
else
    too_young=false;
end

It is also generally a bad idea to repeat code. If you remove the repetition you will probably clearly see where your mistake is.

[ANSHUMAN SARTHAK MEHER]

function too_young = under_age(age,limit)
if nargin==1
    limit=21;
end
if age<limit
    too_young = true;
else too_young = false;
end
%might be this is the shortest

[Jan]

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]

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"

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.

Answer 1

[John D'Errico]

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.

[John D'Errico]

Please use comments. Moved to a comment by Abhishek singh:

function too_young = under_age(age,limit)
if  age >= 21 || age == limit
    too_young = false;
elseif age < 21 && age < limit
    too_young = true;
elseif age > 21 && age > limit
    too_young  = false;
else 
    too_young = false;
end
    # not getting output for age = 20

[John D'Errico]

If you do not get a result for age = 20, when limit was not provided, then you need to learn about HOW to specify a default for limit. That means you need to provide a new if clause to create the variable limit at the very beginning of your code. They will typically look like this:

if (nargin < 2) || isempty(limit)
  limit = 21;
end

Do you see the difference between what you did initially, when you had ONLY the single line limit=21, with out the if clause around it, and this test?

This test only creates the variable limit when limit either does not exist, or when it was provided as an empty input, so as [].

[Abhishek singh]

getting error for 22

[Amal P]

Why nargain < 2? Why not some other integer?

[Rik]

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.

Answer 2

[Saurabh Kumar]

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

[ARUN KUMAR GUPTA]

function [too_young] = under_age(age,limit)
if nargin<2
    limit=21;
end
if age<limit
      too_young = true;
else too_young = false;
   end

[Rohan Singla]

function [too_young] = under_age(age,limit)
if nargin<2
    limit=21;
end
    too_young =age<limit
end

[Rohan Singla]

why to have a long codes when same function can be performed by small ones

[Andres]

Great question jajajajaj Excellent answer!

[Quyen Le]

Hello, why when i put the first arguement like that it wouldn't go through?

function too_young = under_age(age, limit)
if age < limit;
too_young = true
    if nargin < 2; 
       limit = 21;
   end
else too_young = false
end

Thank you so much. I wonder why?

[Quyen Le]

Ofc I tried your answer and it worked. but i would want to understand clearly

[Walter Roberson]

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.

Answer 3

[mayank ghugretkar]

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

[Ajith Thomas]

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  
if nargin==2
    age<limit
    too_young=true;
elseif age==limit
    too_young=false;
else age>limit
    too_young=false;
end 

why this code is not working?
its not working for 18 vs 18 and 20

[Geoff Hayes]

Ajith - look at your second if statement

if nargin==2
    age<limit
    too_young=true;

Do you really need the nargin check? And the age < limit does nothing. I would just remove this block (the if/elseif/else) and use the one that you have already coded at

if age<limit
    too_young=true;
else age>=limit
    too_young=false;
end  

There should be no reason to use both.

[Ajith Thomas]

thank you goeff :)

[Karina Medina Barzola]

nargin is a variable? if i change it to an another variable, it sent me an error. what nargin is. Thanks

[Rik]

@Karina Medina Barzola

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

Answer 4

[Astr]

function too_young = under_age(age, limit)
if nargin == 1
limit = 21;
end
too_young = gt(limit, age);

Answer 5

[Nijita Kesavan Namboothiri]

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

[Ajith Thomas]

thank you nijita :)

[Chech Joseph]

function [a too_young] = under_age(age,limit)
            a  = true;
     too_young = true;
            
            if nargin < 2
                limit = 21
            end
            
                if nargin == 2 && age <= limit
                    a = true;
                elseif nargin < 2 && age <= limit
                    a = true;
                else
                    a = false;
                end
                
                if nargout == 2 && age < limit
                  too_young = true;
                 else
                  too_young = false;
                end
end
      
    

[Chech Joseph]

Never mind. Its resolved. Used <= instead of just < operator

[Maria Clara Miguel Descendente de Melo Silva]

The problem is the less than or equal to operator. It should be only less than. The right resolution is:

function [a too_young] = under_age(age,limit)
            a  = true;
     too_young = true;
            
            if nargin < 2
                limit = 21
            end
            
                if nargin == 2 && age < limit
                    a = true;
                elseif nargin < 2 && age < limit
                    a = true;
                else
                    a = false;
                end
                
                if nargout == 2 && age < limit
                  too_young = true;
                 else
                  too_young = false;
                end
end

[Kilaru Venkata Krishna]

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]

its taking age >= limit as older

.......and ......age<limit as young

Answer 6

[Siddharth Joshi]

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

Answer 7

[sudeep lamichhane]

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]

correct but how u do it

Answer 8

[Sai Hitesh Gorantla]

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

Answer 9

[mohammad elyoussef]

function c = under_age(a,b)
if nargin < 2
    b = 21;
end
if b > a
    c = true;
else
    c = false;
end

Answer 10

[maha khan]

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]

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.

Answer 11

[Mir Mahim]

function a = under_age(age,limit)
if nargin < 2
    limit = 21;
end
    a = age < limit;
end

Answer 12

[Aasma Shaikh]

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]

Returns a false result regardless of the age if three arguments are passed.

Answer 13

[jaya shankar veeramalla]

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

Answer 14

[AYUSH MISHRA]

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

Answer 15

[Keshav Patel]

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]

This is essentially identical to @Saurabh Kumar's answer, except one of the variable names has been changed.

Answer 16

[saurav Tiwari]

function [a]=under_age(n,m)
a=n<m;
if a==1;
fprintf('true')
end
if nargin<2;
m=21;
end
end

[saurav Tiwari]

awesome

[Rik]

Why did you decide to put the part with nargin at the end?

Answer 17

[AKASH SHELKE]

function too_young = under_age(age,limit)
if nargin<2
    limit = 21;
end
    if age < limit
        too_young = true;
    else 
        too_young = false;
    end
    
   

Answer 18

[Muhammad Akmal Afibuddin Putra]

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

[saurav Tiwari]

Cool bro

[Rik]

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]

    too_young = 1 == 1;

You should

    doc true
    doc false

Answer 19

[Omkar Gharat]

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]

no its not correct for under_age(20)

Answer 20

[Khom Raj Thapa Magar]

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

Answer 21

[Jessica Trehan]

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

[Madan Kc]

no not for under_age(20)

[Rik]

It looks like it works. What line do you think is causing the problem?

Also, if you plan to cheat, you shouldn't complain about how some answers don't work.

[Olha Skurikhina]

function too_young = under_age(age,limit)
    if nargin<2
    limit=21
    end
   too_young = (age>=limit)
end

It should work and it works correctbut still assesment_tool says that "too_young has an incorrect value". I can`t think of idea why.

[Rik]

Look at you code. What happens if age is 40 and limit is 21? What do you expect age>=limit to return?

[Olha Skurikhina]

I expect the value of too_young to be logical 1 and it gives the logical one when I run it in Matlab.

So I do not see why it is an incorrect value.

[Rik]

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]

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.

Answer 22

[amjad ali]

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]

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]

thankx for your comment

i have no idea before that the second switch statement is useless,

Answer 23

[VIGNESH B S]

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

    

[Walter Roberson]

answer = (1>0)

That tells us you already know that comparisons return a value.

What is the reasoning for not just using the value of x<y ?

[VIGNESH B S]

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]

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

Answer 24

[Alexandar]

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]

If you don't understand what nargout does, why did you use it?

That line will also never execute, as the number of output arguments will never be 2.

[Alexandar]

function too_young = under_age(age, limit)
if nargin < 2
    limit = 21
end

too_young = age<limit;

if too_young == true(1);
else too_young == false(0);

end

Somehow what I had before worked, but I changed it to this since it makes much for sense because I only have one output.

[Rik]

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]

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.

Answer 25

[Muhammad]

%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  

[Walter Roberson]

What will your code return if the age is more than the limit? Where do you define your x when the if fails?

[Muhammad]

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]

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

Answer 26

[MURSHED SK]

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]

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.