where is the mistake?

function oo(w) 
   if isstring(w)
   


count =0;

while count<15
  
   
    if strcmp(w(count),'r')
      disp('yes')
        
    
        
  count=count+1;
  
    
    
    end
end
   end

Answer 1

[David Hill]

Why not just:

w='abcderfrgrhhr';
l=w=='r'
l =

  1×13 logical array

   0   0   0   0   0   1   0   1   0   1   0   0   1

Answer 2

[VBBV]

while count<15
  if strcmp(w(count),'r')
    disp('yes')    
    count=count+1; 
  else 
    break;% this will exit from while 
  end 
end

Given while condition is good when the count is incremented continuously inside if statement for given input string. If you provide input string such that it doesn't enter if statement then your program becomes frozen without further increment of count variable.

[abdullah elsalem]

I add the break but still it doesn't work

[VBBV]

function oo(w) 
if isstring(w)  
  count =0;
while count<15
if strcmp(w(count),'r')
  disp('yes')    
  count=count+1; 
else 
  break;% this will exit from 
end 
end
end % end if 
end % end function

Complete closing end for function and if statements

Answer 3

[DGM]

If you want it to do what your code implies:

w = 'a4jkltr6mramlkrffgajr';
oo(w)

function oo(w)
    if isstring(w) % strings are not chars 
        w = char(w);
    end
    if ischar(w)
        for c = 1:numel(w) % don't use a while loop, don't hard-code variable length
            if w(c) == 'r' % don't need strcmp for single char tests
                disp('yes') % if all you print is 'yes'
            else
                disp('no') % then there's no way to know where it occurred
            end
        end
    end
end

If instead you want a function that's actually useful for something, @David Hill already posted a useful answer.

w = 'a4jkltr6mramlkrffgajr';
idx = letterisr(w)

function out = letterisr(w) % use a meaningful name and provide output
    if isstring(w) % strings are not chars 
        w = char(w);
    end
    if ischar(w)
        out = w == 'r';
    end
end