How does the for-cycle check its conditions?

I have a problem with using for cycle. Inside the cycle i want to decrease the maximum iteration for the cycle (see the code below).

for i=1:length(numL)% cell array
        numL{i}=sortrows(numL{i},-1); 
        if size(numL{i},1)>1  % if the matrix inside the cell have more than one row
            sz=size(numL{i},1);
            for j=2:sz
                if numL{1,i}(j-1,1)-numL{1,i}(j,1)<SideSize/2 %if the digits are close enough
                    numL{1,i}(j-1,5)=str2num(strcat(num2str(numL{1,i}(j,5)),num2str(numL{1,i}(j-1,5))));%merge digits
                    numL{1,i}(j,:)=[];%remove the row, where the second digit was
                    sz=sz-1;%decrement the value 'j' can have 
                end
            end
        end

          numLL{i}=fliplr(numL{i}(:,5)');%store the vector of numbers into a new array
      end

The problem happens in the 15. row (where there is a number 1 and 11). Basically, this is a post processing cycle after using OCR. In spite of decrementing the maximum value 'j' can have, it reaches 3 in case of a 'previously 3-row-matirx', but after merging the digits into one number, it should end... however, it does not. Or is the whole cycle wrong?

Answer 1

[David Fletcher]

From the docs: Avoid assigning a value to the index variable within the loop statements. The for statement overrides any changes made to index within the loop.

Whilst you are not explicitly changing the loop variable, I suspect the end condition is set as the loop begins execution and will not recalculate on each iteration. If you need to vary the number of iterations of a loop then a while loop would be a better option.

Answer 2

[Jos (10584)]

You cannot change the parameters of the for-loop within the counter, as demonstrated here:

a = 2 ; b = 6 ;
c = 0 ;
for k=a:b    % executes b-a+1 = 5 times 
  c = c + 1 ;
  disp([c k a b]) ;
  a = 0 ; b = 0 ; k = 0 ;
  disp([c k a b]) ;
end

To be flexible use a while loop:

k = 2 ; b = 6 ;
while k < b
   k = k + 1 
   b = b - 1 
end

or perhaps an if-break statement is an option:

for k=1:10
  disp(k)
  if k > 4
     break
  end
end