How to solve this problem?

I have a doubt in my program logic. I dont know how to access the cells. Can somebody help me.

x1={'1' '0' '0' '0' '1' '0'};
x=hex2dec(x1);
v1={'1' '1' '0' '1' '0' '1'};
v=hex2dec(v1);
y1={'0' '0' '0' '0' '0' '0'};
y=hex2dec(y1);
for j=1
 for i=1
     if x(i) == 0
         Lj = y;
     else x(i) == 1
         Lj = bitxor(y,v);
     end    
 end

Thus, as output I will get Lj= 110101. For i part,x(i), my logic was to access each element in the cell array. For j part,Lj, my logic was to access the entire cell. I want to repeat the steps for each arrays like L1,L2... Can somebody help me to solve this? How can I get access to the complete cell. As output i must have: L1=110101; L2=110100;...and so on

[Bob Thompson]

Basic cell array notation uses parentheses, (), to refer to the cell as a whole, while curly braces, {}, refer to the contents of a cell. If you're looking to move through two for loops where j is cell number, and i is element number your indexing should look something like L{j}(i). This calling says you are interested in the contents of cell j, specifically element i.

If you want to use an entire cell for something L(j) is a possible notation, but most computations do not work with 'cell' variables.

[Darsana P M]

I am still not able to access the entire row of Lj. What is the best option?

[Bob Thompson]

What do you mean by "access"? When I run the code you have written down it seems to recognize and read the cells just fine. It produces output of Lj which is a 5x1 double array, not a cell. If you would like to turn it into a single row (1x5 array) you can do so by transposing Lj = Lj'