How to do xor operation?

How to do xor operation in cell array. Suppose i have bits x={'1' '1' '1' '0' '1' '1' '0' '1'}; and v={'1' '1' '0' '0' '1' '0' '1' '1'}; Should i use cell function? Can somebody help me with the correct code.

[Stephen23]

@Darsana P M: why are you storing this data in cell arrays? Your code would be a lot simpler and more efficient if you stored character data in a char array, or used logical/numeric arrays. Cell arrays do not seem to serve any useful purpose here.

Answer 1

[Birdman]

x={'1' '1' '1' '0' '1' '1' '0' '1'};
v={'1' '1' '0' '0' '1' '0' '1' '1'};
for i=1:1:length(x)
    xx(i)=str2double(cell2mat(x(i)));
    vv(i)=str2double(cell2mat(v(i)));
    if(xx(i)==vv(i))
        XOR(i)=0;
    else
        XOR(i)=1;
    end
end

Hope this helps.

[Stephen23]

See my answer for a much simpler and more efficient solution

[Birdman]

Checked it and saw that it is more efficient but it is an already defined function. If he tries to understand my way, he will learn and understand the algorithm of the xor gate.

[Darsana P M]

thanks a lot sir

[Jan]

@Çevikalp Sütunç: In the line:

xx(i)=str2double(cell2mat(x(i)))

the cell2mat can be replaced by a simpler cell indexing with curly braces:

xx(i) = str2double(x{i})

But str2double works on cell strings directly also:

xx(i) = str2double(x(i))

or better outside the loop:

xx = str2double(x)

Answer 2

[Stephen23]

Ugly loops are not required, this is all you need:

>> x = {'1' '1' '1' '0' '1' '1' '0' '1'};
>> v = {'1' '1' '0' '0' '1' '0' '1' '1'};
>> xor([x{:}]-'0',[v{:}]-'0')
ans =
   0   0   1   0   0   1   1   0

If you had stored your data in a simpler character array, then all you would need is this:

>>  x = '11101101'; v = '11001011';
>> xor(x-'0',v-'0')
ans =
   0   0   1   0   0   1   1   0

Compare with Çevikalp Sütunç's answer: which one is going to be more efficient, is easier to understand the purpose of, and will be easier to debug?

[Darsana P M]

Thanks a lot sir. I think this is much more simpler. I took them as cell array because i was dealing with hexa decimal values. I was trying out this with small values.

So this gives xor of x and v. If suppose i have few vectors as x1={'1' 1' 0 1 1 0 1 0} x2={ 1 1 0 0 1 1 1 0} upto xn same for v1,v2 and so on. So while applying loops how can i access the entire set of elements and find the xor of each. ie xor(x1,v1), xor(x2,v2) etc

Answer 3

[KL]

Another method maybe,

 x = {'1' '1' '1' '0' '1' '1' '0' '1'};
 v = {'1' '1' '0' '0' '1' '0' '1' '1'};
 res = logical(zeros(size(x)));
 res(str2double(x)~=str2double(v)) = logical(1);

As Stephen says, having a cell array here is unnecessary.

[Jan]

Or directly:

res = (str2double(x) ~= str2double(v));

[KL]

Oh yeah! Thanks Jan.

[Darsana P M]

If suppose i have few vectors as x1={'1' 1' 0 1 1 0 1 0} x2={ 1 1 0 0 1 1 1 0} upto xn same for v1,v2 and so on. So while applying loops how can i access the entire set of elements and find the xor of each. ie xor(x1,v1), xor(x2,v2) etc

[Stephen23]

"If suppose i have few vectors as x1={'1' 1' 0 1 1 0 1 0} x2={ 1 1 0 0 1 1 1 0} upto xn..."

Do not do this. Magically accessing variables names is complex, slow, buggy, and hard to debug. Read this to know why:

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval

Much neater, faster, and more efficient would be if you simply stored all of your data in one array (which could be an ND numeric array, or a cell array). Using indexing is simple, fast, easy to debug, and very efficient.

Answer 4

[Jan]

It is strange, that the inputs are char's, but there is no need to convert them:

x = {'1' '1' '1' '0' '1' '1' '0' '1'};
v = {'1' '1' '0' '0' '1' '0' '1' '1'};

L = {'0', '1'};
Result = L(~strcmp(x, v) + 1)

Answer 5

[Christina Christofidi]

Hey, I have an employment about the network coding. I need to run a file with some numbers and do XOR between them. You can help me????

[Jan]

Please do not attach a new question in the section for answers of another question. Create your won question instead. By the way, this seems to be a job for a simple xor comand.