how to do bitxor operation of two 1*255 matrix

h1 =1x255 logical

h3 = 1x255 logical

howto do bitxor of h1 and h3

Answer 1

[Greg]

result = h1 | h3;

Edit: this is logical (bit) or, not xor. As posted elsewhere, simply use the xor function.

[moni sinha]

Error using |

Matrix dimensions must agree.

how to correct it?

and while doing bitxor operation the dimension is changed.

[Guillaume]

Your screenshot clearly shows that h1 is 1x254, and c is 1x256. So yes, you're going to get an error telling you that their size do not match.

Note that |is the or operator, not the xor operator.

[Jan]

The error message is clear: The array sizes are different. Here the variables h1 (logical) and c (double) are concerned. So why do you ask for "h1 =1x255 logical, h3 = 1x255 logical"?

[Greg]

Good catch Guillaume, i kept reading or not xor.

Why are we assuming c is the second argument? The original post explicitly states h1 and h3, both are logical and same size. All following posts are new problems to the original question.

Answer 2

[James Tursa]

It's not entirely clear to me what operation you really want, but if the elements of h1 and h2 represent "bits", then you could just do this:

result = (h1 ~= h2); % equivalent of xor between the elements of h1 and h2

If h1 and h2 don't have the same number of elements, then that is a different problem that you will need to fix before doing the xor operation.

[Guillaume]

Or

result = xor(h1, h2);

[Greg]

Guillaume's comment here should be a separate answer, and accepted.