Bit xor of two binary strings and conversion into decimal

Hi, I have two large binary strings (e.g 256 bits each), then how to perform bit Xor operation and then convert the answer into decimal number.

str1='11010011110111110000110110111111';
str2='00011111001101011010001111100001';
str3=bitxor(str1,str2);
%%getting error in str3

[James Tursa]

To be clear, you want a decimal number that can't be represented in any native integer type? A 256-bit integer? Or did you just mean you want the result as a vector of 0's and 1's but not character?

[lilly lord]

I want to divide the string into 8 bit chunks and need decimal value between 0-255.

[James Tursa]

See my edited answer.

Answer 1

[James Tursa]

The xor part as a logical vector result

result = str1 ~= str2;

Or if you wanted it as char:

str3 = char((str1 ~= str2) + '0');

For 8-bit chunks turned into decimal numbers:

result = 2.^(7:-1:0) * reshape(str1 ~= str2,8,[]);

[lilly lord]

Thanks this part works well. I am using the function below to convert the binary string into dec but error

function dec1 = b2dec(b1,mn)%mn is the dim of matrix i.e 1*8*8 for 8*8 matrix
b2 = reshape(b1,8,mn)';
dec1 = [];
for i=1:mn
    x = b2(i,:);
    y = bin2dec(x);
    dec1 = [dec1 y];
end

[lilly lord]

Thank u sir

[James Tursa]

This assumes the strings str1 and str2 are rows as your example shows.

[lilly lord]

result = 2.^(7:-1:0) * reshape(str1 ~= str2,8,[]);

when I replace str1~=str2 directly with str3 ,I am getting big numbers . Can i use only the output string and get the numbers with in the range of 0-255

[lilly lord]

Is there a way in which i can use the output string only.

[James Tursa]

So, can you tell me what exactly you have as inputs (size and class) and what exactly you want as an output (size and class)?

[lilly lord]

I was workig on random bits generating and try different methods. Your both answer solved my problem . Thanks once again

%%%%Test random bit generation
I=imread('airplane.png');
I2=imresize(I,[128 128]);figure;imshow(I2)
R=I2(:,:,1);G=I2(:,:,2);
R_r=reshape(R,1,[]);
G_row=reshape(G,1,[]);
G_bits=dec2bin(G_row,8);
R_bits=dec2bin(R_r,8);
Red_bits=R_bits(:)';
Gr_bits=G_bits(:)';
%Extract last four bits
R_newbit=R_bits(:,5:end);
R_finabit=R_newbit(:)';
G_newbit=G_bits(:,5:end);
G_finalbit=G_newbit(:)';
s1=length(R_finabit);%check the size of both strings
s2=length(G_finalbit);

%bit Xor g final bit and r final bit and give the answer in dec number
result = 2.^(7:-1:0) * reshape(R_finabit ~= G_finalbit,8,[]);

Answer 2

[David Hill]

Use java BigInteger

import java.math.*;
a=BigInteger(str1,2);
b=BigInteger(str2,2);
c=a.xor(b);%c will be the decimal BigInteger

[lilly lord]

It gives a big number Sir