# ^ - bitwise XOR (exclusive OR)

suggest change
``````#include <iostream>
#include <string>

int main(int argc, char **argv) {
int a = 5;     // 0101b  (0x05)
int b = 9;     // 1001b  (0x09)
int c = a ^ b; // 1100b  (0x0C)

std::cout << "a = " << a << ", b = " << b << ", c = " << c << std::endl;
}``````
``````a = 5, b = 9, c = 12
``````

Why

A bit wise `XOR` (exclusive or) operates on the bit level and uses the following Boolean truth table:

``````true OR true = false
true OR false = true
false OR false = false``````

Notice that with an XOR operation `true OR true = false` where as with operations `true AND/OR true = true`, hence the exclusive nature of the XOR operation.

Using this, when the binary value for `a` (`0101`) and the binary value for `b` (`1001`) are `XOR`’ed together we get the binary value of `1100`:

``````int a = 0 1 0 1
int b = 1 0 0 1 ^
---------
int c = 1 1 0 0``````

The bit wise XOR does not change the value of the original values unless specifically assigned to using the bit wise assignment compound operator `^=`:

``````#include <iostream>

int main(int argc, char **argv) {
int a = 5;  // 0101b  (0x05)
a ^= 9;    // a = 0101b ^ 1001b

std::cout << "a = " << a << std::endl;
}``````
``````a = 12
``````

also in 2015+ compilers variables may be assigned as binary:

``int cn = 0b0111;``