Essential Books

^ - 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;
}
try online 
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;
}
try online 
a = 12

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

int cn = 0b0111;
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Bit operators:
* | - bitwise OR
* ^ - bitwise XOR (exclusive OR)
* & - bitwise AND
* << - left shift
* >> - right shift
* ~ - bitwise NOT (unary complement)
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