Logical XOR operator in C++?

Question

Is there such a thing? It is the first time I encountered a practical need for it, but I don\'t see one listed in Stroustrup. I intend to write:

// Detect wh         


        

Answer 1

The != operator serves this purpose for bool values.

Answer 2

For a true logical XOR operation, this will work:

if(!A != !B) {
    // code here
}

Note the ! are there to convert the values to booleans and negate them, so that two unequal positive integers (each a true) would evaluate to false.

Answer 3

I use "xor" (it seems it's a keyword; in Code::Blocks at least it gets bold) just as you can use "and" instead of && and "or" instead of ||.

if (first xor second)...

Yes, it is bitwise. Sorry.

Answer 4

#if defined(__OBJC__)
    #define __bool BOOL
    #include <stdbool.h>
    #define __bool bool
#endif

static inline __bool xor(__bool a, __bool b)
{
    return (!a && b) || (a && !b);
}

It works as defined. The conditionals are to detect if you are using Objective-C, which is asking for BOOL instead of bool (the length is different!)

Answer 5

Use a simple:

return ((op1 ? 1 : 0) ^ (op2 ? 1 : 0));

Answer 6

There is another way to do XOR:

bool XOR(bool a, bool b)
{
    return (a + b) % 2;
}

Which obviously can be demonstrated to work via:

#include <iostream>

bool XOR(bool a, bool b)
{
    return (a + b) % 2;
}

int main()
{
    using namespace std;
    cout << "XOR(true, true):\t" << XOR(true, true) << endl
         << "XOR(true, false):\t" << XOR(true, false) << endl
         << "XOR(false, true):\t" << XOR(false, true) << endl
         << "XOR(false, false):\t" << XOR(false, false) << endl
         << "XOR(0, 0):\t\t" << XOR(0, 0) << endl
         << "XOR(1, 0):\t\t" << XOR(1, 0) << endl
         << "XOR(5, 0):\t\t" << XOR(5, 0) << endl
         << "XOR(20, 0):\t\t" << XOR(20, 0) << endl
         << "XOR(6, 6):\t\t" << XOR(5, 5) << endl
         << "XOR(5, 6):\t\t" << XOR(5, 6) << endl
         << "XOR(1, 1):\t\t" << XOR(1, 1) << endl;
    return 0;
}