Logical XOR operator in C++?

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.

Proper manual logical XOR implementation depends on how closely you want to mimic the general behavior of other logical operators (|| and &&) with your XOR. There are two important things about these operators: 1) they guarantee short-circuit evaluation, 2) they introduce a sequence point, 3) they evaluate their operands only once.

XOR evaluation, as you understand, cannot be short-circuited since the result always depends on both operands. So 1 is out of question. But what about 2? If you don't care about 2, then with normalized (i.e. bool) values operator != does the job of XOR in terms of the result. And the operands can be easily normalized with unary !, if necessary. Thus !A != !B implements the proper XOR in that regard.

But if you care about the extra sequence point though, neither != nor bitwise ^ is the proper way to implement XOR. One possible way to do XOR(a, b) correctly might look as follows

a ? !b : b

This is actually as close as you can get to making a homemade XOR "similar" to || and &&. This will only work, of course, if you implement your XOR as a macro. A function won't do, since the sequencing will not apply to function's arguments.

Someone might say though, that the only reason of having a sequence point at each && and || is to support the short-circuited evaluation, and thus XOR does not need one. This makes sense, actually. Yet, it is worth considering having a XOR with a sequence point in the middle. For example, the following expression

++x > 1 && x < 5

has defined behavior and specificed result in C/C++ (with regard to sequencing at least). So, one might reasonably expect the same from user-defined logical XOR, as in

XOR(++x > 1, x < 5)

while a !=-based XOR doesn't have this property.

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;
}

The XOR operator cannot be short circuited; i.e. you cannot predict the result of an XOR expression just by evaluating its left hand operand. Thus, there's no reason to provide a ^^ version.

There was some good code posted that solved the problem better than !a != !b

Note that I had to add the BOOL_DETAIL_OPEN/CLOSE so it would work on MSVC 2010

/* From: http://groups.google.com/group/comp.std.c++/msg/2ff60fa87e8b6aeb

   Proposed code    left-to-right?  sequence point?  bool args?  bool result?  ICE result?  Singular 'b'?
   --------------   --------------  ---------------  ---------- ------------  -----------  -------------
   a ^ b                  no              no             no          no           yes          yes
   a != b                 no              no             no          no           yes          yes
   (!a)!=(!b)             no              no             no          no           yes          yes
   my_xor_func(a,b)       no              no             yes         yes          no           yes
   a ? !b : b             yes             yes            no          no           yes          no
   a ? !b : !!b           yes             yes            no          no           yes          no
   [* see below]          yes             yes            yes         yes          yes          no
   (( a bool_xor b ))     yes             yes            yes         yes          yes          yes

   [* = a ? !static_cast<bool>(b) : static_cast<bool>(b)]

   But what is this funny "(( a bool_xor b ))"? Well, you can create some
   macros that allow you such a strange syntax. Note that the
   double-brackets are part of the syntax and cannot be removed! The set of
   three macros (plus two internal helper macros) also provides bool_and
   and bool_or. That given, what is it good for? We have && and || already,
   why do we need such a stupid syntax? Well, && and || can't guarantee
   that the arguments are converted to bool and that you get a bool result.
     Think "operator overloads". Here's how the macros look like:

   Note: BOOL_DETAIL_OPEN/CLOSE added to make it work on MSVC 2010
  */

#define BOOL_DETAIL_AND_HELPER(x) static_cast<bool>(x):false
#define BOOL_DETAIL_XOR_HELPER(x) !static_cast<bool>(x):static_cast<bool>(x)

#define BOOL_DETAIL_OPEN (
#define BOOL_DETAIL_CLOSE )

#define bool_and BOOL_DETAIL_CLOSE ? BOOL_DETAIL_AND_HELPER BOOL_DETAIL_OPEN
#define bool_or BOOL_DETAIL_CLOSE ? true:static_cast<bool> BOOL_DETAIL_OPEN
#define bool_xor BOOL_DETAIL_CLOSE ? BOOL_DETAIL_XOR_HELPER BOOL_DETAIL_OPEN

(A || B) && !(A && B)

The first part is A OR B, which is the Inclusive OR; the second part is, NOT A AND B. Together you get A or B, but not both A and B.

This will provide the XOR proved in the truth table below.

|-----|-----|-----------|
|  A  |  B  |  A XOR B  |
|-----|-----|-----------|
|  T  |  T  |   False   |
|-----|-----|-----------|
|  T  |  F  |   True    |
|-----|-----|-----------|
|  F  |  T  |   True    |
|-----|-----|-----------|
|  F  |  F  |   False   |
|-----|-----|-----------|

Here is how I think you write an XOR comparison in C++:

bool a = true;   // Test by changing to true or false
bool b = false;  // Test by changing to true or false
if (a == !b)     // THIS IS YOUR XOR comparison
{
    // do whatever
}

Proof

XOR TABLE
 a   b  XOR
--- --- ---
 T   T   F
 T   F   T
 F   T   T
 F   F   F

a == !b TABLE
 a   b  !b  a == !b
--- --- --- -------
 T   T   F     F
 T   F   T     T
 F   T   F     T
 F   F   T     F

The proof is that an exhaustive study of inputs and outputs shows that in the two tables, for every input set the result is always the identical in the two tables.

Therefore, the original question being how to write:

return (A==5) ^^ (B==5)

The answer would be

return (A==5) == !(B==5);

Or if you like, write

return !(A==5) == (B==5);

Use a simple:

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

return ((A==5) != (B==5))


+------+------+--------+-----+
| A==5 | B==5 | op(==) | !op |
+------+------+--------+-----+
|    0 |    0 |      1 |   0 |
|    0 |    1 |      0 |   1 |
|    1 |    0 |      0 |   1 |
|    1 |    1 |      1 |   0 |
+------+------+--------+-----+

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.

#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!)