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 when exactly one of A B is equal to five  return  A  5      B  5     But there is no    operator  Can I use the bitwise   here and get the right answer  regardless of machine representation of true and false   I never mix  amp  and  amp  amp   or   and     so I hesitate to do that with   and       I d be more comfortable writing my own bool XOR bool bool  function instead

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if defined   OBJC         define   bool BOOL      include  lt stdbool h gt       define   bool bool  endif  static inline   bool xor   bool a    bool b        return   a  amp  amp  b      a  amp  amp   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

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Proper manual logical XOR implementation depends on how closely you want to mimic the general behavior of other logical operators     and  amp  amp   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  amp  amp   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  amp  amp  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  gt  1  amp  amp  x  lt  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  gt  1  x  lt  5    while a   -based XOR doesn t have this property

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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  lt iostream gt   bool XOR bool a  bool b        return  a   b    2     int main         using namespace std      cout  lt  lt   XOR true  true   t   lt  lt  XOR true  true   lt  lt  endl           lt  lt   XOR true  false   t   lt  lt  XOR true  false   lt  lt  endl           lt  lt   XOR false  true   t   lt  lt  XOR false  true   lt  lt  endl           lt  lt   XOR false  false   t   lt  lt  XOR false  false   lt  lt  endl           lt  lt   XOR 0  0   t t   lt  lt  XOR 0  0   lt  lt  endl           lt  lt   XOR 1  0   t t   lt  lt  XOR 1  0   lt  lt  endl           lt  lt   XOR 5  0   t t   lt  lt  XOR 5  0   lt  lt  endl           lt  lt   XOR 20  0   t t   lt  lt  XOR 20  0   lt  lt  endl           lt  lt   XOR 6  6   t t   lt  lt  XOR 5  5   lt  lt  endl           lt  lt   XOR 5  6   t t   lt  lt  XOR 5  6   lt  lt  endl           lt  lt   XOR 1  1   t t   lt  lt  XOR 1  1   lt  lt  endl      return 0

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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

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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

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Use a simple   return   op1   1   0     op2   1   0

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The    operator serves this purpose for bool values

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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

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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 lt bool gt  b    static cast lt bool gt  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  amp  amp  and    already     why do we need such a stupid syntax  Well   amp  amp  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 lt bool gt  x  false  define BOOL DETAIL XOR HELPER x   static cast lt bool gt  x  static cast lt bool gt  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 lt bool gt  BOOL DETAIL OPEN  define bool xor BOOL DETAIL CLOSE   BOOL DETAIL XOR HELPER BOOL DETAIL OPEN

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A    B   amp  amp    A  amp  amp  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      ----- ----- -----------

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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  amp  amp  and  or  instead of      if  first xor second       Yes  it is bitwise  Sorry

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