How to make bit wise XOR in C

Question

I'm trying to get into C programming, and I'm having trouble writing a bitwise XOR function with only ~ and & operators. Example: bitXor(4, 5) = 1. How can I achieve this?

So far I have this:

int bitXor(int x, int y) {

    return z;
}

Answer 1

Well, let's think about this. What does XOR do?

x   y    XOR
------------
0   0     0
1   0     1
0   1     1
1   1     0

So how do we turn that into a function? Let's think about AND, and the inverse order of AND (~x&~y) (this happens to be NOR):

              (~x&~y)
 x   y   AND    NOR   
 ---------------------
 0 & 0  = 0      1    
 1 & 0  = 0      0 
 0 & 1  = 0      0
 1 & 1  = 1      0

Looking at those two outputs, it's pretty close, all we have to do is just NOR the two previous outputs (x AND y) (x NOR y) and we'd have the solution!

  (a)       (b)    ( a NOR b )
x AND y   x NOR y    ~a & ~b
-------------------------------
   0         1          0
   0         0          1
   0         0          1
   1         0          0

Now just write that out:

a = ( x & y )
b = ( ~x & ~y )
XOR'd result = (~a & ~b)

BINGO! Now just write that into a function

int bitXor(int x, int y) 
{
    int a = x & y;
    int b = ~x & ~y;
    int z = ~a & ~b;
    return z;
}     

Answer 2

Using NAND logic:

int bitNand(int x, int y)
{
    return ~ (x & y);
}

int bitXor(int x, int y)
{
    return bitNand( bitNand(x, bitNand(x, y)),
                    bitNand(y, bitNand(x, y)) );
}

Or:

int bitXor(int x, int y)
{
    return ~( (x & y) | (~x & ~y) );
}

Or:

int bitXor(int x, int y)
{
    return (x & ~y) | (~x & y);
}

Of course this is easier:

int bitXor(int x, int y)
{
    return x ^ y;
}