How to make bit wise XOR in C

我与影子孤独终老i 提交于 2019-11-28 17:19:46

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;
}     

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;
}

It is easily seen that

x ^ y = (x | y) & ~(x & y)

so it remains to express | by only & and ~. De Morgan's laws tell us

x | y = ~(~x & ~y)

I want it to write it only with ~ and &

That counts for NAND gates, right? After studying this circuit diagram:

int z = ~ ((~(a & ~(a & b)) & (~(b & ~(a & b)));

The same applies for the non-bitwise, i. e. logic one as well, just substitute ! instead of the ~.

MVP

You can perform a bitwise XOR operation in c using the ^ operator.

vaibhav
int xorro(a, b)
{
    if (a == b)
        return 0;
    return 1; 
}
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