Can you write a CPP program to displaying XOR truth table using logical AND OR and NOT?

Answer 1

I do not believe this is possible. The only solution I could come up with does not involve either OR or NOT, but it does use the plus and minus operators combined with the logical AND operator (more formally termed addition modulo 2, or addition without carry). The following C++ program demonstrates that the function, XOR(), correctly emulates the logical XOR operator (^).

#include

int XOR( int x, int y )

{

return(( x + y ) - ( x & y ) - (x & y ));

}

int main()

{

int x, y;

std::cout << "Truth table for 0 and 1:\n" << std::endl;

std::cout << "x y ^ XOR" << std::endl;

for( x=0; x<=1; ++x )

for( y=0; y<=1; ++y )

std::cout << x << " " << y << " " << (x^y) << " " << XOR( x,y ) << std::endl;

std::cout << std::endl;

std::cout << "Evaluation of values 0 to 4:\n" << std::endl;

std::cout << "x\ty\t^\tXOR" << std::endl;

for( x=0; x<=4; ++x )

for( y=0; y<=4; ++y )

std::cout << x << "\t" << y << "\t" << (x^y) << "\t" << XOR( x,y ) << std::endl;

std::cout << std::endl;

return(0);

}

Output

Truth table for 0 and 1:

x y ^ XOR

0 0 0 0

0 1 1 1

1 0 1 1

1 1 0 0

Evaluation of values 0 to 4:

x y ^ XOR

0 0 0 0

0 1 1 1

0 2 2 2

0 3 3 3

0 4 4 4

1 0 1 1

1 1 0 0

1 2 3 3

1 3 2 2

1 4 5 5

2 0 2 2

2 1 3 3

2 2 0 0

2 3 1 1

2 4 6 6

3 0 3 3

3 1 2 2

3 2 1 1

3 3 0 0

3 4 7 7

4 0 4 4

4 1 5 5

4 2 6 6

4 3 7 7

4 4 0 0

As you can see, the third and fourth columns are all equal, thus proving the function correctly emulates the logical XOR for values 0 to 4. The same is true for all permutations of x and y.

Note that while it is possible to use AND, OR and NOT when both x and y are guaranteed to be in the range 0 to 1 (and therefore capable of producing the truth table shown above), the function does not correctly emulate the XOR operator when x or y is neither 0 nor 1. You can test this by swapping the XOR function above with the following implementation:

int XOR( int x, int y )

{

return(( x | y ) & !( x & y ));

}