about bitwise

[muhammad92]

who can explain about these:
(6*6+6)^(2-9%4) = 43 ?

who can explain to me what is exclusive bitwise OR ^

(6*6+6) = 42
(2-9%4) = 1
why the answer become 43 ??

and what is the differences between bitwise inclusive OR (|) andn bitwise exclusive OR (^) ?

[vlad from moscow]

To understand how the exclusive OR operator works try this code

#include <iostream>

int main()
{
	std::cout << "0 ^ 0 = " << ( 0 ^ 0 ) << std::endl;
	std::cout << "1 ^ 0 = " << ( 1 ^ 0 ) << std::endl;
	std::cout << "0 ^ 1 = " << ( 0 ^ 1 ) << std::endl;
	std::cout << "1 ^ 1 = " << ( 1 ^ 1 ) << std::endl;
}

Value of sub-expression (6*6+6) is equal to 42.
Value of sub-expression (2-9%4) is equal to 1

So in the last bits of the values we will have 0 (for 42) and 1 ( for 1 )

0 ^ 1 gives 1

All other bits of result value will not be changed. So we will get 43.

As for the second question then try the code I suggested with the operator | and look the result.

[Chervil]

Understanding bitwise operations on decimal values such as 42 is hard to visualise. It helps to first convert it to binary, for example using the online tool here: http://www.mathsisfun.com/binary-decimal-hexadecimal-converter.html

As for the operations, there's a tutorial here: http://www.cprogramming.com/tutorial/bitwise_operators.html

[Stewbond]

The bitwise or will return the following (just a normal or)

   0011
or 1001
   1011

The bitwise xor will return:

    0011
xor 1001
    1010

Effectively when you xor two bits, the output is 1 if they are different and 0 if they are the same. When you use the ^ operator, you are xor'ing every bit in the numbers.

42 is 0010 1010

1 is 0000 0001

So in your equation
42 ^ 1 = 43
becomes

    0010 1010
xor 0000 0001
    0010 1011

0010 1011 is binary for 43

Note, the ^ operator is NOT the exponent operator.

[Lowest0ne]

The bitwise operators deal with bits, so it is better to visualize bits rather than integers.

For example, the number 5 is: 0101
and the number 6 is : 0110

The result of OR (|) has every bit that is 1 "turned on"

0101 | 0110 = 0111 -- 5 | 6 = 7

The result of XOR (^) has every bit that is 1 that is 0 in the other "turned on"

0101 ^ 0110 = 0011 -- 5 ^ 6 = 3

[muhammad92]

so, for dealing with ^ and |
we ust convert give values to base 2 and then make the operation right?