/*A PROGRAM TO FIND WHAT HAPPENS WHEN n & (-n)....n*-n....n&(1-n) like operations*/

/*A PROGRAM TO FIND WHAT HAPPENS WHEN n & (-n)....n*-n....n&(1-n) like operations*/

#include<stdio.h>

int main()

{

int a=2;

int n=2;

printf("n=%d\n",n);//2

printf("-n=%d\n",-n);//-2

printf("nxn=%d\n",n*-n);//-4

printf("n & n=%d\n",n & n);//2 & 2=0000 0010 & 0000 0010=0000 0010

printf("n & -n=%d\n",n & (-n));//n=2 means 0000 0010 & n=-2 means internally it will take -ve no. as its 2's complement....means invert 2 +1 ....2=0000 0010 ~2=1111 1101 ....~2+1=1111 1110..when it is made bitwise & with 2...i.e.///

//1111 1110 & 0000 0010=0000 0010....so answer is 2..//

//BCZ & MULTIPLIES THE BITS//

if((n & -n)==n)//2 & -2 ==2

printf("I AM RIGHT\n");//CONDITION WILL BE TRUE//

else

printf("YOU ARE WRONG\n");

if((n & (1-n))==n)

printf("1-N  TRUE\n");//CONDITION IS TRUE BCZ,,,n=2 & (1-2)=-1 it takes 2's complement of -ve value ...so ~1+1=-1

//~1=1111 1110 ~1+1=1111 1110+1=1111 1111.....so this 1111 1111 & 2= 1111 1111 & 0000 0010=0000 0010..bcz it multiplies the bits..so ((2 & (1-2))==2) is true

else

printf("FALSE\n");

if((n & 1-n)==n)//CONDITION WILL BE FALSE ..BCZ PRIORITY OF '==' SIGN IS MORE THAN '&' OPERATOR....//

printf("1-N  TRUE\n");

else

printf("FALSE\n");

if((n & (2+~n))==n)//THIS CONDITION IS TRUE BCz...n=2=0000 0010 ~n=1111 1101 +2=1111 1101+0000 0010=1111 1111 &0000 0010==

//======0000 0010==2..//

printf("TRUE\n");

else

printf("FALSE\n");

return 0;

}