CareerCup

bool isSet = (num & (1<<8)

((a>>8)>> 1) & 1

Doesnt sound like amazon question........2 easy

((a>>8)&1)

a & 2^8

set(int x)
{
int n=9;
n=1<<(9-1);
if((x&n))
printf("Bit not set:1");
else
printf("Bit set:0");
}

int n;
n=1<<8;
if (x&n) then true

return num & 0x00000100

For solving the above algorithm we need to shift the given binary representation of the number 8 to the right and then while we reach on the 9th bit we will check for the condition that if(n&1 == 1) then return true else return false.

#include <stdio.h>

int main()
{
	int number = 0;

	printf("enter any number :");
	scanf("%d", &number);

	printf("\n");
	if (number & (1<<8))
		printf("9th bit set\n");
	else 
		printf("9th bit unset\n");

	return 0;
}