Answer / santhi perumal

#include<stdio.h>
#include<conio.h>

int main()
{
int number;
int position;
int result;

printf("Enter the Number\n");
scanf("%d",&number);
printf("Enter the Position\n");
scanf("%d",&position);

result = (number >>(position-1));

if(result & 1)
printf("Bit is Set");
else
printf("Bit is Not Set");
}

Answer / valli

#include<stdio.h>
main()
{
int n,p;
printf("enter number");
scanf("%d",&n);
printf("enter the position");
scanf("%d",&p);
if(n&(1<<p))
printf("the bit is set");
else
printf("the bit is clear");
}

Answer / kishora v

#include<stdio.h>
#include<conio.h>
main()
{
int n,p,r;
printf("entr the number\n");
scanf("%d",&n);
printf("enter the pos\n");
scanf("%d",&p);
r=(n&(1<<(p-1)));
if(r)
printf("bit is set");
else
printf("bit is not set");
getch();
}


if the question is check the bit if the bit is not set then
set that bit then we use the following code


#include<stdio.h>
#include<conio.h>
main()
{
int p,r;
static int n;
printf("entr the number\n");
scanf("%d",&n);
printf("enter the pos\n");
scanf("%d",&p);
r=(n&(1<<(p-1)));
if(r)
printf("bit is set");
else
{
printf("bit is not set");
n=n|(1<<p-1);
printf("number %d",n);
}
getch();
}

Answer / vignesh1988i

#include<stdio.h>
#include<conio.h>
void main()
{
int m,n,o,p;
printf("enter the number :");
scanf("%d",&m);
printf("enter the bit position for checking it is set or reset :");
scanf("%d",&n);
o=m;
p=o>>(n-1);
p=p&1;
if(p==1)
printf("the bit is set");
else
printf("the bit is reset");
getch();
}

thank u

Answer / abdur rab

#include <stdio.h>

int main ()
{
int variable = 6;
int position = 3;
char array [2][10] = {"NOT SET", "SET"};

// check the wheather the second bit is set
printf ( "\n The bit is %s",
array [ ( variable & ( 1 << (
position - 1 ) ) ) / position ] );
return ( 0 );
}

18)struct base {int a,b; base(); int virtual function1(); } struct derv1:base{ int b,c,d; derv1() int virtual function1(); } struct derv2 : base {int a,e; } base::base() { a=2;b=3; } derv1::derv1(){ b=5; c=10;d=11;} base::function1() {return(100); } derv1::function1() { return(200); } main() base ba; derv1 d1,d2; printf("%d %d",d1.a,d1.b) o/p is a)a=2;b=3; b)a=3; b=2; c)a=5; b=10; d)none 19) for the above program answer the following q's main() base da; derv1 d1; derv2 d2; printf("%d %d %d",da.function1(),d1.function1(),d2.function1 ()); o/p is a)100,200,200; b)200,100,200; c)200,200,100; d)none 20)struct { int x; int y; }abc; you can not access x by the following 1)abc-->x; 2)abc[0]-->x; abc.x; (abc)-->x; a)1,2,3 b)2&3 c)1&2 d)1,3,4

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