Even this is one of the most frequently asked interview questions. I really dont know whats so great in it. Nevertheless, here is a C program
Method1
if(!(num & (num - 1)) && num)
{
// Power of 2!
}
Method2
if(((~i+1)&i)==i)
{
//Power of 2!
}
I leave it up to you to find out how these statements work.
we can do it as
ReplyDeleteint findpower(unsigned char x)
{
return !((x-1)&x)
}
if its true then its power of 2 else its not....
if((double)((int)(log(n)*1.0/log(2))) == (log(n)*1.0/log(2)))
ReplyDelete{
// power of 2
}
above should be good for any power-of number, not just 2
ReplyDeleteI give u best solution.
ReplyDeleteLogic- when u convert a decimal number to its binary number then u get series of 1 and 0
Now if a number is power of two then u will observe it has only one (bit 1) and it u do binary anding of ot it with just a number previous to it ,then u will get Zeros
For example
8=1000(binary form ) here u can see only one (bit 1) is there rest are zeros
Now the number just previous to 8 is 7
so, 7=0111 and if u perform anding operation on it as
1000(8)
(AND) 0111(7)
---------
0000(0)
---------
So if we perform AND operation between the number (which we want to check is a power of two or not) and the number just previous to it and result comes as zeros then its power of two otherwise it s not.
PROGRAM(IN C_LANGUAGE)
#include
#include
main()
{
int num, num1,sum;
clrscr();
printf("Enter the number");
scanf("%d",&num);
printf("\nEntered number=%d",num);
num1=num-1;
sum=num & num1;
if(sum==0)
printf("\n Entered number is power of two");
else
printf("\n Entered number is not power of two");
getch();
This really answered my problem, thanks! gsn casino slots
ReplyDelete