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.

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## 8 comments:

we can do it as

int 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)))

{

// power of 2

}

above should be good for any power-of number, not just 2

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Is this possible?

I give u best solution.

Logic- 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();

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