Detecting a number if it is a power of 2

If(a&-a==a)

Then the number is a power of 2.

It can be proved easily.

Let a=4, then a=00000100 (in 8 bit)

-a is the 2’s complement of a, then, -a=11111100

If we ANDing them then it only produce 1 in the 6^th bit and all other is 0. It will be 00000100, that is the number a, and we know 4  it is a power of 2.

But if the number is otherwise, then it can’t results the same number. Because it has more 1 in it’s bit representation, that will be changed in 2’s complement and doesn’t produce the same number while ANDing. Try it yourself J

There is another way to find whether a number is a power of 2 or not. If there is only 1 in a bit representation of a positive number then it is a power of two.

Like,

2= 10

4=100

8=1000

16=10000

See, there is only one 1 in bit representation of this numbers.

We can detect it by this sample code:

#include<bitset>

bitset<16>num;

                               

scanf(“%d”,&num);

ans=num.count();

if(ans==1)

printf("yes");

else 
           printf("no");


I found another process from zobayer vai's blog.
The process is: (just pasting from the post of zobayer vai )


Is a number a power of 2?
How can we check this? of course write a loop and check by repeated division of 2. But with a simple bit-wise operation, this can be done fairly easy.
We, know, the binary representation of p = 2ⁿ is a bit string which has only 1 on the n^thposition (0 based indexing, right most bit is LSB). And p-1 is a binary number which has 1 on 0 to n-1 th position and all the rest more significant bits are 0. So, by AND-ing p and (p-1) will result 0 in this case:

p   = ....01000 ⇒ 8

p-1 = ....00111 ⇒ 7

-----------------------

AND = ....00000 ⇒ 0

No other number will show this result, like AND-ing 6 and 5 will never be 0.