To find whether a number is in power of 2 using pow()

A number is in power of 2, when 2 raised to some power is equal to that number.

Is 32 in power of 2? yes
2 raised to power 5 equals 32.

Is 256 in power of 2? yes
2 raised to power 8 equal 256

Is 24 in power of 2? no

In C, header file math.h provide us pow() function
pow(x,y) -- signifies x raised to power y

So we can easily check if number is in power of two
Steps
1.Enter number
2.Set flag=0
3.Set check=0
4.Set i=0
5.loop until check < number
   1.check = pow(2,i) [ 2^i]
   2. if check equals number then Set flag equal to 1 and break from loop goto step 6
   3. increment i = i+1
6.if flag equals to 1 then number is 2 raised to power i
   else it is not.

For example
number=4
flag=check=i=0

check< 4 [ 0<4  yes. Go inside loop]
check = pow(2,0) [2^0=1]
check == number [no]
i=i+1 [0+1=1]

check< 4 [ 1<4  yes Go inside loop]
check = pow(2,1) [2^1=2]
check == number [no]
i=i+1 [1+1=2]

check< 4 [ 2<4  yes Go inside loop]
check = pow(2,2) [2^2=4]
check == number [yes] set flag=1, break from loop-goto step 6

is flag == 1 [yes]
print(number is 2 raised to power 2( value of i=2 when we break from loop)


#include<stdio.h>
#include<math.h>

int main()
{

    int i, num,check, flag;
    printf("\nEnter number:");
    scanf("%d",&num);

    i=0;
    check=0;
    flag=0;

    while(check<num)
    {
        check=pow(2,i);
        if(check==num)
        {
            flag++;
            break;
        }
        i++;
    }
    if(flag==1)
        printf("%d is 2 raised to power %d",num,i);
    else
        printf("Number is not in power of 2 ");

    return 0;
}