1)
if(!(x& (x - 1)) && x)
{
printf("GIVEN NUMBER IS POWER OF 2");
}
EXPLANATION:
Let n be the position of the leftmost 1 bit if x. If x is a power of two, its lone 1 bit is in position n. This means x – 1 has a 0 in position n. To see why, recall how binary subtraction works. When subtracting 1 from x, the borrow propagates all the way to position n; bit n becomes 0 and all lower bits become 1. Now, since x has no 1 bits in common with x – 1, x & (x – 1) is 0, and !(x & (x – 1)) is true.
2)
if(((~x+1)&x)==i)
{
printf("GIVEN NUMBER IS POWER OF 2");
}
EXPLANATION:
The two’s complement of x is computed with ~x + 1, which inverts the bits of x and adds 1 (~x + 1 is equivalent to -x, but negation is technically illegal for an unsigned integer
Let n be the position of the leftmost 1 bit if x. If x is a power of two, its lone 1 bit is in position n. This means ~x has a 0 in position n and 1s everywhere else. When 1 is added to ~x, all positions below n become 0 and the 0 at position n becomes 1. In other words, the carry propagates all the way to position n. So what happens is this: negating x inverts all its bits, but adding 1 inverts them back, from position n on down. So, (x & (~x + 1)) == x is true.
