Write one line C function to find whether a no is power of two | GeeksforGeeks
Write one line C function to find whether a no is power of two 1.
1. A simple method for this is to simply take the log of the number on base 2 and if you get an integer then number is power of 2.
2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2
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Write one line C function to find whether a no is power of two 1.
If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit become unset.
So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero.
The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) bool isPowerOfTwo (int x){ return x && (!(x&(x-1)));}1. A simple method for this is to simply take the log of the number on base 2 and if you get an integer then number is power of 2.
2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2
bool isPowerOfTwo(int n){ if (n == 0) return 0; while (n != 1) { if (n%2 != 0) return 0; n = n/2; } return 1;}
3. All power of two numbers have only one bit set. So count the no. of set bits and if you get 1 then number is a power of 2. http://www.geeksforgeeks.org/count-set-bits-in-an-integer/int countSetBits(int n)
{
unsigned int count = 0;
while (n)
{
n &= (n-1) ;
count++;
}
return count;
}
