For an integer n: If n is less than 1, return false. If the bitwise & of n and n-1 is 0, return true. Otherwise, return false.

I think the main idea is to use recursion function, for the integer which is larger than 0, if it is 1 return true, else return function(n-1)

function int power2(int i); if (i%2==1) return 0; else if (i/2==1) return 1; else return power2(i/2); endfunction

See if the sum of all bits is 1. If that's the case then the number is a power of 2.

first check if no is 1 then return false else write the number in binary and then check number of ones in that.if only one 1 is there then its a power of 2