Determines whether a number is a power of 2

Given an integer n, determine if it is a power of 2.

Idea: 2,4,8,16,32 .... are 2 of the power of n times

Conversions to binary are:

10 100 1000) 10000 100000

These numbers minus 1 after the bitwise and with themselves, if the result is 0, indicating that the number is 2 of the power of N

01 011 0111) 01111 011111

10&01 = 0 100&011 = 0 1000&0111 = 0 10000&01111 = 0 100000&011111 = 0

Source:

#include <stdio.h>/

* Determines whether an integer is a power of 2 *
/bool Fun (int v)
{
	bool flag = 0;
	if ((v>0) && (v& (V-1)) ==0)
		flag = 1;
	return flag;
}

int main (void)
{
	int A;
	printf ("Please enter a 32-bit integer:");
	scanf ("%d", &a);
	if (fun (a))
		printf ("This number is 2 power \ n");
	else
		printf ("This number is not 2 power \ n");
	return 0;
}

Execution Result: