[C exercise] find the number of 1 of an integer in binary

The first one is

In the decimal system, if we want to get every digit of a number , This number can be iterated （%10） and （/10）.

According to this conclusion , We can know , If you want to get a number of each 2 Hexadecimal number , Then you can cycle this number （%2） and （/2）.

int main()
{
	unsigned int num = 0;
	int count = 0;
	scanf("%u", &num);
	while (num)
	{
		if (num % 2 == 1)
			count++;
		num /= 2;
	}
	printf("%d", count);
	return 0;
}

Because negative numbers are stored in memory in a different way from positive numbers , So we'd better use unsigned shaping to receive and input , In this way, negative numbers can also be found .

The second kind

We can also use the displacement operator to solve the problem

int main()
{
	int num = 0;
	int count = 0;
	scanf("%d", &num);
	for (int i = 0; i < 32; i++)
	{
		if (((num >> i) & 1) == 1)
			count++;
	}
	printf("%d", count);
	return 0;
}

The third kind of

int main()
{
	unsigned int num = 0;
	int count = 0;
	scanf("%u", &num);
	while (num)
	{
		num &= (num - 1);
		count++;
	}
	printf("%d", count);
	return 0;
}

We can also use the third algorithm to judge whether a number is 2 Of n Power .

int main()
{
	int num = 0;
	scanf("%d", &num);
	if ((num & (num - 1)) == 0)
		printf("%d yes 2 Of n Power \n", num);
	else
		printf("%d No 2 Of n Power \n", num);
	return 0;
}

When we take one 2 Of n Second power -1 When , Its binary will produce borrow , It will result in the sum of each digit of the original number -1 Everyone behind is different , This is the time for & Operation will get 0.