一个函数返回参数二进制中1的个数

 一个函数返回参数二进制中1的个数

第一种境界

int count_one_bits(unsigned int value)
{
	int count = 0;
	while (value)
	{
		if (value % 2 == 1)
			count++;
		value = value / 2;
	}
	return count;
}

#include<stdio.h>
int main()
{
	int ret = 0;
	unsigned int num = 0;
	printf("请输入一个数:");
	scanf_s("%d",&num);
	ret = count_one_bits(num);
	printf("二进制中1的个数：%d\n",ret);
	system("pause");
}

第二种境界

int count_one_bits(int value)
{
	int count = 0;
	int i = 32;
	while (i)
	{
		if (value & 1 == 1)
			count++;
	}
	value = value >> 1;
	i--;
	return count;
}

#include<stdio.h>
int main()
{
	int ret = 0;
	int num = 0;
	printf("请输入一个数:");
	scanf_s("%d", &num);
	ret = count_one_bits(num);
	printf("二进制中1的个数：%d\n", ret);
	system("pause");
}

第三种境界

#include<stdio.h>

int count_one_bits(int value)
{
	int count = 0;
	while (value)
	{
		count++;
		value = value&(value - 1);
	}
	return count;
}

int main()
{
	int ret = 0;
	int num = 0;
	printf("请输入一个数:");
	scanf_s("%d", &num);
	ret = count_one_bits(num);
	printf("二进制中1的个数：%d\n", ret);
	system("pause");
}