一个函数返回参数二进制中 1 的个数

方法一：

#include<stdio.h>

int bit_count(unsigned int n)

{

int count;

for (count = 0; n; n &= n - 1)

{

count++;

}

return count;

}

int main()

{

int y;

int c;

printf("请输入一个数：");

scanf("%d", &c);

y = bit_count(c);

printf("%d\n", y);

system("pause");

return 0;

}

方法二：

#include<stdio.h>

int bit_count(unsigned int n)

{

int count=0;

int i = 32;

while (i)

{

if (n & 1 == 1)

{

count++;

n >>= 1;

}

i--;

}

return count;

}

int main()

{

int y;

int c;

printf("请输入一个数：");

scanf("%d", &c);

y = bit_count(c);

printf("%d\n", y);

system("pause");

return 0;

}