C语言实现计算二进制数字1的个数

Posted 2021-01-15 Duikerdd

#include<stdio.h>
#include<stdlib.h>
int print_one_bits01(unsigned int value){                   //0000 1111  
                                                            //&0000 0001   value右移i次 每次和1 &运算
    int count = 0;
    int i = 0;
    for (; i <32; i++)
    {
        if (((value >> i) & 1) == 1)
            count++;
    }
    return count;
}
int print_one_bits02(unsigned int value){                //移位操作符不会改变值大小,需要用=赋值
    int count = 0;                                         //思路:  i=0000 0001
    int i = 1;                                             //       value=0000 1111  &1

    for (i = 1; i != 0;i<<=1)                              //       value>>=0000 0111
    {                                                      //        i=0000 0010 i只是测定是否溢出  value自己右移和1比较
        if ((value & 1) == 1)
        {
            count++;
        }
        value = value >> 1;
    }

    return count;
}
int print_one_bits03(unsigned int value)                                      //每次运算n消去二进制最后一位1
{
    int count = 0;
    while (value!=0){
        value = value&(value - 1);
        count++;
    }
    return count;
}
int main(){
    printf("%d\n", print_one_bits01(15));
    printf("%d\n", print_one_bits02(15));
    printf("%d\n", print_one_bits03(15));
    system("pause");
    return 0;
}