Question

我想编写一个程序来比较两个数字得到1位数。我想比较任意两个数字之间的比特，以找出二进制数在1和0和0中的不同之处。换句话说，＆＃34;独家OR＆＃34; （异）关系。

喜欢22（具有10110二进制）并将其与15（具有01111二进制）进行比较

第一个是10110。第二个是01111。

结果：11001。

答案是25，但我想要的是3，其中有三个1和0 s不同。

Answer 1

要查找不同的位，您需要XOR值：

unsigned int first  = 22;
unsigned int second = 15;
unsigned int result = first ^ second; // only the bits that are different
                                      // will be set to 1 in result

要计算result中的1位，您可以使用：

unsigned int CountBits(unsigned int n) {

   unsigned int count = 0;
   while(n) {
       count += n & 0x01; // checks the least significant bit of n
                          // if the bit is 1, count is incremented
       n >>= 1; // shift all bits of n one to the right
                // if no 1 bits are left, n becomes 0 and the loop ends
   }
   return count;
}

unsigned int count = CountBits(result);

在一步中执行此操作：

unsigned int count = CountBits(first ^ second);

在某些系统上，可以使用POPCNT指令。

更新 - 完整示例：

#include <stdio.h>

unsigned int CountBits(unsigned int n) {

    unsigned int count = 0;
    while(n) {
        count += n & 0x01;
        n >>= 1;
    }
    return count;
}

int main(void) {

    unsigned int first  = 22;
    unsigned int second = 15;
    unsigned int result = first ^ second;

    unsigned int count = CountBits(result);

    printf("result: %u - count: %u\n", result, count);

    return 0;
}

打印：

result: 25 - count: 3

或者，具有额外功能：

#include <stdio.h>

unsigned int CountBits(unsigned int n) {

    unsigned int count = 0;
    while(n) {
        count += n & 0x01;
        n >>= 1;
    }
    return count;
}

unsigned int CountDifferentBits(unsigned int n1, unsigned int n2) {

    return CountBits(n1 ^ n2);
}

int main(void) {

    unsigned int count = CountDifferentBits(22, 15);

    printf("Different bits count: %u\n", count);

    return 0;
}

Answer 2

int count_bit(unsigned int x, unsigned int y){
int x1[31], y1[31];

unsigned int x_count = sizeof(x) * 8 - 1; /*x is of "unsigned int" type*/
unsigned int y_count = sizeof(y) * 8 -1;

for (int i = 0; i<=x_count; i++){
    x1[i] = x << i;}
for (int j = 0; j<= y_count;j++){
    y1[j] = y << j;}

while( (i<= x_count) && (j<= y_count)){
    if( x1[i] != y1[j]){
        count++;
        i++;
        j++;
        return count;}
    else
        return 0;
}

计算两个数字的位数并比较[C]

2 个答案: