C实现奇偶校验

奇偶校验原理(来自百度百科):奇偶校验(Parity Check)是一种校验代码传输正确性的方法。根据被传输的一组二进制代码的数位中“1”的个数是奇数或偶数来进行校验。采用奇数的称为奇校验,反之,称为偶校验。

C代码实现如下:

#include <stdio.h>

#define uint32_t unsigned int
#define uint16_t unsigned short
#define uint8_t  unsigned char

/* 从data中获取第n bit的值 注:data只能为uint8*类型指针 */
#define GET_BIT_N_VAL(data, n)  \
(0x1 & (( *((data) + (n)/8) & (0x1 << ((n)%8)) ) >> ((n)%8)))

uint8_t soc_gen_even_parity_common(uint8_t *entry_data, uint16_t entry_len)
{
    uint32_t i = 0;
    uint32_t even_parity = 0;
    for(i = 0; i < entry_len; i++)                  
    {                                                                              
        even_parity += GET_BIT_N_VAL((entry_data), i);                 
    }

    return (even_parity & 0x1);
}

uint8_t soc_gen_old_parity_common(uint8_t *entry_data, uint16_t entry_len)
{
    uint32_t i = 0;
    uint32_t odd_parity = 0;

    for(i = 0; i < entry_len; i++)                  
    {                                                                              
        odd_parity += odd_parity+ GET_BIT_N_VAL((entry_data), i);                 
    }

    if (odd_parity % 2 == 0)
        return 1;
    else
        return 0;
}

int main(void)
{
    uint8_t data[] = {0xff, 1, 0xff, 1};
    uint8_t ret = 0;
    uint8_t i = 0;

    printf("original data: ");
    for (i = 0; i < sizeof(data); i++)
        printf("0x%x ", data[i]);
    printf("\n");
    
    ret = soc_gen_even_parity_common(data, sizeof(data) * 8);
    printf("even parity result: %u\n", ret);
    ret = soc_gen_old_parity_common(data, sizeof(data) * 8);
    printf(" odd parity result: %u\n", ret);

    return 0;
}

编译运行测试:

$ ./a.out           
original data: 0xff 0x1 0xff 0x1 
even parity result: 0
 odd parity result: 1

分析运行结果,输入的原始数据为:0xff 0x1 0xff 0x1,则"1"的个数为8+1+8+1=18,采用偶校验,"1"的个数为偶数,所以结果为1;采用奇校验,“1”的个数为奇数,所以结果为1。

posted @ 2021-08-01 15:44  zhengcixi  阅读(143)  评论(0编辑  收藏  举报
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