如何在 C++ 中实现模 2 二进制除法

Question

现在我想用c++实现循环冗余校验所以我想知道怎么做我可以执行 mod 2 二进制 division.
好吧，虽然我知道 mod-2 二进制 div 有一个字符串算法，但我想知道是否有任何 int 算法。抱歉我的菜鸟解释。

谢谢。

Answer 1

这是 CRC 的代码示例，使用模 2 二进制除法):

/*
 * The width of the CRC calculation and result.
 * Modify the typedef for a 16 or 32-bit CRC standard.
 */
typedef uint8_t crc;

#define WIDTH  (8 * sizeof(crc))
#define TOPBIT (1 << (WIDTH - 1))

crc
crcSlow(uint8_t const message[], int nBytes)
{
    crc  remainder = 0; 


    /*
     * Perform modulo-2 division, a byte at a time.
     */
    for (int byte = 0; byte < nBytes; ++byte)
    {
        /*
         * Bring the next byte into the remainder.
         */
        remainder ^= (message[byte] << (WIDTH - 8));

        /*
         * Perform modulo-2 division, a bit at a time.
         */
        for (uint8_t bit = 8; bit > 0; --bit)
        {
            /*
             * Try to divide the current data bit.
             */
            if (remainder & TOPBIT)
            {
                remainder = (remainder << 1) ^ POLYNOMIAL;
            }
            else
            {
                remainder = (remainder << 1);
            }
        }
    }

    /*
     * The final remainder is the CRC result.
     */
    return (remainder);

}   /* crcSlow() */

来自 https://barrgroup.com/Embedded-Systems/How-To/CRC-Calculation-C-Code.

Answer 2

#include <iostream>
using namespace std;

int main(){
      int bits[100],generator[20];

      int sizebits , sizeGen ;
      cin >> sizebits >> sizeGen ;

      for(int i = 0; i < sizebits ; i++)
           cin >> bits[i];

      for(int i = 0; i < sizeGen ; i++)
          cin >> generator[i];

      for(int i = 0; i < sizebits-sizeGen;){
            for(int j = 0; j < sizeGen; j++)
                 bits[i+j] = (bits[i+j] == generator[j]? 0 : 1) ;
            for(; i < sizebits-sizeGen && bits[i] != 1 ;i++);
      }

      for(int i = 0; i < sizebits ; i++)
          cout << bits[i];

      return 0;
}

可以这样做！

Answer 3

模2除法使用异或运算而不是减法，下面是c++实现

#include <iostream>
#include <math.h>
#include <cstring>
using namespace std;

char exor(char a,char b)                                      // perform exor operation
{
if(a==b)
return '0';
else
return '1';
}

void modulo2div(char data[], char key[], int datalen, int keylen)
{

char temp[20],rem[20];

for(int i=0;i<keylen;i++)
rem[i]=data[i];                    //considering keylen-1 bits of data for each step of binary division/EXOR 

for(int j=keylen;j<=datalen;j++)
{
    for(int i=0;i<keylen;i++)
    temp[i]=rem[i];                // remainder of previous step is divident of current step

    if(rem[0]=='0')                //if 1's bit of remainder is 0 then shift the rem by 1 bit
    {
        for(int i=0;i<keylen-1;i++)
            rem[i]=temp[i+1];
    }
    else                    //else exor the divident with generator polynomial
    {    
        for(int i=0;i<keylen-1;i++)
            rem[i]=exor(temp[i+1],key[i+1]);
            
    }
        rem[keylen-1]=data[j];        //appending next bit of data to remainder obtain by division
    
}
    
cout<<"CRC="<<rem<<"\nDataword="<<data;        //remainder obtain is crc

}

访问Implementation of CRC in C++了解更多详情

如何在 C++ 中实现模 2 二进制除法

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