I am trying to resolve this exercise: Given a sequence of bits, knowing that every n bit (of data) there is a parity bit (1 if the number of bits at 1 among the previous n is odd), check if there are errors. I have done it in C.

#include <stdio.h>

void main()
{
    // Input
    unsigned char vet[] = { 2,4,67,2,2,58,99 };
    unsigned int len = 55;    // lunghezza (numero di bit)
    unsigned char n = 4;    // numero di bit dati

    // Output
    unsigned char errori = 0;    // 1 = errori; 0 = no errori

    unsigned int i, j, cnt_1;

    for (i = 0; i < len; i += 9)
    {
        cnt_1 = 0;
        for (j = 0; j < 8 && i + j < len; j++)
        {
            unsigned char mask = 1 << j;
            if (vet[i + j] & mask)
                cnt_1++;
        }

        if (cnt_1 % 2 != 0)
        {
            if (!(vet[i + 8] & 1))
            {
                errori = 1;
                break;
            }
        }
        else
        {
            if (vet[i + 8] & 1)
            {
                errori = 1;
                break;
            }
        }
    }


    // Stampa su video
    printf("La sequenza di bit %scontiene errorin", (errori ? "" : "non "));
}

The problem is that I need to translate it in inline assembly:

#include <stdio.h>

void main()
{
    // Input
    unsigned char vet[] = { 2,4,67,2,2,58,99 };
    unsigned int len = 55;  // lunghezza (numero di bit)
    unsigned char n = 4;    // numero di bit dati

    // Ouput
    unsigned char errori = 0;   // 1 = errori; 0 = no errori

    __asm
    {
    }

    // Stampa su video
    printf("La sequenza di bit %scontiene errorin", (errori ? "" : "non "));
}

I have tried to resolve it but didnt make it. If anyone can help me to translate this part, I will appreciate it so much.

for (i = 0; i < len; i += 9)
    {
        cnt_1 = 0;
        for (j = 0; j < 8 && i + j < len; j++)
        {
            unsigned char mask = 1 << j;
            if (vet[i + j] & mask)
                cnt_1++;
        }

        if (cnt_1 % 2 != 0)
        {
            if (!(vet[i + 8] & 1))
            {
                errori = 1;
                break;
            }
        }
        else
        {
            if (vet[i + 8] & 1)
            {
                errori = 1;
                break;
            }
        }
    }