如何在C中将二进制编码的十进制转换为十进制？

Question

So my task is to convert BCD to decimal, but the problem is I don't know how to go from the first tetrad to the second one. For example the BCD number is 10010011 (93 in decimal), my code works for the first tetrad(1001), but how do I convert the other ones? Here is a code I tried to work with:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void main(void) {
    int bcd, bcd1, bcd2, bcd3, bcd4, zero, one, dec = 0;
    char bcdstr[40];
    printf("Type 2-10 number\n");
    scanf_s("%d", &bcd);
    snprintf(bcdstr, 40, "%d", bcd);
    bcd1 = (bcd / 1000);
    printf("%d\n", bcd1);
    bcd2 = (bcd / 100) % 10;
    printf("%d\n", bcd2);
    bcd3= (bcd % 100) / 10;
    printf("%d\n", bcd3);
    bcd4 = ((bcd % 1000) % 100) % 10;
    printf("%d\n", bcd4);
    if (bcd1 == 1) dec = 8;
    if (bcd2 == 1) dec = dec + 4;
    if (bcd3 == 1) dec = dec + 2;
    if (bcd4 == 1) dec = dec + 1;
    if (dec > 9) printf("Not BCD!");
    else printf("Decimal is %d", dec);
    printf("\nstrlen is %d", strlen(bcdstr));
    getchar();
}

Answer 1

You just take each digit of the BCD individually and add it to a number:

unsigned bcd_to_decimal(unsigned bcd)
{
    unsigned result = 0;
    unsigned short exp = 1;
    unsigned char ctr = 4, tmp;

    while(ctr--)
    {
        tmp = bcd & 0xf;
        if(tmp > 9)
            printf("Not BCD!");
        result += tmp * exp;
        exp *= 10;
        bcd >>= 4;
    }
}