Conversion from Integer to BCD

Question

I want to convert the integer (whose maximum value can reach to 99999999) in to BCD and store in to array of 4 characters. Like for example: Input is : 12345 (Integer) Output should be = "00012345" in BCD which is stored in to array of 4 characters. Here 0x00 0x01 0x23 0x45 stored in BCD format. I tried in the below manner but didnt work

int decNum = 12345;
long aux;
aux = (long)decNum;
cout<<" aux = "<<aux<<endl;

char* str = (char*)& aux;
char output[4];
int len = 0;
int i = 3;
while (len < 8)
{
cout <<"str: " << len << " " << (int)str[len] << endl;
unsigned char temp = str[len]%10;
len++;
cout <<"str: " << len << " " << (int)str[len] << endl;
output[i] = ((str[len]) << 4) | temp;
i--;
len++;
}

Any help will be appreciated

Answer 1

str points actually to a long (probably 4 bytes), but the iteration accesses 8 bytes. The operation str[len]%10 looks as if you are expecting digits, but there is only binary data. In addition I suspect that i gets negative.

Answer 2

First, don't use C-style casts (like (long)a or (char*) ). They are a bad smell. Instead, learn and use C++ style casts (like static_cast<long>(a) ), because they point out where you are doing things that are dangeruos, instead of just silently working and causing undefined behavior.

char* str = (char*)& aux; gives you a pointer to the bytes of aux -- it is actually char* str = reinterpret_cast<char*>(&aux); . It does not give you a traditional string with digits in it. sizeof(char) is 1, sizeof(long) is almost certainly 4, so there are only 4 valid bytes in your aux variable. You proceed to try to read 8 of them.

I doubt this is doing what you want it to do. If you want to print out a number into a string, you will have to run actual code, not just reinterpret bits in memory.

std::string s; std::stringstream ss; ss << aux; ss >> s; will create a std::string with the base-10 digits of aux in it.

Then you can look at the characters in s to build your BCD.

This is far from the fastest method, but it at least is close to your original approach.

Answer 3

First of all sorry about the C code, I was deceived since this started as a C questions, porting to C++ should not really be such a big deal.

If you really want it to be in a char array I'll do something like following code, I find useful to still leave the result in a little endian format so I can just cast it to an int for printing out, however that is not strictly necessary:

#include <stdio.h>

typedef struct
{
    char value[4];
} BCD_Number;

BCD_Number bin2bcd(int bin_number);

int main(int args, char **argv)
{
    BCD_Number bcd_result;

    bcd_result = bin2bcd(12345678);

    /* Assuming an int is 4 bytes */
    printf("result=0x%08x\n", *((int *)bcd_result.value));

}

BCD_Number bin2bcd(int bin_number)
{
    BCD_Number bcd_number;  

    for(int i = 0; i < sizeof(bcd_number.value); i++)
    {
        bcd_number.value[i] = bin_number % 10;
        bin_number /= 10;
        bcd_number.value[i] |= bin_number % 10 << 4;
        bin_number /= 10;
    }

    return bcd_number;
}