Converting a character array to be handled as a hex in C

Question

I've just recently been required to work with C—I normally work with Python and a bit of Java—and I've been running into some issues.

I created a function that converts a base-10 unsigned int into a character array that represents the equivalent hex. Now I need to be able to set a variable with type uint32_t to this 'hex'; what can I do to make sure this char[] is treated as an actual hex value?

The code is below:

int DecToHex(long int conversion, char * regParams[])
{    
    int hold[8];

    for (int index = 0; conversion > 0; index++)
    {
        hold[index] = conversion % 16;
        conversion = conversion / 16;
    }

    int j = 0;

    for (int i = 7; i > -1; i--)
    {
        if (hold[i] < 10 && hold[i] >= 0)
        {
            regParams[j] = '0' + hold[i];
        }
        else if (hold[i] > 9 && hold[i] < 16)
        {
            regParams[j] = '7' + hold[i];
        }
        else
        {
            j--;
        }
        j++;
    }
    return 0;
}

Answer 1

You should just use snprintf :

int x = 0xDA8F;
char buf[9];
snprintf(buf, sizeof(buf), "%X", x);   // Use %x for lowercase hex digits

To convert a hex representation of a number to an int , use strtol (the third argument to it lets you specify the base), or, since you want to assign it to an unsigned data type, strtoul .

The code would look something like this:

char* HexStr = "8ADF4B";
uint32_t Num = strtoul(HexStr, NULL, 16);
printf("%X\n", Num);        // Outputs 8ADF4B