Convert long to 6 letter hex string

Question

I have a long value and I would like to convert it into 6 digit hex string.

My Try:

long ID = some value;

char * hex = (char *) calloc(n, sizeof(char)); // dynamic memory required.

int c = sprintf(hex, n+1, "%x", ID);

assert(c == ID);

1) how would I find the value of 'n' to give to calloc?

2) how do I make the string only 6 hex characters?

Thanks.

Answer 1

Number of hex digits can be computed like this:

int numberOfHexDigits(long ID)
{
    int n=0;
    if (ID==0) {
        return 1;
    }
    while (ID!=0) 
    {
        ID >>= 4;
        n++;
    }
    return n;
}

Using calloc (or malloc) to allocate memory for just 7 bytes is wasteful. I assume you have a reason to do that. You are better off just doing this:

const int nDigits = 6;
char hex[nDigits+1];
snprintf(hex, sizeof(hex), "%lx", ID);

If you really want to allocate memory on the heap:

int bufSize = numberOfHexDigits(ID)+1;
char *hex = (char *) malloc(bufSize);
snprintf(hex, bufSize, "%lx", ID);