Question

I have:

uint8 buf[] = {0, 1, 10, 11};

I want to convert the byte array to a string such that I can print the string using printf:

printf("%s\n", str);

and get (the colons aren’t necessary):

"00:01:0A:0B"

Any help would be greatly appreciated.

Answer 1

printf("%02X:%02X:%02X:%02X", buf[0], buf[1], buf[2], buf[3]);

For a more generic way:

int i;
for (i = 0; i < x; i++)
{
    if (i > 0) printf(":");
    printf("%02X", buf[i]);
}
printf("\n");

To concatenate to a string, there are a few ways you can do this. I’d probably keep a pointer to the end of the string and use sprintf. You should also keep track of the size of the array to make sure it doesn’t get larger than the space allocated:

int i;
char* buf2 = stringbuf;
char* endofbuf = stringbuf + sizeof(stringbuf);
for (i = 0; i < x; i++)
{
    /* i use 5 here since we are going to add at most 
       3 chars, need a space for the end '\n' and need
       a null terminator */
    if (buf2 + 5 < endofbuf)
    {
        if (i > 0)
        {
            buf2 += sprintf(buf2, ":");
        }
        buf2 += sprintf(buf2, "%02X", buf[i]);
    }
}
buf2 += sprintf(buf2, "\n");