Thread: converting unsigned char to char* for concatenation

Hi I have a function that finds my mac address, I am trying to store it in a char * variable.
This is my function to display it:

Code:

void PrintMACaddress (unsigned char *addr)
{
    int i;
    unsigned char * macaddress;
    char temp[80];
   for (i = 0; i < 6; i++)
   {
    printf("%x:", *addr);
    sprintf(temp[i], "%x", *addr++);
      //strcat(macaddress, *addr);
   }
   printf("%s\n", temp);

}

So I can roll through the 6 parts of the mac address just fine, using the printf statement and outputting it in the hex format, however I cant get it to sprintf properly, and my attempt at using strcat crashes the console. I would like to return a char * if possible.

Any help greatly appreciated.
Thanks

Hi Sorry for my confusion, I have changed the code to:

Code:

    int i = 0;
char temp[25];
char *cs = temp;
int used = 0;
do {
   sprintf(cs, "%x", *addr++);
    printf("%s", cs);
   cs += used;
   i++;
   if( i < 6){
   printf(":");
   }
}while (i < 6);

printf(" cs: %s\n", cs);

the printf occuring inside the loop displays the correct thing, i.e cs temporarily holds the correct part of my mac address, however I cannot seem to get it to reset to the beginning of temp so that I can display the mac address after the loop - i.e cs or temp holds the entire mac address that I can then return to the method that called it.
I tried doing something like:

Code:

*cs = temp;
for (i = 0; i < 25; i++){
printf("\n%s\n", cs++);
}

but that crashes the terminal,
so does

Code:

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

and so does even looping through the temp array displaying each character.

Your sprintf:

Code:

sprintf(cs, "%x", *addr++);

whiteflags' sprintf:

Code:

sprintf(cs, "%x%n", *addr++, &used);

Notice anything different?

So this gives me:

Code:

int PrintMACaddress (unsigned char *addr)
{
    int i = 0;
char temp[25];
char *cs = temp;
int used = 0;
do {
   sprintf(cs, "%x%n", *addr++, &used);
    printf("%s", cs);
   cs += used;
   i++;
   if( i < 6){
   printf(":");
   }
}while (i < 6);
printf(" cs: %s\n", cs);


return 1;

}

cs: 46

The printf in the while gives the right thing which tells me at the time cs is the right character. How do I move it back so that the printf outside the while loop is back at the beginning of the mac address?

Originally Posted by a.mlw.walker

The printf in the while gives the right thing which tells me at the time cs is the right character. How do I move it back so that the printf outside the while loop is back at the beginning of the mac address?

You moved cs through the array (by doing cs += used), so it no longer points to the start of the sprintf'ed MAC address. You already have a pointer to the beginning of the sprintf'ed MAC address however, so just print that. Yes, I'm intentionally being vague, I want you to give a little thought to what your code is doing (and should be doing), rather than just copy-pasting whatever we post here.

Originally Posted by vart

why not to check return value of sprintf? %n usage in this case is overkill IMHO

Selecting one option among two that do the same thing is not overkill.

So I see what you mean now Andruil463, in theory:

Code:

int PrintMACaddress (unsigned char *addr)
{
    int i = 0;
char temp[25];
char *cs = temp;
int used = 0;
do {
   sprintf(cs, "%x:%n", *addr++, &used);
    //printf("%s", cs);
   cs += used;
   i++;
}while (i < 6);
*cs = temp;
printf("cs: %s\n", cs);
return 1;

}

used is keeping tabs on how many characters have been printed and therefore moves cs along that many places in the array temp. cs starts at the beginning of temp on this line:

*cs = temp;
so by my understanding I should be able to set cs back to *cs = temp; and then print cs again.
However that outputs:
cs: 46

So Im struggling to understand how to set cs back to the beginning to print it out completely.

*cs affects the character pointed to by cs. If you want to move the pointer itself, you have to assign to cs.

You need to assign cs to temp if you want to use cs to print. But you don't need to do that. cs was not the variable I was referring to.

What variable contains the MAC address (i.e. where did you sprintf it)?
Do you have a pointer to the beginning of that array?
What do you get if you use the name of an array without any [ ] brackets?

Originally Posted by Click_here

"used" is declared as 0 and never changes from 0

Code:

int used = 0;
...
cs += used;

Therefore cs never changes from one loop to the next.

used is set by the line

Code:

sprintf(cs, "%x%n", *addr++, &used);