passing a char array by reference in C

Question

Hi I really can't get my head around this. I'm basically trying to return a char array from a function by passing the output array in as a parameter. Here is what I have so far:

The function:

int GetComputerName(char *name, char *ip_address){
    *name = "testing";
    return 0;
}

And calling it:

char comp_name[50];
GetComputerName(&comp_name, serverIP);
printf("\n\n name: %s\n\n", comp_name);

I have tried switching and swapping the * and & to see what will work and have read up on pointers and stuff yet what I think should be happening an what actually does happen is two very different things and now I think I have confused myself more than when I started!! lol

Can someone please help me out and explain what the correct way of doing this is?!

Thanks in advance =)

Answer 1

This line:

*name = "testing"; 

is invalid, as you assign the pointer to "testing" into a char pointed by name. You need to copy the data into the buffer. It should be:

int GetComputerName(char *name, char *ip_address){
    strcpy(name,"testing");
    return 0;
}

or even better (to avoid overflows):

int GetComputerName(char *name, size_t buff_len, char *ip_address){
    strncpy(name,"testing", buff_len);
    name[buff_len - 1] = '\0';
    return 0;
}

And call it:

GetComputerName(comp_name, sizeof(comp_name), serverIP);