Question

I’ve tried everything Google-able, but can’t seem to wrap my head around this. I’m trying to return a char and pass to another function with no luck. Do I need to do some sort of memory location?

char hostname[] = "www.XXXXX.com";
uint8 ipaddr[]  = {XXX,XXX,XXX,XXX};
char uri[]      = "/api/Login/";  
char key[] = API_KEY;  //3490324032-asdfasd-234234
const int port  = 80;

//function to combine uri and key
char combine(char key[], char uri[]) {
  int i = 0;
  int x = 0;
  char long_s[sizeof(uri) + sizeof(key)];

  while(uri[i]) {
      long_s[i] = uri[i];
      ++i;
  }
  while(key[x]) {
      long_s[i] = key[x];
      ++i;
      ++x;
  }
  long_s[i] = '\0';
  //Serial.print(long_s);
  return long_s;  //pointer?
} 

char lon = combine (key, uri);
char* newURI = &lon;

// A get request
GETrequest getSession(ipaddr, port, hostname, newURI);

Answer 1

Your function return value is char – that means that it returns a single char (which can only represent one character). A string is represented by an array of chars, which is usually referred to by a pointer (char *).

So you need to return a char * – but this leads to another problem. The pointer must be pointing at an array that outlives the function call. Your array long_s will be destroyed as soon as the function returns, so this is not suitable. The usual alternative is to allow the caller to supply a buffer of sufficient size (note also that your loops can be replaced with the strcpy() and strcat() functions):

char *combine(char result[], size_t result_sz, const char key[], const char uri[])
{
  /* Test to see if there is sufficient space */

  if (strlen(uri) + strlen(key) + 1 > result_sz)
  {
    return NULL;
  }

  strcpy(result, uri);
  strcat(result, key);

  return result;
} 

Then your caller must use:

char lon[1024];
char *newURI = combine (lon, sizeof lon, key, uri);

if (newURI == NULL)
{
    /* Error - URL too long */
}
else
{
    GETrequest getSession(ipaddr, port, hostname, newURI);
}