Question

I have the following code:

#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#include <unistd.h>

int main()
{
  time_t result;
  char   hostname[10] = "ws45new";
  char   graphite[10] = "127.0.0.1";
  char   buf[1024];
  int    n = 0;
  int    num = 0;
  int    graphiteport = 2003;

  while (1 == 1)
  {
    result = time(NULL);
    n = n + 1;
    if (n >= 100)
      n = 0;
    printf("\n %d", n);
    // snprintf(buf, sizeof(buf), "echo \"stats.ws45new.test %d %d \" | nc 127.0.0.1 2003", n, result);
    num = sprintf(buf, "echo \"stats.ws45new1.test %d %d \" | nc 127.0.0.1 2003", n, result);
    system(buf);
    printf("\n %s", buf);
    sleep(2);
  }
  return 0;
}

This is supposed to generate something like a sawtooth signal, which is only for testing. My problem is that when i run the program it doesn’t do anything.

If i hit Ctrl-C to stop it, the program iterates once through the while loop. It is supposed to send some statistics to a graphite server and I’m using it to understand how graphite is working. From what i’ve seen it has something to do with the concatenation function sprintf (snprintf also acts the same way).

If I comment out that line it works correctly and generates the numbers I want (and also the epoch), but i need the concatenation to be able to send the required dynamic info to graphite.

If anyone has any idea why these concatenation functions don’t respect the while function please let me know as I’m really curious. Also I’m open to other suggestions but I’d rather not go in to deep, like create sockets and not use nc and system, because i’m not that good with the C language.

Answer 1

The program most likely gets stuck on the system call.

You are printing out text without the trailing newline, so it will be kept in the buffer, until a newline is printed, which you do after the system call.