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cfunc

Function call syntax

I'm analyzing a C program in which I find a strange fucntion call here is the function definition :

static void endSignal (int32_t dummy)
{
  if (nTerminating) return;
  nTerminating=1;
  printf("terminating....\n");
  terminateDLNAsystem();
  sleep(1);
  exit (0);
}

This function takes an int32_t parameter ! Now this the main function calling "endSignal"

int32_t main (int32_t argc, char **argv)
{
/*Statements
.
.
*/
signal(SIGINT, endSignal);
signal(SIGABRT, endSignal);
signal(SIGQUIT, endSignal);
signal(SIGTERM, endSignal);

return 0;
}

the main function call endSignal without any parameter, what happen in this case ?

Solution

  • Main function calls signal functions and not endSignal.

    endSignal is parameter acting as callback.

    This is passing function pointers as arguments.

    How do you pass a function as a parameter in C?