Programming

struct arguments
{
  int *Z_DEBUG, *M_DEBUG;
  char * interf;
  char * outfile;            /* Argument for -o */
};

int main(int argc, char** argv) {

    struct arguments arguments;
    outstream = stdout;
    arguments.M_DEBUG=&MAIN_DEBUG;
    arguments.Z_DEBUG=&ZMQ_DEBUG;
    strcpy( arguments.interf, "eth0" );
    arguments.outfile = NULL;
    s_catch_signals();

    argp_parse(&argp, argc, argv, 0, 0, &arguments);

char *mystring="useless phrase";

char *mystring;
strcpy(mystring, "useless phrase");

  char * interf;
  char * outfile;

  char interf[256];
  char outfile[256];

malloc()

  interf=malloc(256);
  outfile=malloc(256);
  if(interf==NULL || or outfile==NULL)
  {
      perror("Cannot allocate memory");
      exit(1);
  }

arguments.interf="eth0";
arguments.outfile=NULL;

1 - the OS  creates a stack frame for main.
2 - the os simply overlays that stack on top of existing garbage in memory
3 - it does this for efficiency reasons and because that memory is no longer part of any process.
4 - when your program ran,  those pointers were parked on top of memory that had some existing values in it.
5 - what was in the memory depends on the program that lived in that exact memory before
6 - it could be all 00000000, it could literally be anything.
7 - since it could be anything, it is possible that the memory pointed to (0xfaaa0000) - let's pretend.
8 - 0xfaaa0000 just HAPPENED to be by random chance an OS allocated  location on your existing stack frame.
9 - Now we can use the memory for our program - no crash.
10 Why? because the memory is part of the process so you can what you want to it
11 What if 0xfaaa0000 was NOT part of allocated memory?  Boom, program crash.
12 Therefore there is no known way to predict the behavior of the code, it is undefined