CS360 Final -- December 11, 1999. Question 2: Answer and grading

15 points

You will need to create two pipes. One for prog2 and one for prog3. Standard output will be dup'd to the first pipe, and f1 will be fdopen'd on the second pipe. Then you call fork()/execlp() for the two programs. Actually, the ordering is different than the above. You need to:

Call pipe() for one pipe.
Call fork()
In the child, you:
- Call dup2() so that standard input comes from the pipe.
- Close both ends of the pipe (note, the read end will still be open as standard input.
- Call execlp() on prog2.
Meanwhile, we're back in the parent.
Call dup2() so that standard output goes to the pipe.
Close both ends of the pipe (note, the write end will still be open as standard output.
Call pipe() again.
Call fork() again.
In the child, you:
- Call dup2() so that standard input comes from the pipe.
- Close both ends of the pipe (note, the read end will still be open as standard input.
- Call execlp() on prog3.
Back in the parent:
Call f1 = fdopen() on the write end of the pipe.
Close the read end of the pipe.
At the end of the program call wait() for both children.

Here's the code with all error checking.

main(int argc, char **argv)
{
  /* Declarations: */

  FILE *f1;
  int p1[2];
  int p2[2];
  int i;

  if (argc != 1) { fprintf(stderr, "usage: prog1\n"); exit(1); }

  if (pipe(p1) < 0) { perror("pipe"); exit(1); }
  i = fork();
  if (i < 0) { perror("fork"); exit(1); }

  if (i == 0) { /* Child */
    if (dup2(p1[0], 0) < 0) { perror("dup2"); exit(1); }
    close(p1[0]);
    close(p1[1]);
    execlp("prog2", "prog2", NULL);
    perror("execlp"); 
    exit(1);
  }

  if (dup2(p1[1], 1) < 0) { perror("dup2"); exit(1); }
  close(p1[0]);
  close(p1[1]);

  if (pipe(p2) < 0) { perror("pipe"); exit(1); }
  i = fork();
  if (i < 0) { perror("fork"); exit(1); }

  if (i == 0) { /* Child */
    if (dup2(p2[0], 0) < 0) { perror("dup2"); exit(1); }
    close(p2[0]);
    close(p2[1]);
    execlp("prog3", "prog3", NULL);
    perror("execlp"); 
    exit(1);
  }
  
  f1 = fdopen(p2[1], "w");
  if (f1 == NULL) { perror("fdopen"); exit(1); }
  close(p2[0]);

  ....   /* Various code that writes to both standard output and f1 */

  wait(&i);
  wait(&i);

}

Grading

Points were allocated as follows:

Call pipe() for one pipe: 1 point
Call fork() for prog2: 1 point
Call dup2() correctly in the child: 1 point
Call close() on both ends of the pipe in the child: 2 points
Call execlp() on prog2 in the child: 1 point
Call exit() after execlp(): 1 point
Call dup2() in the parent so that standard output goes to the pipe: 1 point.
Close the read end of the pipe in the parent: 1 point
(You should close the write end too, but since that won't kill the code too badly, I didn't take off for it).
Call pipe() again: 1 point
Call fork() again: 1 point
Have the child's code match the code for the first child: 1 point
Call f1 = fdopen() on the write end of the pipe: 1 point
Close the read end of the pipe: 1 point
At the end of the program call wait() for both children: 1 point

That assumes that you got the ordering correct. I took off points (up to 10) for when the flow of control was confused, since that is extremely important for getting the program right.

I didn't take off for error checking, except for execlp().