CS360 Final -- December 10, 2002

Answers to Question 1

• File descriptor 0: Coming from f3.
• File descriptor 1: Going to fout-3.
• File descriptor 2: Going to the screen.
• File descriptor 3: Coming from f1.
• File descriptor 4: Coming from f2.
• File descriptor 5: Going to fout-1.
• File descriptor 6: Going to fout-2.
• File descriptor 7: Coming from the pipe.
• File descriptor 8: Going to the pipe.

So, the answers:

• C1: f3 is written to fout-3:
  

  fout-1: a

  fout-2: a

  fout-3: d

  screen: a

• C2:File descriptor 3 (f1) is written to file descriptor 5 (fout-1):
  

  fout-1: b

  fout-2: a

  fout-3: a

  screen: a

• C3: f2 is written to fout-3:
  

  fout-1: a

  fout-2: a

  fout-3: c

  screen: a

• C4: f3 is written to the screen:
  

  fout-1: a

  fout-2: a

  fout-3: a

  screen: d

• C5: f1 is written to the pipe. Since all of f1 will fit into the pipe's buffer, the write()'s will complete just fine. No files will be altered.

  fout-1: a

  fout-2: a

  fout-3: a

  screen: a

• C6: f2 is written to the read-end of the pipe. This will cause an error on write(), which will be printed to the screen:

  fout-1: a

  fout-2: a

  fout-3: a

  screen: w

• C7: This is going to hang forever reading from the pipe, which has an open write end, but nothing in it.

  fout-1: z

  fout-2: z

  fout-3: z

  screen: z

• C8: This dup's file descriptor 3 (f1) onto file descriptor 9 (the lowest unused one), and then writes that to fout-2:

  fout-1: a

  fout-2: b

  fout-3: a

  screen: a

• C9: This dup's file descriptor 1 (fout-3) onto file descriptor 3, and then tries to read from it. This will cause an error, because the file descriptor is not write-only:

  fout-1: a

  fout-2: a

  fout-3: a

  screen: i

• C10: This dup's file descriptor 0 (f3) onto file descriptor 3, and then dup's file descriptor 4 (f2) onto file descriptor 1. Then file descriptor 3 (now f3) is written to file descriptor 5 (f1).

  fout-1: d

  fout-2: a

  fout-3: a

  screen: a

• C11:dup's file descriptor 6 (fout-2) onto standard error, then closes file descriptor 8, and then tries to read from it. That will generate an error, but since standard error is now going to fout-2, the error message will go to fout-2:

  fout-1: a

  fout-2: n

  fout-3: a

  screen: a

• C12: Now, two processes are forked. f3 goes to fout-3 in the first, and f1 goes to fout-2 in the other:

  fout-1: a

  fout-2: b

  fout-3: d

  screen: a

• C13: Data now flows from f2 through the pipe to fout-1:

  fout-1: c

  fout-2: a

  fout-3: a

  screen: a

• C14: Again, two processes are forked. In the first, an error will be generated trying to read from the closed file descriptor 3. That message will go to the screen. In the other process, f3 will be written to fout-1:

  fout-1: d

  fout-2: a

  fout-3: a

  screen: i

• C15: Now, file descriptor 3 is closed again, which will generate a read error. However, this time, the error will go to fout-3, since it has been dup'd onto standard error.

  fout-1: a

  fout-2: a

  fout-3: i

  screen: a

• C16: The same dup's as before are done. However, there's no error this time. f2 will be written to f1 (which has been dup'd onto file descriptor 1).

  fout-1: c

  fout-2: a

  fout-3: a

  screen: a

• C17: Finally, standard error is closed, and fout-2 is dup'd onto it. Therefore, when the program tries to read from file descriptor 1, which is write-only, it will generate an error message which is printed to fout-2:

  fout-1: a

  fout-2: g

  fout-3: a

  screen: a

If you want to try these out yourself, grab wcat.c and p1.c from this page.

Grading: 18 points

If there was an error message, I gave lots of partial credit on which error message was produced. If the actual error was a read error, then you received .25 for putting down any read error, even if you didn't put down the correct one. If you put down a write error, then you received .15 points, and if you answered e, then you received .1 points. Finally, if you put down the wrong kind of error, but the correct file descriptor, you received .3 points.

Four questions had special grading.

• In C5, the answer should have been a for all outputs -- if you answered that, you received .5 points. Otherwise zero.
• In C7, the answer should have been z somewhere. If you put a z down, you received full credit (.5 points). If you put down all a's (since there was no output), you received .3 points.
• In C12, there were two inputs and outputs, so the problem was worth 2 points.
• In C14, there were two inputs and outputs, so the problem was worth 2 points.

Oh, and the random lottery extra credit was answering i, w, g, a for C18.