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                            LAB 04 QUESTIONS
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- Name: Michael Zhang
- NetID: zhan4854

Answer the questions below according to the lab specification. Write
your answers directly in this text file and submit it to complete Lab01.


PROBLEM 1 `parent_listen.c'
===========================

A
~

  Compile and run the program in `parent_listen.c'. Show it's output
  below.

Child wrote 17 bytes
Parent read 17 bytes
Child said: 'Send $$$ please!'


B
~

  Consider the call to pipe(). Do some research to figure out what a
  pipe does and explain in a few sentences.

According to the manpage, a pipe is "a unidirectional data channel that can be
used for interprocess communication." Simply put, it's a way for processes to
communicate with each other through a file-like object, with a read end and a
write end that can be used by separate processes.


C
~

  Observe the calls to read() and write(). Their meaning should be
  self-evident (read and write data) but the parameters to these
  functions are interesting. Do some reading and explain the 3
  parameters to each of them. Also explain their return value.

The arguments for read and write are: int fd, void *buf, and size_t count. count
is obvious, and *buf is the buffer that the function is writing to, but the
argument of interest is fd. fd is a numerical file descriptor that can be used
to ask the kernel to make read/write transfers. File descriptors can be linked
to tty IO (stdin, stdout) or files or even sockets or pipes.

Also, the return value is simply the number of bytes that have been transferred.


D
~

  If you run the program a number of times, you may see output in
  different orders: the child may report writing data before the parent
  has read it.  Adjust the position of the wait() call to guarantee that
  the order is always
  - Child wrote
  - Parent read
  - Child said
  Paste your code below.

The wait() call will guarantee that the rest of the code will not run until the
child has terminated, or, more specifically, until after the print statement has
been executed. The code does not need to be changed.


PROBLEM 2 `capture_stdout.c'
============================

A
~

  Compile and run the program in `capture_stdout.c'. Show its output.

Process 28530 Piping
28530 Read from the my_pipe
28530 Read: '28530 In the pipe, five by five'


B
~

  The calls `dup()' and `dup2()' are used in this program to manipulate
  file descriptors. Explain the effects of the lines below.
  ,----
  | int stdout_bak = dup(STDOUT_FILENO);
  | dup2(my_pipe[PWRITE], STDOUT_FILENO);
  | ...
  | dup2(stdout_bak, STDOUT_FILENO);
  `----

The first line makes a copy of STDOUT_FILENO and stores it into stdout_bak. The
second line changes my_pipe[PWRITE] to be the fd of STDOUT_FILENO, and finally
the last line changes STDOUT_FILENO back to the stored value that was created in
the first of the three lines.


C
~

  The use of `printf()' normally puts output directly on the
  screen. Explain why the statement
  ,----
  | printf("%d In the pipe, five by five",
  |        getpid());
  `----
  does not print to screen as usual.

This statement does not print because by this time, the output tty STDOUT_FILENO
has had its fd changed such that it will be writing to a pipe rather than the
standard tty output.


D
~

  Modify the code so that the `In the pipe...' expression is printed by
  a child process.
  - Add a `fork()' AFTER `dup2()' redirects standard output but before
    the print
  - Add an `if()' to distinguish between parent and child
  - The child should print then exit
  - The parent should restore stdout then read from the pipe
  - Add a `wait()' to guarantee the parent waits for the child to
    complete prior to reading from the pipe
  Paste your completed code below