If you don't want to work on this lab over the break, I would urge you to get started early. On the other end, it won't be due until 18:00 on Tuesday July 5, which is two full days after classes have resumed.
In lab 06 we modified it to take one mandatory file name argument, and to read from there instead of from stdin.
Let's now make it function like a normal "zero or more file names" unix tool. If argc is 1 it processes the standard input (as originally), otherwise it loops through the arguments: fopen(), process it, fclose().
So instead of getchar() in your loop, you use getc(fp). You still use putchar().
Avoid duplicating code. When you go to process stdin, do so with the same code as you use for processing a file by filename, probably in a separate function, by making fp be the value stdin (defined in stdio.h).
1b (optional): Next, change it to take a (mandatory) rotation amount as argv[1] rather than always rotating by 13.
So in this new version, argc is required to be at least 2, and argv[1] is the rotation amount; but after that, if argc == 2 it processes the standard input (as before), otherwise it loops through the remaining arguments (fopen(), process it, fclose()).
(So this is more like grep's command-line processing, in that grep similarly has a mandatory first argument.)
Background required for this exercise:
Your task for submission in this lab is to write a program which does most of the work of "pwd", as follows:
First, call lstat() on "." to find out the inode number of the current directory (the "st_ino" member of the struct stat). (Remember that we looked at stat() and lstat() in lab 06.)
Then we change our current directory to the parent directory, with chdir("..").
Then we want to read the parent directory with opendir()/readdir() (using "opendir(".")" because we've now cd'd there), but unlike in the assignment two directory traversal problem, we are mostly interested in the d_ino value in the struct returned by readdir(). We are looking for the original directory in the parent directory. When we find it, we print its name.
Then call chdir("..") again, and repeat until the inode number is 2. At this point, we're at the root directory. Please print "[root]" to indicate this.
So if the real pwd would output "/home/ajr/whatever/example", your pwd will output
example
whatever
ajr
home
[root]
Except that on systems like teach.cs, the above algorithm will often stop short of the real root directory, because there are multiple filesystems which are "mounted" in the directory tree, as we discussed when discussing the unix filesystem; and the root directory of any filesystem has inode number 2. To avoid this problem, pwd has to compare "major and minor device numbers" too, but we won't do that for this exercise today. So your program will show you only part of the pwd data, until it gets to a filesystem mount point. All this is to say that you might have to cd into some subdirectories to make the output interesting for testing.
Remember to include all relevant error checking, with appropriate use of perror() (although the appropriate argument to perror() isn't always clear, but just try to do something appropriate there).
The "real" pwd in linux these days actually uses a totally different algorithm, which is a linux-specific interface for querying the current working directory. But the above is the traditional algorithm. But this means, especially given the fact that mount points cause this simplified version to stop short of the root directory, that you might want to compare your program's behaviour to mine. So I've put a compiled solution in /u/csc209h/summer/pub/lab/07/pwd
(To be clear: You may not use getcwd(). You are to use the above algorithm. An algorithm similar to the above traditionally appears in getcwd(). Your task is to implement this algorithm.)
(Note that you don't need to declare the argc and argv parameters to main (and don't need to check usage).)
Submit with the command
submit -c csc209h -a lab07 pwd.c
and the other submit commands are as before.