I'm new on StackOverflow and learning about pipes in C. I'm trying to make a shell interpeter that allows introducing 2 commands linked by pipe like
/bin/ls -l | /bin/grep a
I have a main which have a function that reads a command, then main calls a function that searchs for a "|" and if it finds it, the function makes 2 commands from it, then that function calls another one which executes the commands. My problem is that it just works once. If I introduce a command with pipes it executes it perfectly, but if I do it again it doesn't work, like the directory had change. It doesn't matter that the first command has pipes or not, the second command (if it has pipes) wont execute correctly. I think it's because of the pipes, the code of the function that execute the pipes:
void execute_pipes(char *** command1, char *** command2){
pid_t son;
int tube2[2];
pipe(tube2);
son=fork();
if(son==0){//son1
dup2(tube2[1], STDOUT_FILENO);
close(tube2[0]);
if(execv((*command1)[0],*command1)==-1){
printf("Error");
exit(0);
}
}else{//father
son=fork();
if(son==0){//son2
dup2(tube2[0], STDIN_FILENO);
close(tube2[1]);
if(execv((*command2)[0],*command2)==-1){
printf("Error");
exit(0);
}
}else{//father
close(tube2[0]);
close(tube2[1]);
waitpid(son, NULL, 0);
}
}
}
I'm sure that the parameters are ok, ending with a NULL parameter (if not I guess it wouldnt execute one time). In each new read, a new pipe is created. I guess that the last pipe won't affect this new one but I don't know...
An example of this:
/home/user/Desktop$ /bin/ls -l | /bin/grep -
-rw-rw-r-- 1 user user 435 dec 18 02:33 filename
drwxrwsr-x 9 user user 4096 dec 20 19:23 filename
-rwxrwxr-x 1 user user 14464 dec 20 20:29 filename
-rw-rw-r-- 1 user user 3580430 dec 5 03:24 filename
-rw-rw-r-- 1 user user 6833 dec 20 20:05 filename
-rw-rw-r-- 1 user user 6772 dec 20 18:48 filename
-rw-rw-r-- 1 user user 1226 dec 19 21:48 filename
-rwxrwxr-x 1 user user 8704 dec 18 16:23 filename
-rw-rw-r-- 1 user user 33673847 oct 17 20:50 filename
/home/user/Desktop$ /bin/ls -l | /bin/grep -
-rwxr-xr-x 1 root root 126584 feb 18 2016 /bin/ls
/home/user/Desktop$
NOTE: I've changed my username for 'user' and the names of the files for 'filename'.
Thanks whoever who reads this.
Summarising: my code only executes a command correctly one time, the second time it doesn't work properly and I don't know why, but I guess the problem are the pipes.
I can't comment yet, so I'm posting it as an answer. Welcome to SO. It will be easier for people to help you if you provide a working example, which I think you can. Here is a guide How to create a Minimal, Complete, and Verifiable example
Unfortunately I can't get your code to work. Maybe looking at an example implementation of a lightweight shell will help you. Here is the source code for xv6 shell. Search for PIPE keyword.
xv6 is a re-implementation of Dennis Ritchie's and Ken Thompson's Unix
Version 6 (v6).
Oke, I solved the problem, it was a problem of the initialization of the parameters. It was in a function I didn't write because I supposed it worked.
The only doubt I have now is why was even able to execute...
Thanks everyone for answering me .
Related
For pedagogical purposes, I want to set up a basic command injection in C. I have the following code :
#include <stdio.h>
#include <unistd.h>
int main(int argc, char **argv) {
char cat[] = "cat ";
char *command;
size_t commandLength;
commandLength = strlen(cat) + strlen(argv[1]) + 1;
command = (char *) malloc(commandLength);
strncpy(command, cat, commandLength);
strncat(command, argv[1], (commandLength - strlen(cat)) );
system(command);
return (0);
}
I compile it, set the binary as owned by root and set the SUID to 1, as follows :
gcc injectionos.c -o injectionos
sudo chown root:root injectionos
sudo chmod +s injectionos
I obtain the following result :
ls -la
total 40
drwxr-xr-x 2 olive olive 4096 Jan 6 13:17 .
drwxr-xr-x 3 olive olive 4096 Jan 6 12:15 ..
-rwsr-sr-x 1 root root 16824 Jan 6 13:17 injectionos
-rw-r--r-- 1 olive olive 415 Jan 6 13:17 injectionos.c
-rwx------ 1 root root 9 Jan 6 12:43 titi.txt
-rw-r--r-- 1 olive olive 9 Jan 6 12:16 toto.txt`
So, basically, with the SUID set to 1, i should be able to open both toto.txt and titi.txt files by performing the following injection :
./injectionos "toto.txt;cat titi.txt"
But executing this command, I got a permission denied when accessing titi.txt. Finally, when I add a setuid(geteuid()); in my code, the injection is working and I can access to titi.txt file.
Given that injectionos is ran as root and titi.txt belong to root, I supposed that it was enough, but apparently no. What am I missing here?
The privileges are being dropped by /bin/sh executed as part of the system() call. See the man page for bash and the -p option
If the shell is started with the effective user (group) id not equal
to the real user (group) id, and the -p option is not supplied, no
startup files are read, shell functions are not inherited from the
environment, the SHELLOPTS, BASHOPTS, CDPATH, and GLOBIGNORE
variables, if they appear in the environment, are ignored, and the
effective user id is set to the real user id. If the -p option is
supplied at invocation, the startup behavior is the same, but the
effective user id is not reset.
Well, technically debian uses dash by default, but it does the same thing.
So the default behavior of the shell has been adjusted to mitigate this injection at least somewhat.
I'm trying to implement my own ps command, called psmod.
I can use linux system call and all utilities of the /proc directory.
I discovered that all directory in /proc directory with a number as their name are the processes in the system. My question is: how can I select only those processes which are active when psmod is called?
I know that in /proc/<pid>/stat there's a letter representing the current status of the process; anyway, for every process in /proc, this letter is S, that is sleeping.
I also tried to send a signal 0 to every process, from 0 to the maximumnumberofprocesses (in my case, 32768), but in this way it discovers far more processes than the ones present in /proc.
So, my question is, how does ps work? The source is a little too complicated for me, so if someone can explain me, I would be grateful.
how does ps work?
The way of learning standard utils - is to check their source code. There are several implementations of ps: procps and busybox; and busybox is smaller and it will be easier to begin with it. There is sources for ps from busybox: http://code.metager.de/source/xref/busybox/procps/. Main loop from ps.c:
632 p = NULL;
633 while ((p = procps_scan(p, need_flags)) != NULL) {
634 format_process(p);
635 }
Implementation of procps_scan is in procps.c (ignore code from inside ENABLE_FEATURE_SHOW_THREADS ifdefs for first time). First call to it will open /proc dir using alloc_procps_scan():
290 sp = alloc_procps_scan();
100 sp->dir = xopendir("/proc");
Then procps_scan will read next entry from /proc directory:
292 for (;;) {
310 entry = readdir(sp->dir);
parse the pid from subdirectory name:
316 pid = bb_strtou(entry->d_name, NULL, 10);
and read /prod/pid/stat:
366 /* These are all retrieved from proc/NN/stat in one go: */
379 /* see proc(5) for some details on this */
380 strcpy(filename_tail, "stat");
381 n = read_to_buf(filename, buf);
Actual unconditional printing is in format_process, ps.c.
So, busybox's simple ps will read data for all processes, and will print all processes (or all processes and all threads if there will be -T option).
how can I select only those processes which are active when psmod is called?
What is "active"? If you want find all processes that exists, do readdir of /proc. If you want to find only non-sleeping, do full read of /proc, check states of every process and print only non-sleeping. The /proc fs is virtual and is it rather fast.
PS: for example, normal ps program prints only processes from current terminal, usually two:
$ ps
PID TTY TIME CMD
7925 pts/13 00:00:00 bash
7940 pts/13 00:00:00 ps
but we can strace it with strace -ttt -o ps.log ps and I see that ps does read every process directory, files stat and status. And the time needed for this (option -tt of strace gives us timestamps of every syscall): XX.719011 - XX.870349 or just 120 ms under strace (which slows all syscalls). It takes only 20 ms in real life according to time ps (I have 250 processes in total):
$ time ps
PID TTY TIME CMD
7925 pts/13 00:00:00 bash
7971 pts/13 00:00:00 ps
real 0m0.021s
user 0m0.006s
sys 0m0.014s
"My question is: how can I select only those processes which are active when psmod is called?"
I hope this command will help you:
top -n 1 | awk "NR > 7" | awk {'print $1,$8,$12'} | grep R
I am on ubuntu 12.
When you do this:
cp file1 file2
(file2 already exists)
What actually happens behind the scene?
1) Does the content of file1 actually get copied to file2?
2) Or is a new file created with the name file2 (overriding the old one) which has same content of file1?
1) Since you're using "cp", I assume the OS is Linux.
2) On Linux, a "file" is referenced by "inodes". Here are two example files:
$ ls -li 1 2
245728 -rw-r--r-- 1 paulsm users 8 Aug 14 14:52 1
245729 -rw-r--r-- 1 paulsm users 8 Aug 14 14:52 2
$ cat 1
Hello 1
$ cat 2
Hello 2
3) Here is the result after "cp"
$ cp 1 2
$ ls -li 1 2
245728 -rw-r--r-- 1 paulsm users 8 Aug 14 14:52 1
245729 -rw-r--r-- 1 paulsm users 8 Aug 14 14:55 2
$ cat 2
Hello 1
You see:
a) the contents of "1" completely replace "2"
b) there is no "new file" - the inode for "2" remains unchanged from before the copy
c) the file date is changed along with the file contents
'Hope that helps .. PSM
Usually the first. Both an index-entry as well as the file's data are written.
Yet it would help to know on what (file-)system you are (guessing linux flavour).
You would probably be aware if you were creating a junction point or symbolic/hard LINK.
Think of it like this:
Hardlink is a pointer/name, that points to a data; i.e. it's just an alternative filename; it has same inode number as the file it was created from.
Copy obviously, copy of the data; point to a different direction that file it was copyed from; has different inode number.
Also difference is in system calls, but that`s somewhat deep-diving into issue
After I found a file on the disk , I now need to print out all its details , for example :
-rwxr-xr-x 1 1000 1000 8296 2010-01-06 22:29 ./Documents/exer4
-rwxr-xr-x 1 1000 1000 8517 2009-12-30 11:30 ./Documents/os/exer4
lrwxrwxrwx 1 1000 1000 8 2010-01-10 13:10 ./Documents/cs/2012/exer4 -> ../a.out
I need to print a file details without using ls -ln .Any idea how to do that ?
Thanks
You want the stat() function.
Here's a web page that documents *NIX file functions including stat():
http://rabbit.eng.miami.edu/info/functions/unixio.html
You can use this function:
int lstat(const char *path, struct stat *buf);
This works for me from the shell.
ls -l | grep -v "\->"
It simply filters out any line that has a -> in it.
Note however, that if you have any files/directories that for some reason have -> in their names, they will also be filtered out. Having said that, I've never seen that, nor would it be a good idea.
Ok, I have been working with Solaris for a 10+ years, and have never seen this...
I have a directory listing which includes both a file and subdirectory with the same name:
-rw-r--r-- 1 root other 15922214 Nov 29 2006 msheehan
drwxrwxrwx 12 msheehan sysadmin 2048 Mar 25 15:39 msheehan
I use file to discover contents of the file, and I get:
bash-2.03# file msheehan
msheehan: directory
bash-2.03# file msh*
msheehan: ascii text
msheehan: directory
I am not worried about the file, but I want to keep the directory, so I try rm:
bash-2.03# rm msheehan
rm: msheehan is a directory
So here is my two part question:
What's up with this?
How do I carefully delete the file?
Jonathan
Edit:
Thanks for the answers guys, both (so far) were helpful, but piping the listing to an editor did the trick, ala:
bash-2.03# ls -l > jb.txt
bash-2.03# vi jb.txt
Which contained:
-rw-r--r-- 1 root other 15922214 Nov 29 2006 msheehab^?n
drwxrwxrwx 12 msheehan sysadmin 2048 Mar 25 15:39 msheehan
Always be careful with the backspace key!
I would guess that these are in fact two different filenames that "look" the same, as the command file was able to distinguish them when the shell passed the expanded versions of the name in. Try piping ls into od or another hex/octal dump utility to see if they really have the same name, or if there are non-printing characters involved.
I'm wondering what could cause this. Aside from filesystem bugs, it could be caused by a non-ascii chararacter that got through somehow. In that case, use another language with easier string semantics to do the operation.
It would be interesting to see what would be the output of this ruby snippet:
ruby -e 'puts Dir["msheehan*"].inspect'
You can delete using the iNode
If you use the "-i" option in "ls"
$ ls -li
total 1
20801 -rw-r--r-- 1 root root 0 2010-11-08 01:55 a?
20802 -rw-r--r-- 1 root root 0 2010-11-08 01:55 a\?
$ find . -inum 20802 -exec rm {} \;
$ ls -li
total 1
20801 -rw-r--r-- 1 root root 0 2010-11-08 01:55 a?
I've an example (in Spanish) how you can delete a file using then iNode on Solaris
http://sparcki.blogspot.com/2010/03/como-eliminar-archivos-utilizando-su.html
Urko,
And a quick answer to part 2 of my own question...
I would imagine I could rename the directory, delete the file, and rename the directory back to it's original again.
... I would still be interested to see what other people come up with.
JB
I suspect that one of them has a strange character in the name. You could try using the shell wildcard expansion to see that: type
cat msh*
and press the wildcard expansion key (in my shell it's Ctrl-X *). You should get two names listed, perhaps one of which has an escape character in it.
To see if there are special characters in your file, Try the -b or -q options to ls,
assuming solaris 8 has those options.
As another solution to deleting the file you can bring up the graphical file browser
(gasp!) and drag and drop the unwanted file to the trash.
Another solution might be to move the one file to a different name (the one without the unknown special character), then delete the special character directory name with wildcards.
mv msheehan temp
rm mshee*
mv temp msheehan
Of course, you want to be sure that only the file you want to delete matches the wildcard.
And, for your particular case, since one was a directory and the other a file, this command might have solved it all:
rmdir msheeha*
One quick-and-easy way to see non-printing characters and whitespace is to pipe the output through cat -vet, e.g.:
# ls -l | cat -vet
Nice and easy to remember!
For part 2, since one name contains two extra characters, you can use:
mv sheehan abc
mv sheeha??n xyz
Once you've done that, you've got sane file names again, that you can fix up as you need.