I'm trying to build a minishell and I've mostly everything but pipes working. I've read a lot of answers in this site and many others, but I'm not able to find my particular problem.
The sequence should end when the right-most process ends, this is: sleep 3 | sleep 3 | sleep 0 should immediately end, and not wait for any of the sleep 3 processes.
*argvv is a pointer to the first process (*argvv would be sleep 3, *(argvv + 1) is the next sleep 3, etc. I don't know what's going on, but something is wrong with file descriptors and everything breaks after a test.
For instance, when the input is sh> ls | wc | wc, the output is
sh> ls | wc | wc
sh> 1 3 25
, then it waits for some input. It's not until I press enter again it finishes. After that, sleep 3 does not work anymore.
Any help would be welcome, thanks a lot!
Edit:
Ok, I edited the code and everything seems to work fine, except sleep 3 | sleep 0 lasts 3 seconds, instead of immediately ending. I don't know how to fix this.
void execute(char ***argvv)
{
pid_t pid;
int pd[2];
int in = 0;
while (*argvv != NULL) {
pipe(pd);
if ((pid = fork()) < 0)
{
perror();
exit(1);
}
else if (pid == 0) {
dup2(in, 0);
if (*(argvv + 1) != NULL)
dup2(pf[1], 1);
close(pd[0]);
close(pd[1]);
execvp((*argvv)[0], *argvv);
perror();
exit(1);
}
else
{
close(pd[1]);
in = pd[0];
if (*(argvv + 1) == NULL)
close(pd[0]);
argvv++;
}
wait(&pid);
}
}
I think part of the problem may be in the use of wait(). It's pid parameter is an OUTPUT, reporting which fork()ed process ended. Every forked processed ends, and needs to be wait()ed for. When you run a second pipeline, all but one piece of the preceding pipeline is still ready to be wait()ed for.
Generally, you have several choices:
wait() in a loop until the process you want to end ends.
wait() in a loop until all the subprocesses end.
Use waitpid() to look for a particular process ending.
Arrange that only the last process in the chain is a child of your process, making the rest children of that. (This is done by some shell, though I don't remember which.) The problem with this is that it leaves extra children for the last process in the chain, which may confuse it.
Track all the child processes, and which have ended. When you background a pipeline, this allows you to display which parts of the pipeline are still running. (tcsh does this. bash does half.)
fork() twice for the earlier pipeline members, execvp() in the grandchild, then exit() in the child, and have the parent wait() for the child to finish before going on. This disconnects the grandchild from the parent, so it never needs to be waited for.
Do things involving ignoring SIGCHLD, but this will probably cause you a lot of grief, as it is all or nothing.
Related
The output of the program are not obviously contents from the printf()s in teh code. Instead it looks like characters in irregular sequence. I know the reason is because the parent process and child process are running
at the same time, but in this program I only see pid=fork(), which I think means pid is only the id of child process.
So why can the parent process print?
How do the two processes run together?
// fork.c: create a new process
#include "kernel/types.h"
#include "user/user.h"
int
main()
{
int pid;
pid = fork();
printf("fork() returned %d\n", pid);
if(pid == 0){
printf("child\n");
} else {
printf("parent\n");
}
exit(0);
}
output:
ffoorrkk(()) rreettuurrnende d 0
1c9h
ilpda
rent
I focus my answer on showing how the observed output can result from the shown program. I think that it will already clear things up for you.
This is your output.
I edited it to use a good guess of what is parent (p) and child (c):
ffoorrkk(()) rreettuurrnende d 0\n
cpcpcpcpcpcpcpcpcpcpcpcpccpcpcppccc
1 c9h\n
pccpcpp
ilpda\n
ccpcpcc
rent
pppp
If you only use the chars with a "c" beneath, you get
fork() returned 0
child
If you only use the chars with a "p" beneath, you get
fork() returned 19
parent
Split that way, it should match what you know about how fork() works.
Comments already provided the actual answer to the three "?"-adorned questions in title and body of your question post.
Lundin:
It creates two processes and they are executed just as any other process, decided by the OS scheduler.
Yourself:
each time fork() is called it will return twice, the parent process will return the id of child process, and child process will return 0
Maybe for putting a more obvious point on it:
The parent process receives the child ID and also continues executing the program after the fork().
That is why the output occurs twice, similarily, interleaved, with differences in PID value and the selected if branch.
Relevant is also that in the given situation there is no line buffering. Otherwise there would be no character-by-character interleaving and everthing would be much more readable.
I've been working on this school assignment forever now, and I'm super close to finishing.
The assignment is to create a bash shell in C, which sounds basic enough, but it has to support piping, IO redirect, and flags within the piped commands. I have it all working except for one thing; the | piping child isn't getting any of the data written to the pipe by the user command process child. If I were to remove the child fork for pipechild, and have everything from if(pipe_cmd[0] != '\0') run as the parent, it would work just fine (minus ending the program because of execlp). If I were to use printf() inside the pipe section, the output would be in the right file or terminal, which just leaves the input from the user command process child not getting to where it needs to be as a culprit.
Does anyone see an issue on how I'm using the pipe? It all felt 100% normal to me, given the definition of a pipe.
int a[2];
pipe(a);
//assume file_name is something like file.txt
strcat(file_name, "file.txt");
strcat(pipe_cmd, "wc");
if(!fork())
{
if(pipe_cmd[0] != '\0') // if there's a pipe
{
close(1); //close normal stdout
dup(a[1]); // making stdout same as a[1]
close(a[0]); // closing other end of pipe
execlp("ls","ls",NULL);
}
else if(file_name[0] != '\0') // if just a bare command with a file redirect
{
int rootcmd_file = open(file_name, O_APPEND|O_WRONLY|O_CREAT, 0644);
dup2(rootcmd_file, STDOUT_FILENO);
execlp("ls","ls",NULL); // writes ls to the filename
}
// if no pipe or file name write...
else if(rootcmd_flags[0] != '\0') execlp("ls","ls",NULL)
else execlp("ls","ls",NULL);
} else wait(0);
if(pipe_cmd[0] != '\0') // parent goes here, if pipe.
{
pipechild = fork();
if(pipechild != 0) // *PROBLEM ARISES HERE- IF THIS IS FORKED, IT WILL HAVE NO INFO TAKEN IN.
{
close(0); // closing normal stdin
dup(a[0]); // making our input come from the child above
close(a[1]); // close other end of pipe
if(file_name[0] != '\0') // if a filename does exist, we must reroute the output to the pipe
{
close(1); // close normal stdout
int fileredir_pipe = open(file_name, O_APPEND|O_WRONLY|O_CREAT, 0644);
dup2(fileredir_pipe, STDOUT_FILENO); //redirects STDOUT to file
execlp("wc","wc",NULL); // this outputs nothing
}
else
{
// else there is no file.
// executing the pipe in stdout using execlp.
execlp("wc","wc",NULL); // this outputs nothing
}
}
else wait(0);
}
Thanks in advance. I apologize for some of the code being withheld. This is still an active assignment and I don't want any cases of academic dishonesty. This post was risky enough.
} else wait(0);
The shown code forks the first child process and then waits for it to terminate, at this point.
The first child process gets set up with a pipe on its standard output. The pipe will be connected to the second child process's standard input. The fatal flaw in this scheme is that the second child process isn't even started yet, and won't get started until the first process terminates.
Pipes have limited internal buffering. If the first process generates very little output chances are that its output will fit inside the tiny pipe buffer, it'll write its output and then quietly terminate, none the wiser.
But if the pipe buffer becomes full, the process will block and wait until something reads from the pipe and clears it. It will wait as long as it takes for that to happen. And wait, and wait, and wait. And since the second child process hasn't been started yet, and the parent process is waiting for the first process to terminate it will wait, in vain, forever.
This overall logic is fatally flawed for this reason. The correct logic is to completely fork and execute all child processes, close the pipe descriptors in the parent (this is also important), and then wait for all child processes to terminate. wait must be the very last thing that happens here, otherwise things will break in various amazing and mysterious ways.
I am working on a tiny shell(tsh) implemented in C(it's an assignment). One part of assignment belongs to PIPING. I have to pipe a command's output to another command. e.g:ls -l | sort
When I run the shell, every command that I execute on it, is processed by a child process that it spawns. After the child finishes the result is returned. For piping I wanted to implement a harcoded example first to check how it works. I wrote a method, that partially works. The problems is when I run the pipe command, after child process finishes, the whole program quits with it! Obviously I am not handling the child process signal properly(Method code below).
My Question:
How does process management with pipe() works? if i run a command ls -l | sort does it create a child process for ls -l and another process for sort ? From the piping examples that I have seen so far, only one process is created(fork()).
When the second command (sort from our example) is processed, how can i get its process ID?
EDIT: Also while running this code I get the result twice. don't know why it runs twice, there is no loop in there.
Here is my code:
pid_t pipeIt(void){
pid_t pid;
int pipefd[2];
if(pipe(pipefd)){
unix_error("pipe");
return -1;
}
if((pid = fork()) <0){
unix_error("fork");
return -1;
}
if(pid == 0){
close(pipefd[0]);
dup2(pipefd[1],1);
close(pipefd[1]);
if(execl("/bin/ls", "ls", (char *)NULL) < 0){
unix_error("/bin/ls");
return -1;
}// End of if command wasn't successful
}// End of pid == 0
else{
close(pipefd[1]);
dup2(pipefd[0],0);
close(pipefd[0]);
if(execl("/usr/bin/tr", "tr", "e", "f", (char *)NULL) < 0){
unix_error("/usr/bin/tr");
return -1;
}
}
return pid;
}// End of pipeIt
Yes, the shell must fork to exec each subprocess. Remember that when you call one of the execve() family of functions, it replaces the current process image with the exec'ed one. Your shell cannot continue to process further commands if it directly execs a subprocess, because thereafter it no longer exists (except as the subprocess).
To fix it, simply fork() again in the pid == 0 branch, and exec the ls command in that child. Remember to wait() for both (all) child processes if you don't mean the pipeline to be executed asynchronously.
Yes, you do need to call fork at least twice, once for each program in the pipeline. Remember that exec replaces the program image of the current process, so your shell stops existing the moment you start running sort or (tr).
I am working on a tiny shell(tsh) implemented in C(it's an assignment). One part of assignment belongs to PIPING. I have to pipe a command's output to another command. e.g:ls -l | sort
When I run the shell, every command that I execute on it, is processed by a child process that it spawns. After the child finishes the result is returned. For piping I wanted to implement a harcoded example first to check how it works. I wrote a method, that partially works. The problems is when I run the pipe command, after child process finishes, the whole program quits with it! Obviously I am not handling the child process signal properly(Method code below).
My Question:
How does process management with pipe() works? if i run a command ls -l | sort does it create a child process for ls -l and another process for sort ? From the piping examples that I have seen so far, only one process is created(fork()).
When the second command (sort from our example) is processed, how can i get its process ID?
EDIT: Also while running this code I get the result twice. don't know why it runs twice, there is no loop in there.
Here is my code:
pid_t pipeIt(void){
pid_t pid;
int pipefd[2];
if(pipe(pipefd)){
unix_error("pipe");
return -1;
}
if((pid = fork()) <0){
unix_error("fork");
return -1;
}
if(pid == 0){
close(pipefd[0]);
dup2(pipefd[1],1);
close(pipefd[1]);
if(execl("/bin/ls", "ls", (char *)NULL) < 0){
unix_error("/bin/ls");
return -1;
}// End of if command wasn't successful
}// End of pid == 0
else{
close(pipefd[1]);
dup2(pipefd[0],0);
close(pipefd[0]);
if(execl("/usr/bin/tr", "tr", "e", "f", (char *)NULL) < 0){
unix_error("/usr/bin/tr");
return -1;
}
}
return pid;
}// End of pipeIt
Yes, the shell must fork to exec each subprocess. Remember that when you call one of the execve() family of functions, it replaces the current process image with the exec'ed one. Your shell cannot continue to process further commands if it directly execs a subprocess, because thereafter it no longer exists (except as the subprocess).
To fix it, simply fork() again in the pid == 0 branch, and exec the ls command in that child. Remember to wait() for both (all) child processes if you don't mean the pipeline to be executed asynchronously.
Yes, you do need to call fork at least twice, once for each program in the pipeline. Remember that exec replaces the program image of the current process, so your shell stops existing the moment you start running sort or (tr).
While going thorough fork() command i got struck with a question .
how many no.of processes are created by the end of 12th second, if
time starts from 0th second? Process id's start from 0.
Pseudo code
while(true)
{
sleep 1second;
if( getpid() % 2 == 0 )
{
fork();
printf("Hello\n");
}
}
when i run above code on my system it is not showing output on konsole. Is no . of process at end of 12 sec is dependent on OS ?Need suggestion as i am not good in fork()
Since when do process IDs "start at 0"? Not even when the system boots; the first process has the id 1 :-)
You're only fork()ing when your own process ID is even; so if it happens to be odd then nothing will happen... which means that if you run the program several times, sometimes it will do something and sometimes it won't.
Add this after your printf:
fflush(stdout);
But you have a fundamental problem with your logic. fork() returns 0 in the child, and the child pid in the parent. You don't check, so both the parent and the child continue doing the loop, which happens again, and again, and again, forever. You need to change the loop body to this:
if(fork() == 0)
{
printf("Hello!\n");
fflush(stdout);
}