I've just recently learned about pipes and I would like to emulate the "|" gimmick provided by shells.
In the code below, the parent process spawns 2 child processes, after which they do their piping and get replaced by ls and grep. While that happens the parent process waits patiently. The problem is that the child processes never finish although they manage to send some data though the pipe and onto the screen.
There are other posts regarding pipes on SO, but I've never seen the setup in which the parent process launches 2 children. I've only seen the parent communicating with one child.
int p0[2];
pipe(p0); //creating pipe
if(fork() == 0) { //child 1
dup2(p0[0], STDIN_FILENO);
close(p0[0]); close(p0[1]);
execlp("grep","grep","a",NULL);
}
else { //parent
if(fork() == 0) { //child 2
dup2(p0[1], STDOUT_FILENO);
close(p0[0]); close(p0[1]);
execlp("ls","ls",NULL);
}
else { //parent
wait(NULL);
wait(NULL); //waiting for c1 and c2
close(p0[0]); close(p0[1]);
printf("parent exit\n");
}
}
My questions are: Why don't the child processes finish? Is fork-pipe structure sound or am I doing it completely wrong?
Close the pipe before starting to wait in the last section.
Related
I'm working on my own little shell program in C. When I run the child process as a background process, I would like to ignore the input from the user coming from the parent process. I am currently trying to pipe it then close stdin for the child, but the input still goes to the child.
else // A process
{
pid_t child_pid;
char lastArgument = args[currArgsIndex-1][0];
if (lastArgument != '&'){ //Normal process
if((child_pid = fork()) == 0) {
execvp(filepath, args);
exit(0);
}
else
{
while(wait(NULL) != child_pid);
}
}
else { // Background
args[currArgsIndex-1] = NULL;
int process_pipe[2];
pipe(process_pipe); // Piping
if((child_pid = fork()) == 0) {
close(process_pipe[0]); // Ignore stdin for child
execvp(filepath, args);
exit(0);
}
}
}
You create a pipe and close the read end, but you never say that the pipe should be stdin.
It sounds like your intention was instead to 1. open the pipe only in the child, 2. close the write end so that no data can be read, 3. set the read end as stdin:
else { // Background
args[currArgsIndex-1] = NULL;
if((child_pid = fork()) == 0) {
int process_pipe[2];
pipe(process_pipe); // Piping
dup2(process_pipe[0], 0); // Copy read end as stdin
close(process_pipe[0]); // Close FD that is now unused
close(process_pipe[1]); // Close write end so no data can be read
execvp(filepath, args);
perror("execvp failed");
exit(1); // exit with error
}
}
There's no point having a pipe though. You can more easily open /dev/null for reading and setting that as stdin. Alternatively, simply close stdin entirely (some programs will complain):
else { // Background
args[currArgsIndex-1] = NULL;
if((child_pid = fork()) == 0) {
close(0); // Close stdin
execvp(filepath, args);
/* error handling */
}
Be aware that real shells allow redirecting to backgrounded processes, in which case none of the above will work:
wc -l < myfile &
Real shells will in fact not close or redirect stdin at all, but will put the command in its own process group that's not controlling the terminal. The process will then receive a SIGTSTP when it tries to read from stdin, and you can then use fg to bring it to the foreground to start typing data.
I came across a stripped down shell program in Tannenbaum's book on MINIX.
while(1) {
read_command(command, parameters);
if (fork() != 0)
wait(&status); /* parent code */
}
else {
execve(command, parameters, 0); /* child code */
}
}
When the infinite loop executes its first iteration, fork() will return 0 indicating it created a child process, when it executes the second time, wont fork() create a new child process ? How wait(&status) will ever execute?
I am new to understanding how an OS works/is built.
Thanks!
fork creates a new process immediately, so both processes see fork return, but with different return values. In the parent, the return value is the non-zero process ID of the child, so the parent executes wait. In the child, fork returns 0, so the child executes execve.
I'm coding a basic shell in C, and I'm working on suspending a child process right now.
I think my signal handler is correct, and my child process is suspending, but after that, the terminal should return to the parent process and that's not happening.
The child is suspended, but my shell isn't registering any input or output anymore. tcsetpgrp() doesn't seem to be helping.
Here's my signal handler in my shell code for SIGTSTP:
void suspend(int sig) {
pid_t pid;
sigset_t mask;
//mpid is the pgid of this shell.
tcsetpgrp(STDIN_FILENO, mpid);
tcsetpgrp(STDOUT_FILENO, mpid);
sigemptyset(&mask);
sigaddset(&mask, SIGTSTP);
sigprocmask(SIG_UNBLOCK, &mask, NULL);
signal(SIGTSTP, SIG_DFL);
//active.pid is the pid of the child currently in the fg.
if (active.pid != 0) {
kill(active.pid, SIGTSTP);
}
else{
//if this code is being run in the child, child calls SIGTSTP on itself.
pid = getpid();
if (pid != 0 && pid != mpid){
kill(pid, SIGTSTP);
}
}
signal(SIGTSTP, suspend);
}
Can anyone tell me what I'm doing wrong?
Am I suspending my shell along with the child, and do I need to return stdin and stdout to the shell somehow? How would I do this?
Thanks!
It's an old question but still I think I found an answer.
You didn't write your parent's code but I'm assuming its looks something like:
int main(){
pid_t pid = fork();
if(pid == 0) //child process
//call some program
else //parent process
wait(&status); //or waitpid(pid, &status, 0)
//continue with the program
}
the problem is with the wait() or waitpid(), it's look like if you run your program on OS like Ubuntu after using Ctrl+Z your child process is getting the SIGTSTP but the wait() function in the parent process is still waiting!
The right way of doing that is to replace the wait() in the parent with pause(), and make another handler that catch SIGCHLD. For example:
void sigHandler(int signum){
switch(signum){
case SIGCHLD:
// note that the last argument is important for the wait to work
waitpid(-1, &status, WNOHANG);
break;
}
}
In this case after the child process receive Ctrl+Z the parent process also receive SIGCHLD and the pause() return.
tcsetpgrp is to specify what is the foreground job. When your shell spawns a job in foreground (without &), it should create a new process group and make that the foreground job (of the controlling terminal, not whatever's on STDIN). Then, upon pressing CTRL-Z, that job will get the TSTP. It's the terminal that suspends the job, not your shell. Your shell shouldn't trap TSTP or send TSTP to anyone.
It should just wait() for the job it has spawned and detect when it has been stopped (and claim back the foreground group and mark the job as suspended internally). Your fg command would make the job's pgid the foreground process group again and send a SIGCONT to it and wait for it again, while bg would just send the SIGCONT
i used folk with signals for make process pause and resume with ctrl+c
video while is running : link
Code:
#include <stdio.h>
#include <unistd.h>
#include <signal.h>
void reverse_handler(int sig);
_Bool isPause=0;
_Bool isRunning=1;
int main()
{
int ppid;
int counter=0;
//make parent respond for ctrl+c (pause,resume).
signal(SIGINT,reverse_handler);
while(isRunning){
while(isPause==0)
{
/*code exec while process is resuming */
printf("\nc:%d",counter++);
fflush(stdout);
sleep(1);
}
//close parent after child is alive.
if((ppid=fork())==0){ exit(0); }
//make child respond for ctrl+c (pause,resume).
signal(SIGINT,reverse_handler);
//keep child alive and listening.
while(isPause==1){ /*code exec while process is pausing */ sleep(1); }
}
return 0;
}
//if process is pause made it resume and vice versa.
void reverse_handler(int sig){
if(isPause==0){
printf("\nPaused");
fflush(stdout);
isPause=1;
}
else if(isPause==1){
printf("\nresuming");
fflush(stdout);
isPause=0;
}
}
i hope that's be useful.
please comment me if there's any questions
I might be late to answer the question here but this is what worked when I was stuck with the same problem. According to the man pages for tcsetpgrp()
The function tcsetpgrp() makes the process group with process group ID
pgrp the foreground process group on the terminal associated to fd,
which must be the controlling terminal of the calling process, and
still be associated with its session. Moreover, pgrp must be a
(nonempty) process group belonging to the same session as the calling
process.
If tcsetpgrp() is called by a member of a background process group in
its session, and the calling process is not blocking or ignoring
SIGTTOU, a SIGTTOU signal is sent to all members of this background
process group.
So, what worked for me was ignoring the signal SIGTTOU in the shell program, before I created the processes that would come to the foreground. If I do not ignore this signal, then the kernel will send this signal to my shell program and suspend it.
So I've got a problem with my code and it's driving me nuts since I can't seem to figure out what's causing this. Basically, I'm trying to write to a file that I have open, using multiple child processes with fork(). Before I start forking, I can write to it just fine, but once I fork and THEN do an if statement to see if it's the child process, it won't write.
Basically, what I have is this:
FILE *output = NULL;
output = fopen(...); // Done successfully
fprintf(output, "This writes okay\n");
// Fork n processes
for (i = 0; i <= n; n++)
{
pid[i] = fork();
fprintf(output, "We can still write\n");
if (pid[i] == 0) // Child process
{
fprintf(output, "This won't write to output\n");
printf("I can still write and calculate stuff otherwise\n");
...
}
}
Can anyone figure out why it's not able to write after checking to see if it's a child process? There's definitely no error with forking from what I've seen.
Opening a file descriptor and using it in more than one process (for example in both a parent and child process) is a race condition and will most likely cause problems and unpredictable behavior.
fork() is used to create a child process...and you see this call appear in the child process as well. I don't understand what it means when they say that 'calls to fork actually return twice'.
And what does this mean...
if (fork() == 0)
/* the child process's thread executes here*/
else
/*the parent process's thread executes here*/
Is the above code part of parent or child. Can you explain in plain English what's going on?
Also, why use a fork()? It says all processes in unix run by this system call? How do you fork() so other programs can run? Do you specify the name of the program?
It's a bit like this:
Process 1 Process 2
int main() {
...
int x = goo();
...
int y = fork();
// fork() returns... // ... but also here!
// here y = 123 // here y = 0
if (y) { if (y) {
// this happens // false
} else { } else {
// false // this happens
} }
int z = baz(); int z = baz();
... ...
return 0; return 0;
} }
When Process 2 comes to life, the program exists twice, and the second process starts with the return of fork(). Since the program is the same in both processes, the only way to distinguish which process you're in is by the return value of fork().
What the mean when they say it returns twice is that the call returns once in the parent process (which called it), and once in the child process (which didn't, although you could probably argue that the child inherited the act of calling fork from the parent just like it inherited so much else).
The code snippet takes advantage of the fact that get a different return value from fork depending on whether you're the parent process or the child process.
The child process gets zero and the parent process gets the non-zero process ID of the child.
You can also get back -1 if the fork fails for some reason, in which case the child won't be running. That's something you should probably check as well.
And, while fork is used to create new processes, it's the exec family of calls which load new programs into those processes: fork on its own cannot do that.
A good overview of the process can be found here.