Background process killing with Parent process in C - c

I have the following code in my main function
pid_t pid;
pid = fork(); //Two processes are made
if (pid > 0 && runBGflag==0) //Parent process. Waits for child termination and prints exit status
{
int status;
if (waitpid(pid, &status, 0) == pid && WIFEXITED(status))
{
printf("Exitstatus [");
for (int i = 0; i < noOfTokens; i++)
{
printf("%s ", commands[i]);
}
printf("\b] = %d\n", WEXITSTATUS(status));
}
}
else if (pid == 0) //Child process. Executes commands and prints error if something unexpected happened
{
if (runBGflag==1) insertElement(getpid(),ptr);
execvp(commands[0], commands);
printf ("exec: %s\n", strerror(errno));
exit(1);
}
In a nutshell, a child process is made and if the runBackGround flag is set, the parent process will not wait for the child process to exit, but rather continue running. If a background process is made, the PID of the background process is stored in a list. At a later point, this function is called
void delete_zombies(void)
{
pid_t kidpid;
int status;
char buffer[1337];
while ((kidpid = waitpid(-1, &status, WNOHANG)) > 0)
{
removeElement(kidpid,buffer,1337);
printf("Child %ld terminated\n", kidpid);
printf("its command was %s\n",buffer);
}
}
This function simply checks if any child processes have died and in that case deletes them. It will then search for the childs PID in the list, remove it and print it out.
The problem is, the delete_zombies function will find that a child has died and will then try to remove it from the list, but it only finds an empty list, as if the child process never inserted its PID into the list.
This is really strange, because delete_zombies only finds a dead child process, when there was one created with the background flag set, so we know insertElement must have been called, but strangely when the parent checks in the list nothing is there
Is the cause for that, that child process and parent process have seperate lists, or is the PID maybe wrong?

Related

Setting parent to child process group with setpgid fails

I'm trying to change the pgrp of the processes to that of the child's so i can setsid on the parent process. The only thing is I keep getting an EPERM error code. Both processes have the same session group, according to htop.
I'm basing this off of this blog post, so I can change which terminal output gets directed to.
void sig_exit(int signum)
{
_Exit(0);
}
pid_t change_process_group()
{
pid_t child_pid;
if ((child_pid = fork()) < 0)
{
perror("fork failed while attaching to term");
exit(1);
}
if (child_pid == 0)
{
pid_t parent = getppid();
setpgid(0, getpid());
signal(SIGUSR1, sig_exit); // wait till parent tells child to exit
//sleep(5);
//kill(parent, SIGUSR2);
pause();
printf("Shouldn't reach this\n");
}
//sleep(5);
//signal(SIGUSR2, sig_wait);
//pause();
int parent_pid = getpid();
int code = setpgid(parent_pid, child_pid); // need child process group
printf("%s\n", strerror(errno));
setsid();
return child_pid;
}
main()
{
pid_t child = change_process_group();
kill(child, SIGUSR1);
}
The commented out lines were from me thinking the processes might not be executing in the correct order, but those don't appear to fix the problem.
How may I correctly use setpgid to change the pgrp of the parent process to the child's?
This is a race condition and it works if you uncomment the sleep(5) line in the parent. When you call setpgid(parent_pid, child_pid), the child_pid process group must exist. It isn't enough that there exists a process with the PID child_pid: setpgid needs an existing process group unless the process is putting itself into its own group. If setpgid(parent_pid, child_pid) in the parent runs after setpgid(0, getpid()) in the child, it works.
Sleeping is both inefficient and fragile, so instead of that the parent should wait for a notification from the child. Signals are fragile because there aren't many different signals and they could come from anywhere. A good way to communicate between related processes is a pipe. Since all you need here is a one-time notification, you can set up a pipe and read from it in the parent (with the write end closed in the parent). The parent will wait until the child writes to the pipe or closes it. In the child, just close the write end of the pipe when you've finished the preparations. The parent's read call (or select if you need to do other things at the same time) will return.
Proof-of-concept code:
pid_t change_process_group()
{
pid_t child_pid;
int child_ready_pipe[2];
if (pipe(child_ready_pipe) < 0)
{
perror("pipe");
exit(1);
}
if ((child_pid = fork()) < 0)
{
perror("fork failed while attaching to term");
exit(1);
}
if (child_pid == 0)
{
close(child_ready_pipe[0]);
sleep(1); // mimic slow start of the child
if (setpgid(0, 0))
perror("child setpgid to create group");
close(child_ready_pipe[1]);
signal(SIGUSR1, sig_exit); // wait till parent tells child to exit
pause();
printf("Shouldn't reach this\n");
}
close(child_ready_pipe[1]);
int parent_pid = getpid();
char ignored;
read(child_ready_pipe[0], &ignored, 1);
close(child_ready_pipe[0]);
if (setpgid(parent_pid, child_pid) < 0) // need child process group
perror("parent setpgid");
if (setsid() < 0)
perror("parent setsid");
return child_pid;
}

Only 1 child after 5 forks (C)

I am trying to make a processor farm in C. I start with opening message queues, and afterwards try to make worker processes: (note that NROF_WORKERS is 5)
static void
makechildren (void) {
// Only the parent should fork. Any children created will become workers.
pid_t processID;
pid_t farmerPID = getpid(); // To identify who the farmer is
// Loop creating processes, indexed by NROF_WORKERS
int i = 0;
while (i < NROF_WORKERS){
if (getpid() == farmerPID){
i++;
printf ("Parent is creating a child!%d\n", getpid());
processID = fork();
}
}
if (processID < 0){
perror("fork() failed");
exit(1);
}
else {
// If parent, start farming
if (processID == farmerPID) {
printf("Parent reporting in!%d\n");
}
// If child, become a worker
if (processID == 0) {
printf("Child reporting in!%d\n", getpid());
join();
}
}
}
As you can see, I want the parent to report any time a child is created, and afterwards I want the parent and all children to report. However, this is all I get:
Parent is creating a child!11909
Parent is creating a child!11909
Parent is creating a child!11909
Parent is creating a child!11909
Parent is creating a child!11909
Child reporting in!11914
Now, I do notice the difference in 11909 and 11914 is 5. So my question: are the other processes created? If so, how come they don't report? And if not, what am I doing wrong? Also, the parent is not reporting at all, how is this caused?
All of the children are created, but will loop forever in the while loop, as i is incremented only for the parent:
int i = 0;
while (i < NROF_WORKERS){
if (getpid() == farmerPID){
i++; // <---- This is happening for the parent process only.
printf ("Parent is creating a child!%d\n", getpid());
processID = fork();
}
}
The only child to terminate is the last one, for which the i is equal to NROF_WORKERS.
Also parent is "not reporting" since the processID you are checking to be equal to the parent PID is never equal to it, as it is equal to the latest fork result, i.e. the latest created child PID:
.........
processID = fork();
.........
.........
if (processID == farmerPID) {
printf("Parent reporting in!%d\n");
}
You always print the farmerPid! But as the message is printed 5 times, you effectively created 5 processes:
while (i < NROF_WORKERS){
if (getpid() == farmerPID){
i++;
printf ("Parent is creating a child!%d\n", getpid());
processID = fork();
}
}
If you want to print the children pids then your code must makes a difference in between parent and child, as in:
while (i < NROF_WORKERS){
if (getpid() == farmerPID){
i++;
printf ("Parent is creating a child!\n");
processID = fork();
if (processID==0) { // child
printf("I am the child %d\n",getpid());
} else { // parent
printf("Parent just created child %d\n",processID);
}
}
}

How to properly fork() a process

I'm trying to understand how to properly used fork() and execvp() to execute a command. So far I have the following code:
When I run ./test vim myFile.c it correctly opens myFile.c but I get strange behavior. It seems as though there are two processes running because whenever I enter anything it seems to happen twice. Why is this?
int main (int argc, char* argv[]) {
int fdin, pid, w, status;
fdin = 0;
if ((pid = fork()) < 0)
errorExit (EXIT_FAILURE);
execvp(argv[0],argv);
do {
w = waitpid(cpid, &status, WUNTRACED | WCONTINUED);
if (w == -1) {
perror("waitpid");
exit(EXIT_FAILURE);
}
if (WIFEXITED(status)) {
printf("exited, status=%d\n", WEXITSTATUS(status));
} else if (WIFSIGNALED(status)) {
printf("killed by signal %d\n", WTERMSIG(status));
} else if (WIFSTOPPED(status)) {
printf("stopped by signal %d\n", WSTOPSIG(status));
} else if (WIFCONTINUED(status)) {
printf("continued\n");
}
} while (!WIFEXITED(status) && !WIFSIGNALED(status))
}
When you call fork(), you create two almost completely identical processes, the parent process and the child process. The only difference between these two processes is the return value of fork(), which returns 0 to the child process and the pid of the child to the parent.
Hence, assuming fork succeeded, fork will return a nonnegative integer to both the parent and child process in line 4. Then, both the parent and child process will execute line 6, the execvp, and hence, you end up with two different processes running your vim myFile.c, causing all the problems you described.
The standard idiom is something like:
if ((pid = fork()) < 0) {
// Handle fork error
}
else if (pid == 0) {
// Child process
execvp(...);
}
else {
// Parent process
w = waitpid(pid, ...);
}
Since the return value for fork is 0 for the child, after fork succeeds, the test (pid == 0) will be true for the child, so execvp will be called.
For the parent, fork returns the pid of the child, so the check (pid == 0), which still get executed, is false, so the else condition is executed, causing the parent to wait for the child.
Both parent and child in your program get execvp():
if ((pid = fork()) < 0)
errorExit (EXIT_FAILURE);
execvp(argv[0],argv);
You should check, if you are in parent with pid != 0, and if you are in child otherwise.
You should look at the return value of fork, after a successful fork you will have two running processes at the same position in your program. The child process will get a return value of 0, the parent will get a return value which is the pid of the child. Most likely you want to do different things in the child process and the parent process.
You might also want to think again about how execvp is called. Do you really want to give "./test" as the first argument to execvp?

How to make parent wait for all child processes to finish?

I'm hoping someone could shed some light on how to make the parent wait for ALL child processes to finish before continuing after the fork. I have cleanup code which I want to run but the child processes need to have returned before this can happen.
for (int id=0; id<n; id++) {
if (fork()==0) {
// Child
exit(0);
} else {
// Parent
...
}
...
}
pid_t child_pid, wpid;
int status = 0;
//Father code (before child processes start)
for (int id=0; id<n; id++) {
if ((child_pid = fork()) == 0) {
//child code
exit(0);
}
}
while ((wpid = wait(&status)) > 0); // this way, the father waits for all the child processes
//Father code (After all child processes end)
wait waits for a child process to terminate, and returns that child process's pid. On error (eg when there are no child processes), -1 is returned. So, basically, the code keeps waiting for child processes to finish, until the waiting errors out, and then you know they are all finished.
POSIX defines a function: wait(NULL);. It's the shorthand for waitpid(-1, NULL, 0);, which will suspends the execution of the calling process until any one child process exits.
Here, 1st argument of waitpid indicates wait for any child process to end.
In your case, have the parent call it from within your else branch.
Use waitpid() like this:
pid_t childPid; // the child process that the execution will soon run inside of.
childPid = fork();
if(childPid == 0) // fork succeeded
{
// Do something
exit(0);
}
else if(childPid < 0) // fork failed
{
// log the error
}
else // Main (parent) process after fork succeeds
{
int returnStatus;
waitpid(childPid, &returnStatus, 0); // Parent process waits here for child to terminate.
if (returnStatus == 0) // Verify child process terminated without error.
{
printf("The child process terminated normally.");
}
if (returnStatus == 1)
{
printf("The child process terminated with an error!.");
}
}
Just use:
while(wait(NULL) > 0);
This ensures that you wait for ALL the child processes and only when all have returned, you move to the next instruction.

How to use Fork() to create only 2 child processes?

I'm starting to learn some C and while studying the fork, wait functions I got to a unexpected output. At least for me.
Is there any way to create only 2 child processes from the parent?
Here my code:
#include <sys/types.h>
#include <stdio.h>
#include <unistd.h>
#include <sys/wait.h>
int main ()
{
/* Create the pipe */
int fd [2];
pipe(fd);
pid_t pid;
pid_t pidb;
pid = fork ();
pidb = fork ();
if (pid < 0)
{
printf ("Fork Failed\n");
return -1;
}
else if (pid == 0)
{
//printf("I'm the child\n");
}
else
{
//printf("I'm the parent\n");
}
printf("I'm pid %d\n",getpid());
return 0;
}
And Here is my output:
I'm pid 6763
I'm pid 6765
I'm pid 6764
I'm pid 6766
Please, ignore the pipe part, I haven't gotten that far yet. I'm just trying to create only 2 child processes so I expect 3 "I'm pid ..." outputs only 1 for the parent which I will make wait and 2 child processes that will communicate through a pipe.
Let me know if you see where my error is.
pid = fork (); #1
pidb = fork (); #2
Let us assume the parent process id is 100, the first fork creates another process 101. Now both 100 & 101 continue execution after #1, so they execute second fork. pid 100 reaches #2 creating another process 102. pid 101 reaches #2 creating another process 103. So we end up with 4 processes.
What you should do is something like this.
if(fork()) # parent
if(fork()) #parent
else # child2
else #child1
After you create process , you should check the return value. if you don't , the seconde fork() will be executed by both the parent process and the child process, so you have four processes.
if you want to create 2 child processes , just :
if (pid = fork()) {
if (pid = fork()) {
;
}
}
You can create n child processes like this:
for (i = 0; i < n; ++i) {
pid = fork();
if (pid > 0) { /* I am the parent, create more children */
continue;
} else if (pid == 0) { /* I am a child, get to work */
break;
} else {
printf("fork error\n");
exit(1);
}
}
When a fork statement is executed by the parent, a child process is created as you'd expect. You could say that the child process also executes the fork statement but returns a 0, the parent, however, returns the pid.
All code after the fork statement is executed by both, the parent and the child.
In your case what was happening was that the first fork statement created a child process. So presently there's one parent, P1, and one child, C1.
Now both P1 and C1 encounter the second fork statement. The parent creates another child (c2) as you'd expect, but even the child, c1 creates a child process (c3). So in effect you have P1, C1, C2 and C3, which is why you got 4 print statement outputs.
A good way to think about this is using trees, with each node representing a process, and the root node is the topmost parent.
you can check the value as
if ( pid < 0 )
process creation unsuccessful
this tells if the child process creation was unsuccessful..
fork returns the process id of the child process if getpid() is used from parent process..
You can create a child process within a child process. This way you can have 2 copies of the original parent process.
int main (void) {
pid_t pid, pid2;
int status;
pid = fork();
if (pid == 0) { //child process
pid2 = fork();
int status2;
if (pid2 == 0) { //child of child process
printf("friends!\n");
}
else {
printf("my ");
fflush(stdout);
wait(&status2);
}
}
else { //parent process
printf("Hello ");
fflush(stdout);
wait(&status);
}
return 0;
}
This prints the following:
Hello my friends!

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