I have some doubt regarding the following code.
#include <stdio.h>
#include <sys/types.h>
int main(void)
{
int pid=fork();
if(pid==0) sleep(5);
printf("Hello World %d %d\n",getpid(),pid);
if(pid>0) while(1){sleep(1);}
if(pid==0) printf("In child process!\n");
return 0;
}
Will the child process ever terminate? Will it remain in zombie state?
In your example, the child process dies but the parent doesn't know about it's exit status. As such, the child (now zombie) is left in the process table.
Moreover, the parent continues to wait for the child and keeps running.
Yes it is zombie ... your Child process dies and your parent does not know the exit status of child process. and parent process is running even after child dies.
ZOMBIE process: when child process dies parent process still running. in this case parent process does not know the exit status of child.
To avoid this parent process uses wait or waitpid to know the child status.
Adding on to other responses.. If the opposite were to happen i.e. Parent dies before the child process completes, then the child would become an orphan and would later be "adopted" by a special system process called init.
Related
if (fork() == 0) for(;;) sleep(1);
else exit(0);
I know what is zombie and orphan process but dont know how to anwser this question. Please explain me how many zombie and orphan processes can be generated by executing this code . Thanks!
There is only one fork, so there is only 1 child. The parent exits without waiting for a child process. The child enter in an endless loop and shortly thereafter the parent is exists, so it becomes an orphaned process. So: 0 zombies, 1 orphaned.
However if the fork fails, then there is no child process at all and the parent exists anyway.
if (fork() == 0) {
printf("[Child] I'm exiting\n");
}
else {
wait(NULL);
printf("[Parent] child exited\n");
}
wait(NULL) waits for any child to complete, and then it will execute everything after it. So in this case, what would be the output of the program?
We might go into the parent first, and then we will have to wait for the child to complete, so its [Child] ..... , [Parent] .... printed.
We can also go into child, print its stuff, and then parent, in which case wait(NULL) will wait forever, since the child process is already completle. So in this case would an output be just [Child].....?
I ran 1000 times but it looks like the output its always child ... parent
Thanks
If the child hasn't exited, the parent will wait for it until it exits.
If the child has exited, the parent will obtain its status information and then continue executing. In this case, the function wait() will return immediately. See this.
So this program will always print:
[Child] I'm exiting
[Parent] child exited
Looks like the underlaying question you are actually asking is how fork works.
When you do fork(), operating system duplicates the calling process (the parent). It creates a child, which is an exact duplicate of the parent (minus some details). The return value of the fork() function is 0 in child process and child's PID in parent process.
So, after your fork, child process will execute the code under the if clause, and the parent will execute the else clause. Keep in mind you now have 2 processes executing in parallel. The parent process, though, is blocked due to wait(NULL) until the child exits. When child (prints the message) and exits, parent continues to execute the code after wait, printing its message and exiting.
That's why you'll always see:
[Child] I'm exiting
[Parent] child exited
I am new to processes in linux and c.
I am using this straightforward example:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main(int argc, const char * argv[]) {
pid_t child_pid_or_zero = fork(); //fork returns twice
if(child_pid_or_zero < 0)
{
//if fork returns a number smaller than zero, something wrong happened
perror("Something wrong happened\n");
exit(-1);
}
if(child_pid_or_zero > 0)
{
//if fork returns a number greater than zero, this is the parent process
printf("I'm the parent, my pid is: %d\t My child pid is %d\n", getpid(), child_pid_or_zero);
wait(NULL);
}
else
{
//this means that fork now returned 0, the child process is running
printf("I am the child with pid: %d\t My parent pid is: %d\n",child_pid_or_zero, getppid());
}
return 0;
}
If I were to omit the wait() method in the
if(child_pid_or_zero > 0)
What would happen? I tried this myself, and apparently, there was no immediate difference. Do we always need to use a wait(), or does this only apply when the child is supposed to perform heavy calculations etc ?
Thanks in advance.
Wait is for listening to state changes and obtaining information about the child. A state change is child termination, stopping or resuming by a signal. Wait allows the system to release the resources associated with the child. If a wait is not performed, then the terminated child remains in a "zombie" state.
The kernel maintains a minimal set of information about the zombie
process (PID, termination status,
resource usage information) in order to allow the parent to later perform a wait to obtain information about the child. As long
as a zombie is not removed from the system via a
wait, it will consume a slot in the kernel process table, and if this table fills, it will not be possible to create further
processes. If a parent process terminates, then its
"zombie" children (if any) are adopted by init(1), which automatically performs a wait to remove the zombies.
The system call wait(2) is typically used to find if the child process's state has changed (i.e. whether it's still running, exited, etc).
Another purpose is to avoid "zombie" processes. If parent process doesn't wait on the child process and the child process exits before the parent process then it becomes a "zombie" process. So, a wait() call is used to "reap" the process and release the system resources associated with the process.
Imagine if the parent process is a long running one and creates several child processes at regular intervals then all the zombie processes will have entries in the process table which is an unncessary use of system resources.
after the fork, you'll have two independent process.
whit the wait() call, you tell the parent process to wait the child process to terminate.
In this example, nothing changes since the two process are not interacting with each other, so the parent can just exit after creating the child and printing the string, but in the scenario where the parent has to wait the child to do some operations and then maybe return some values to the parent, it becomes useful!
A Zombie is created when a parent process does not use the wait system call after a child dies to read its exit status, and an orphan is child process that is reclaimed by init when the original parent process terminates before the child.
In terms of memory management and the process table how are these processes handled differently, specifically in UNIX?
What is an example or extreme case when the creation of zombies or orphans can be detrimental to the greater application or system?
When a child exits, some process must wait on it to get its exit code. That exit code is stored in the process table until this happens. The act of reading that exit code is called "reaping" the child. Between the time a child exits and is reaped, it is called a zombie. (The whole nomenclature is a bit gruesome when you think about it; I recommend not thinking about it too much.)
Zombies only occupy space in the process table. They take no memory or CPU. However, the process table is a finite resource, and excessive zombies can fill it, meaning that no other processes can launch. Beyond that, they are bothersome clutter, and should be strongly avoided.
If a process exits with children still running (and doesn't kill its children; the metaphor continues to be bizarre), those children are orphans. Orphaned children are immediately "adopted" by init (actually, I think most people call this "reparenting," but "adoption" seems to carry the metaphor better). An orphan is just a process. It will use whatever resources it uses. It is reasonable to say that it is not an "orphan" at all since it has a parent, but I've heard them called that often.
init automatically reaps its children (adopted or otherwise). So if you exit without cleaning up your children, then they will not become zombies (at least not for more than a moment).
But long-lived zombies exist. What are they? They're the former children of an existing process that hasn't reaped them. The process may be hung. Or it may be poorly written and forgets to reap its children. Or maybe it's overloaded and hasn't gotten around to it. Or whatever. But for some reason, the parent process continues to exist (so they aren't orphans), and they haven't been waited on, so they live on as zombies in the process table.
So if you see zombies for longer than a moment, then it means that there is something wrong with the parent process, and something should be done to improve that program.
When a process terminates, its resources are deallocated by the operating
system. However, its entry in the process table must remain there until the
parent calls wait(), because the process table contains the process’s exit status.
A process that has terminated, but whose parent has not yet called wait(), is
known as a zombie process. All processes transition to this state when they
terminate, but generally they exist as zombies only briefly. Once the parent
calls wait(), the process identifier of the zombie process and its entry in the
process table are released.
Now consider what would happen if a parent did not invoke wait() and
instead terminated, thereby leaving its child processes as orphans. Linux and
UNIX address this scenario by assigning the init process as the new parent to orphan processes. The init process periodically
invokes wait(), thereby allowing the exit status of any orphaned process to be
collected and releasing the orphan’s process identifier and process-table entry.
Source: Operating System Concepts by Abraham, Peter, Greg
An orphan process is a computer process whose parent process has finished or terminated, though it (child process) remains running itself.
A zombie process or defunct process is a process that has completed execution but still has an entry in the process table as its parent process didn't invoke an wait() system call.
Orphan -
Parent exit , Init process becomes the parent of child process.
Whenever child is terminated, process table gets deleted by os.
Zombie -
When the child terminates it gives exit status to parent.
Meanwhile time suppose your parent is in sleep state and unable to receive any status from child.
Though the child exit but the process occupies space in process table
check out this command in linux ubuntu >>ps -eo pid,ppid,status,cmd
If you found something like defunc at the end i.e your process is zombie and occupying space.
Zombie Process:
A process that has finished the execution but still has an entry in the process table to report to its parent process is known as a zombie process. A child process always first becomes a zombie before being removed from the process table. The parent process reads the exit status of the child process which reaps off the child process entry from the process table.
Orphan Process:
A process whose parent process no more exists i.e. either finished or terminated without waiting for its child process to terminate is called an orphan process.
There are no orphans but the process using PID 1.
From the running process' point of view it makes no difference whether it was started directly and therefore has PID 1 as parent or got inherited by PID 1 because its original parent (being different from PID 1) ended.
It is handled like any other process.
Each process goes through some sort of zombie state, when ending, namely the phase between announcing its end by issuing SIGCHLD and having its processing (delivery or ignorance) acknowledged.
When the zombie state had been entered the process is just an entry in the system's process list.
The only significant resource a zombie is exclusively using is a valid PID.
I would like to add 2 code snippets featuring an orphan and a zombie process. But first, I will post the definition of these processes as stated in the book "Operating System Concepts" by Silberschatz, Galvin and Gagn:
If no parent waiting (did not invoke wait()) process is a zombie
If parent terminated without invoking wait , process is an orphan
Orphan
// A C program to demonstrate Orphan Process.
// Parent process finishes execution while the
// child process is running. The child process
// becomes orphan.
#include <stdio.h> //printf
#include <stdlib.h> //exit
#include <sys/types.h> //fork
#include <unistd.h> //fork and sleep
int main()
{
// Fork returns process id
// in parent process
pid_t child_pid = fork();
// Parent process didn't use wait and finished before child
// so the child becomes an orphan process
// Parent process
if (child_pid > 0) {
printf("I finished my execution before my child");
}
else // Child process
if (child_pid == 0) {
sleep(1); //sleep for 1 second
printf("This printf will not be executed");
}
else{
//error occurred
}
return 0;
}
Output
I finished my execution before my child
Zombie
// A C program to demonstrate Zombie Process.
// Child becomes Zombie as parent is not waiting
// when child process exits.
#include <stdio.h> //printf
#include <stdlib.h> //exit
#include <sys/types.h> //fork
#include <unistd.h> //fork and sleep
int main()
{
// Fork returns process id
// in parent process
pid_t child_pid = fork();
// Parent process didn't use wait
// so the child becomes a zombie process
// Parent process
if (child_pid > 0){
sleep(1); //sleep for 1 second
printf("\nI don't wait for my child");
}
else // Child process
if(child_pid == 0){
printf("My parent doesn't wait me");
exit(0);
}
else{
//error occurred
}
return 0;
}
Output
My parent doesn't wait me
I don't wait for my child
Edit: Source and inspiration taken from here
A process which has finished the execution but still has the entry in the process table to report to its parent process is known as a zombie process.
A process whose parent process no more exists i.e. either finished or terminated without waiting for its child process to terminate is called an orphan process
Here is one summary
Zombie Process
Orphan Process
A Zombie is a process that has completed its task but still, shows an entry in a process table.
A child process that remains running even after its parent process is terminated or completed without waiting for the child process execution is called an orphan.
Zombie process states always indicated by Z
The orphan process was created unknowingly due to a system crash.
The zombie process is treated as dead they are not used for system processing
An orphan process is a computer process even after its parent terminates init it becomes a parent and continues the remaining task.
wait() system call is used to deal with zombie processes
Kernel allocates a new process as parent process to orphan process. Mostly the new parent is the init process (pid=1).
To remove the zombie process executes the kill command.
Terminate the Orphan process using the SIGHUP signal.
Source:
Zombie vs. Orphan Processes
Difference between zombie orphan and daemon processes
Zombie and Orphan Process in OS
This question already has answers here:
What is the reason for performing a double fork when creating a daemon?
(9 answers)
Closed 8 years ago.
Nagios lets me configure child_processes_fork_twice=<0/1>.
The documentation says
This option determines whether or not Nagios will fork() child processes twice when it executes host and service checks. By default, Nagios fork()s twice. However, if the use_large_installation_tweaks option is enabled, it will only fork() once.
As far as I know fork() will spawn a new child process. Why would I want to do that twice?
All right, so now first of all: what is a zombie process? It's a process that is dead, but its parent was busy doing some other work, hence it could not collect the child's exit status. In some cases, the child runs for a very long time, the parent cannot wait for that long, and will continue with it's work (note that the parent doesn't die, but continues its remaining tasks but doesn't care about the child). In this way, a zombie process is created. Now let's get down to business. How does forking twice help here? The important thing to note is that the grandchild does the work which the parent process wants its child to do. Now the first time fork is called, the first child simply forks again and exits. This way, the parent doesn't have to wait for a long time to collect the child's exit status (since the child's only job is to create another child and exit). So, the first child doesn't become a zombie. As for the grandchild, its parent has already died. Hence the grandchild will be adopted by the init process, which always collects the exit status of all its child processes. So, now the parent doesn't have to wait for very long, and no zombie process will be created. There are other ways to avoid a zombie process; this is just a common technique. Hope this helps!
In Linux, a daemon is typically created by forking twice with the intermediate process exiting after forking the grandchild. This has the effect of orphaning the grandchild process. As a result, it becomes the responsibility of the OS to clean up after it if it terminates. The reason has to do with what are known as zombie processes which continue to live and consume resources after exiting because their parent, who'd normally be responsible for the cleaning up, has also died.
Also from the documentation,
Normally Nagios will fork() twice when it executes host and service checks. This is done to (1) ensure a high level of resistance against plugins that go awry and segfault and (2) make the OS deal with cleaning up the grandchild process once it exits.
Unix Programming Faq §1.6.2:
1.6.2 How do I prevent them from occuring?
You need to ensure that your parent process calls wait() (or
waitpid(), wait3(), etc.) for every child process that terminates;
or, on some systems, you can instruct the system that you are
uninterested in child exit states.
Another approach is to fork() twice, and have the immediate child
process exit straight away. This causes the grandchild process to be
orphaned, so the init process is responsible for cleaning it up. For
code to do this, see the function fork2() in the examples section.
To ignore child exit states, you need to do the following (check your
system's manpages to see if this works):
struct sigaction sa;
sa.sa_handler = SIG_IGN;
#ifdef SA_NOCLDWAIT
sa.sa_flags = SA_NOCLDWAIT;
#else
sa.sa_flags = 0;
#endif
sigemptyset(&sa.sa_mask);
sigaction(SIGCHLD, &sa, NULL);
If this is successful, then the wait() functions are prevented from
working; if any of them are called, they will wait until all child
processes have terminated, then return failure with errno == ECHILD.
The other technique is to catch the SIGCHLD signal, and have the
signal handler call waitpid() or wait3(). See the examples section
for a complete program.
This code demonstrates how to use the double fork method to allow the grandchild process to become adopted by init, without risk of zombie processes.
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
int main()
{
pid_t p1 = fork();
if (p1 != 0)
{
printf("p1 process id is %d", getpid());
wait();
system("ps");
}
else
{
pid_t p2 = fork();
int pid = getpid();
if (p2 != 0)
{
printf("p2 process id is %d", pid);
}
else
{
printf("p3 process id is %d", pid);
}
exit(0);
}
}
The parent will fork the new child process, and then wait for it to finish. The child will fork a grandchild process, and then exit(0).
In this case, the grandchild doesn't do anything except exit(0), but could be made to do whatever you'd like the daemon process to do. The grandchild may live long and will be reclaimed by the init process, when it is complete.