The purpose of the wait() in parent c

Question

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.

http://stackoverflow.com/questions/23709888/how-to-use-wait-in-c — Sachith Wickramaarachchi, Oct 11 '16 at 11:48
when compiling, always enable all the warnings, then fix those warnings. — user3629249, Oct 12 '16 at 18:33
for ease of readability and understanding: consistently indent the code. indent after EVERY opening brace '{'. un-indent before EVERY closing brace '}'. — user3629249, Oct 12 '16 at 18:34
a child process should never re-enter the parent code. I.E. end the child process with 'exit();' — user3629249, Oct 12 '16 at 18:37

score 4 · Accepted Answer · answered Oct 11 '16 at 11:56

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.

not quite correct. 1) the user cannot kill a zombie except by re-booting. 2) the `init` process will not perform a `wait()`, so the zombie will be around forever, or until a reboot occurs — user3629249, Oct 12 '16 at 18:41
a zombie process is created when its' parent exits, so the parent (which no longer exists) cannot remove it. A zombie becomes a child of the `init` process and stays around forever — user3629249, Oct 12 '16 at 18:44

score 2 · Answer 2 · answered Oct 11 '16 at 11:57

2

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.

answered Oct 11 '16 at 11:57

P.P

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Can you elaborate a little more? Like providing an example where this could lead do these zombies? – Koen Oct 11 '16 at 12:33
1

It's quite simple to do in your example. In the parent process (`if hild_pid_or_zero > 0)` part), add a `sleep(10);` as the first statement. Then execute your code and in another window in the terminal, if you run `ps -af` you'll see the zombie process (because the parent hasn't reaped it yet and child process exited already). – P.P Oct 11 '16 at 13:08

score 1 · Answer 3 · answered Oct 11 '16 at 11:58

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!

The purpose of the wait() in parent c

3 Answers3

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