So I have this simple program that sleeps for 4 second if the value returned by fork is '0' meaning that the child process is executing, I've tried using sleep in child process but the program is blocked, and flushing standard output isn't working...
code:
#include <stdio.h>
#include <unistd.h>
int main(int argc, char const *argv[]) {
pid_t value = fork();
if (value == 0) {
sleep(4);
}
printf("Value returned by fork: %d\n", value);
printf("I'm the process N°%d\n", getpid());
return 0;
}
I'm running on Ubuntu 20.04.3 LTS.
Output:
Value returned by fork: 12618
I'm the process N°12617\
farouk#farouk-HP-Pavilion-Desktop-TP01-1xxx:~/sysexp$ Value returned by fork: 0
I'm the process N°12618
To allow this question to have an accepted answer.
The child process is not blocking the shell. The shell gave its prompt and the child wrote some output after the prompt, leaving the cursor at the start of a line without a shell prompt visible — because the shell prompt already appeared earlier.
There are a variety of ways around this.
The simplest is just to type a command such as ps and hit return, noting that the shell executes it, and that the ps output does not list the child process. If you type the ps command quick enough, you might see the child listed in the output before its output appears.
Another is to modify the program so that it waits for all child processes to exit before it exits — using wait() or waitpid(). The same code can be used in the child and the parent since the child will have no children of its own. The call to the wait function will return immediately with a 'no more children' status (error).
You can find extensive discussion of all this in the comments — I've chosen to make this a Community Wiki answer since there was a lot of activity in the comments that identified the gist of this answer.
Related
I am trying to learn the fork() and wait() system calls. If I run this code :
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/wait.h>
int main (){
printf("Hi, I am the parent with pid %d \n ",getpid());
int rc = fork();
printf("Fork returned : %d \n ",rc);
printf("I am the process with pid %d \n ",getpid());
wait(NULL);
return 0;
}
I get the output as expected on the terminal :
Hi, I am the parent with pid 3639
Fork returned : 3640
I am the process with pid 3639
Fork returned : 0
I am the process with pid 3640
However , If I remove wait(NULL) , I get a strange output on the terminal :
Hi, I am the parent with pid 3715
Fork returned : 3716
I am the process with pid 3715
John#John-VirtualBox:~/Fork5$ Fork returned : 0
I am the process with pid 3716
I totally understand that , we use wait() to make the parent process waits for the child to end executiion so that we can remove it from the process table and deallocate its PID . But here , if I remove the wait , we see that the terminal is called again :
John#John-VirtualBox:~/Fork5$ Fork returned : 0
I am the process with pid 3716
And even it doesn't return again back . I don't understand what this have to do with the functionality of wait ? Or in other words , why wait will fix this issue ?
The sequence of events appears to be:
The shell is the parent process to your program. When it forks your program your program inherits the standard streams (to the terminal).
Your program forks a child process which also inherits the standard streams (to the terminal).
When your parent process terminates, the shell notices (because it is waiting) and issues a prompt to the terminal.
However, your program’s child has not yet terminated, so after the shell issues its prompt the child prints its output (and then terminates).
You will notice that the shell does not issue a second prompt after the child terminates. (The shell does not know anything about your child process.)
Order of output
The fact that you get complete output lines (instead of anything interleaved) is because the standard streams for all processes are in line oriented mode.
However, there is no guarantee of order between processes. The OS scheduler can order them any way it wants. Your child could have printed before the parent.
:O)
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.
Consider the code:
#include <stdio.h>
#include <errno.h>
#include <sys/types.h>
#include <unistd.h>
/* main --- do the work */
int main(int argc, char **argv)
{
pid_t child;
if ((child = fork()) < 0) {
fprintf(stderr, "%s: fork of child failed: %s\n",
argv[0], strerror(errno));
exit(1);
} else if (child == 0) {
// do something in child
}
} else {
// do something in parent
}
}
My question is from where does in the code the child process starts executing, i.e. which line is executed first??
If it executes the whole code, it will also create its own child process and thing will go on happening continuously which does not happen for sure!!!
If it starts after the fork() command, how does it goes in if statement at first??
It starts the execution of the child in the return of the fork function. Not in the start of the code. The fork returns the pid of the child in the parent process, and return 0 in the child process.
When you execute a fork() the thread is duplicated into memory.
So what effectively happens is that you will have two threads that executes the snippet you posted but their fork() return values will be different.
For the child thread fork() will return 0, so the other branch of the if won't be executed, same thing happens for the father thread.
When fork() is called the operating system assigns a new address space to the new thread that is going to spawn, then starts it, they will both share the same code segment but since the return value will be different they'll execute different parts of the code (if correctly split, like in your example)
The child starts by executing the next instruction (not line) after fork. So in your case it is the assignment of the fork's return value to the child variable.
Well, if i understand your question correctly, i can say to you that your code will run as a process already.When you run a code,it is already a process , so that this process goes if statement anyway. After fork(), you will have another process(child process).
In Unix, a process can create another process, that's why that happens.
Code execution in a child process starts from the next instruction following the fork() system call.
fork() system call just creates a seperate address space for the child process therefore it is a cloned copy of the parent process and the child process has all the memory elements of it's parent's process.
Thus, after spawning a child process through fork(), both processes (the parent process and the child process) resumes the execution right from the next instruction following the fork() system call.
I write the following codes and run it in my linux.Everytime after fork the terminals print two PID, which shows both processes are scheduled by the OS, and then it is time for "scanf" to execute, both processes are blocked waiting for the input.However every time I put a number, and then I get the same PID printed on the terminal. Does it mean the same process is invoked by the OS when a terminal IO meets, or something else happens?
int main(int argc, char* argv[])
{
int num;
if(fork() >= 0)
{
printf("%x\n",getpid());
while(1)
{
if(scanf("%d",&num) != EOF)
{
printf("%x\n",getpid());
}
}
}
printf("\nit is over:%x\n", getpid());
}
As Hunter McMillen already noted in comments you are grouping the cases for the parent and child. Now both of them are scheduled as noted by different PIDs outputted and both of them are now waiting at scanf. As soon as you enter data, you are seeing only one PID, because the input you entered was part of one process. Other process ( can be parent or child ) is still waiting for you to enter something. Now, even though your terminal is flooded by a single PID, continuously outputted by one process, try entering some data again and press enter. Now you can see both PIDs being printed!
This is because, in the if() statement, fork() creates child process and runs in an infinite loop. There's a concept of parent process, the parent of child process created using fork() system call. The statement after if() ends, belongs to parent process and here, it will execute only after child process ends. That's why you get the same process ID, that is of child process.
May be it look childish for most of you but I am unable to understand this small piece of code.
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
int main(int argc, char** argv) {
int i, pid;
pid = fork();
printf("Forking the pid: %d\n",pid);
for(i =0; i<5; i++)
printf("%d %d\n", i,getpid());
if(pid)
wait(NULL);
return (0);
}
Out put of this program is
Forking the pid: 2223
0 2221
1 2221
2 2221
3 2221
4 2221
Forking the pid: 0
0 2223
1 2223
2 2223
3 2223
4 2223
Press [Enter] to close the terminal ...
In the for loop the printf command is used once. Why "Forking the pid" and after that the pid's are printed twice. How this is working? Can anybody explain me this? Thanks in advance.
Can anybody explain me why we have to use wait here? What I understood from the man pages is wait retuns the control to parent process? Is what I understood is correct?Is it necessary to use wait after forking a process?
Operating system : ubuntu, compiler : gcc, IDE : netbeans
But that' exactly what fork does. You forked the process and everything after the fork is done twice because now you have two processes executing the same printing code. You are basically asking why fork forks. fork forks because is is supposed to fork. That's what it's for.
After fork the parent and the child processes are generally executed in parallel, meaning that the nice sequential output you see in your example is not guaranteed. You might have easily ended up with line-interleaved output from two processes.
wait function in your case is executed from the parent process only. It makes it wait until the child process terminates, and only after that the parent process proceeds to terminate as well. Calling wait in this particular example is not really critical, since the program does nothing after that, it just terminates. But, for example, if you wanted to receive some feedback from the child process into the parent process and do some additional work on that feedback in the parent process, you'd have to use wait to wait for the child process to complete its execution.
The fork() call makes a new process. The rest of the code is then executed from each of the 2 processes. (Man page)
You're printing in both processes. Put your printing loop in an else clause of the if (pid):
pid = fork();
if(pid)
{
printf("Child pid: %d\n",pid);
wait(NULL);
}
else
{
for(i =0; i<5; i++)
printf("%d %d\n", i,getpid());
}
You see, fork returns twice, once in the parent process and once in the child process. It returns 0 in the child and the pid of the created process in the parent.
Because both the parent and child process are outputting their results.
See here: http://en.wikipedia.org/wiki/Fork_(operating_system)#Example_in_C for a good example.
fork creates a new process, and returns in both the old process (the parent) and in the new process (the child).
You can tell which one you are in by looking at the return value from fork. In the parent process it returns the PID of the child process. In the child process it return 0.